Answer:
A) [tex]V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}[/tex]
B) [tex] d = 4181.49 kg/m^{3} = 4.18 g/cm^{3} [/tex]
Explanation:
A) Using the Archimedes' force we can find the weight of water displaced:
[tex] W_{d} = W_{a} - W_{w} [/tex]
Where:
[tex]W_{a}[/tex]: is the weight of the block in the air = 20.1 N
[tex]W_{w}[/tex]: is the weight of the block in the water = 15.3 N
[tex] W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N [/tex]
Now, the mass of the water displaced is:
[tex] m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg [/tex]
The volume of the block can be found using the mass of water displaced and the density of the water:
[tex]V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}[/tex]
B) The density of the block can be found as follows:
[tex] d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3} [/tex]
I hope it helps you!
why does a hot-air ballon rise? a the volume of the air dispalced by the balloon is less than the volume of the balloon. b the weight of the air dispaced by the ballon is less than the volue of the ballon. c the wight of the balloon is less than the weight of the air displced by the balloon.
Answer:
the answer is c
Explanation:
the balloon is lighter than the air around displaced by it
If the string is 7.6 m long, has a mass of 34 g , and is pulled taut with a tension of 15 N, how much time does it take for a wave to travel from one end of the string to the other
Answer:
The wave takes 0.132 seconds to travel from one end of the string to the other.
Explanation:
The velocity of a transversal wave ([tex]v[/tex]) travelling through a string pulled on both ends is determined by this formula:
[tex]v = \sqrt{\frac{T\cdot L}{m} }[/tex]
Where:
[tex]T[/tex] - Tension, measured in newtons.
[tex]L[/tex] - Length of the string, measured in meters.
[tex]m[/tex] - Mass of the string, measured in meters.
Given that [tex]T = 15\,N[/tex], [tex]L = 7.6\,m[/tex] and [tex]m = 0.034\,kg[/tex], the velocity of the tranversal wave is:
[tex]v = \sqrt{\frac{(15\,N)\cdot (7.6\,m)}{0.034\,kg} }[/tex]
[tex]v\approx 57.522\,\frac{m}{s}[/tex]
Since speed of transversal waves through material are constant, the time required ([tex]\Delta t[/tex]) to travel from one end of the string to the other is described by the following kinematic equation:
[tex]\Delta t = \frac{L}{v}[/tex]
If [tex]L = 7.6\,m[/tex] and [tex]v\approx 57.522\,\frac{m}{s}[/tex], then:
[tex]\Delta t = \frac{7.6\,m}{57.522\,\frac{m}{s} }[/tex]
[tex]\Delta t = 0.132\,s[/tex]
The wave takes 0.132 seconds to travel from one end of the string to the other.
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
A wave travels at a consent speed. how does the frequency change if the wavelength is reduced by a factor of 4?
Answer:
The frequency increases by 4 because it is inversely proportional to the wavelength.
A 1300-turn coil of wire that is 2.2 cm in diameter is in a magnetic field that drops from 0.11 T to 0 {\rm T} in 12 ms. The axis of the coil is parallel to the field What is the emf of the coil?
Answer:
The induced voltage is [tex]\epsilon = 4.53 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1300 \ turns[/tex]
The diameter is [tex]d = 2.2 \ cm =0.022 \ m[/tex]
The initial magnetic field is [tex]B_i = 0.11 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
The time taken is [tex]t = 12 \ ms = 12*10^{-3} \ s[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.022}{2}[/tex]
[tex]r = 0.011 \ m[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{d\phi}{dt}[/tex]
Where [tex]d\phi[/tex] is the change in magnetic flux of the wire which is mathematically represented as
[tex]d \phi = dB* A * cos \theta[/tex]
=> [tex]d \phi = (B_f - B_i )* A * cos \theta[/tex]
Here [tex]\theta = 0[/tex]
since the axis of the coil is parallel to the field
Where A is the cross-sectional area of the coil which is mathematically represented as
[tex]A = \pi * r^2[/tex]
[tex]A = 3.142 * 0.011^2[/tex]
[tex]A = 3.80*10^{-4} \ m^2[/tex]
So the induced emf
[tex]\epsilon = - 1300 * \frac{(0- 0.11) * 3.80*10^{-4}}{12*10^{-3}}[/tex] Here we substituted the values of [tex]d \phi[/tex]
[tex]\epsilon = 4.53 \ V[/tex]
The emf induced in the coil at the given magnetic field strength is 4.53 V.
The given parameters;
number of turns, N = 1300 turnsdiameter of the coil, d = 2.2 cminitial magnetic field, B₁ = 0.11 Tfinal magnetic field, B₂ 0time, t = 12 msThe area of the coil is calculated as follows;
[tex]A = \pi r^2 = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times 0.022^2}{4} = 0.00038 \ m^2[/tex]
The emf induced in the coil is calculated as follows;
[tex]emf = -N\frac{d\phi}{dt} \\\\emf = N (\frac{\phi_1 - \phi_2}{t} )\\\\emf = N(\frac{AB_1 - AB_2}{t} )\\\\emf = NA(\frac{B_1 - B_2}{t} )\\\\emf = 1300 \times 0.00038 (\frac{0.11 - 0}{12 \times 10^{-3}} )\\\\emf = 4.53 \ V[/tex]
Thus, the emf induced in the coil at the given magnetic field strength is 4.53 V.
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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength is 4.0 mV/m at a point 1.5 m away from the center of the circle. At what rate is the magnetic field changing?
Answer:
The rate at which the magnetic field changes is [tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 4.0 mV/m = 4.0 *10^{-3} V/m[/tex]
The radius of the circular region where the electric field is induced is
[tex]d = 1.5 \ m[/tex]
Generally the induced electric field is mathematically represented as
[tex]E = - \frac{r}{2} * \frac{\Delta B }{\Delta t }[/tex]
The negative sign show that the induced electric field is acting in opposite direction to the change in magnetic field
Where [tex]\frac{\Delta B }{\Delta t }[/tex] is the change in magnetic field
So
[tex]\frac{\Delta B }{\Delta t } = - \frac{2 * E }{r}[/tex]
substituting values
[tex]\frac{\Delta B }{\Delta t } = - \frac{2 * 4.0 *10^{-3}}{ 1.5 }[/tex]
[tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]
5. The path length difference for the waves exiting the two slits of the double slit experiment must be equal to _____ for a bright fringe to appear.
Answer:
An integral or whole multiple of the wavelength λ
or d sin θ = mλ,
for m = 0, 1, −1, 2, −2, ...,
Explanation:
In the double slit interference pattern, if we consider how two waves travel from the slits to the screen, we'll see that each slit is a different distance from a given point on the screen hence, they posses different wavelengths. Waves in a double slit experiment will be in phase if they interfere constructively by starting out crest to crest, or trough to trough. If the waves arrive crest to trough, they will interfere destructively, and arrive out of phase. A constructive interference occurs when the path length difference of the waves exiting the two slits forms an integral multiple of wavelength at the screen. A destructive interference occurs if the path length differs by half a wavelength. Constructive interference forms the bright fringes, while the dark fringes are formed by destructive interference.
Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?
Answer:
0.11m
Explanation:
let's assume the boat is of uniform construction
Ignoring friction losses
Also assume the origin is at the end of the boat originally with the heavier person
the center of mass of the whole system will not change relative to the water when the two swap ends
Originally, the center of mass is
85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin
after the swap, the center of mass is
50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin
The center of mass has shifted
1.14-1.030 = 0.11m
as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat
Someone help find centripetal acceleration plus centripetal force!
Answer:Centripetal force that acts an object keep it along a moving circular path.
Explanation:Centripetal force along a path circular of radius(r) with velocity(V) acceleration the center of the path.
a=v/r
object will along moving continue a straight path unless by the external force.External force is the centripetal force.
Centripetal force is to moving in horizontal circle,Centripetal force is not a fundamental force.Gravitational force satellite and orbit of centripetal force.
Centripetal acceleration and centripetal force are used to calculate the motion of objects in circular motion. The main answer to the question is given below:The centripetal force is given by:F = mv²/rwhere m is the mass of the object, v is the speed of the object and r is the radius of the circle. The unit of centripetal force is Newtons (N).The centripetal acceleration is given by:a = v²/rThe unit of centripetal acceleration is meters per second squared
(m/s²).Explanation:When an object moves in a circular motion, there is a force that acts upon it. This force is called the centripetal force. This force always points towards the center of the circle. It is responsible for keeping the object moving in a circular motion.The centripetal force is related to the centripetal acceleration.
The centripetal acceleration is the acceleration of an object moving in a circle. It is always directed towards the center of the circle.The magnitude of the centripetal force is given by:F = mv²/rwhere F is the force, m is the mass of the object, v is the speed of the object and r is the radius of the circle.The magnitude of the centripetal acceleration is given by:a = v²/rwhere a is the acceleration, v is the speed of the object and r is the radius of the circle.
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BIO A trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. A 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms; these are all typical values. At what speed does it leave the ground
Answer:
Final velocity (v) = 0.509 m/s (Approx)
Explanation:
Ant use impulse power
Given:
Mass of ant = 12 mg = 12 × 10⁻⁶ kg
Average force = 47 mN = 47 × 10⁻³ N
Initial velocity(u) = 0
Time taken = 0.13 ms = 0.13 × 10⁻³ s
Find:
Final velocity (v)
Computation:
Force × Time = change in momentum
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = mv - mu
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = m(v - u)
6.11 × 10⁻⁶ = 12 × 10⁻⁶(v - 0)
6.11 = 12 v
Final velocity (v) = 0.509 m/s (Approx)
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.
Answer:
206.67NExplanation:
The sum of force along both components x and y is expressed as;
[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]
The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
[tex]a_x = \frac{d^2 x }{dt^2}[/tex]
[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]
Similarly,
[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]
[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]
[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
Visible light of wavelength 589 nm is incident on a diffraction grating that has 3500 lines/cm. At what angle with respect to the central maximum is the fifth order maximum observed
Answer:
dsinФ=mΔ
d=1/N
d=1/3500*[tex]10^{-2} m\\[/tex]
d=2.8*[tex]10^{-6}[/tex]
NOw apply all values on formula
dsinФ=mΔ
2.8*[tex]10^{-6\\}[/tex]sinФ=5*589*[tex]10^{-9}[/tex]
sinФ=1.05 error
so due fifth maximum order it cannot be soved by this grating
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is placed between the plates of the capacitor while the capacitor is still connected to the battery. After this is done, we find that
Answer:
The voltage across the capacitor will remain constant
The capacitance of the capacitor will increase
The electric field between the plates will remain constant
The charge on the plates will increase
The energy stored in the capacitor will increase
Explanation:
First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.
The charge on the capacitor is equal to
Q = CV
Since the voltage is constant, and the charge increases, the capacitance will also increase.
The energy in a capacitor is given as
E = [tex]\frac{1}{2}CV^{2}[/tex]
since the capacitance has increased, the energy stored will also increase.
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
Inductance is usually denoted by L and is measured in SI units of henries (also written henrys, and abbreviated H), named after Joseph Henry, a contemporary of Michael Faraday. The EMF E produced in a coil with inductance L is, according to Faraday's law, given by
E=−LΔIΔt.
Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?
Answer:
Explanation:
E= −L ΔI / Δt.
L = E Δt / ΔI
Hence the unit of inductance may be V s A⁻¹
or volt s per ampere .
In the given case
change in current ΔI = - 2.5 A
change in time = .015 s
L = .56 H
E = − L ΔI / Δt.
= .56 x 2.5 / .015
= 93.33 V .
1) What is the highest atomic number element a red dwarf star can produce in its core? a. Carbon b. Oxygen c. Helium d. Iron
2) What is the highest atomic number element that can be produced in the cores of the largest stars?a. Helium b. Oxygen c. Iron d. Carbon
3) If formed at the same time, a red dwarf star is likely to become a white dwarf faster than a Sun-like star would. a) True b) False
Answer:
1) c. Helium
2) Iron
3) False.
Explanation:
1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.
2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.
3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.
if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
Answer:
neither will happen
Explanation:
cause the water is already defreezed
A capacitor is to be constructed to have a capacitance of 100uF.The area of the plates is 6.om by 0.030m and the relative permitivityof dielectric is 7.0 Find the necessary separation of the plates and the electric field strength if a potential difference of 12V is applied across the capacitor.
Answer:
The answer is below
Explanation:
Given that:
The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²
the relative permittivity of dielectric (εr) is 7.0
Permittivity of free space (εo) = 8.854 × 10^(-12)
capacitance of 100uF
potential difference (V) of 12V
d = separation between plate
The capacitance (C) of a capacitor is given by:
[tex]C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m[/tex]
The electric field between plates is given as:
E = V /d
[tex]E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m[/tex]
Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conductor, with respect to the outer conductor, is +600 V.
A) An electron is released from rest at the surface of the outer conductor. The speed of the electron as it reaches the inner conductor is closest to:__________.
B) The maximum electric field magnitude between the cylinders is closest to:_______.
Answer:
a) The speed of the electron as it reaches the inner conductor is closest to:
v = 1.45 × 10⁷m/s
b) The electric field magnitude between the cylinders is
E = 10,000V/m
Explanation:
given
inner radius of the cylinder r₁ = 20mm = 0.02m
outter radius of the cylinder r₂ = 80mm = 0.08m
potential difference V= 600V
mass of electron = 9.1×10⁻³¹kg
charge on electron = 1.6×10⁻¹⁹C
calculating the work done in bringing electron at inner conductor is
[tex]W =\frac{1}{2}mv^{2}[/tex]
note:
[tex]V = \frac{W}{q}[/tex]
∴W = (ΔV)q
(ΔV)q = [tex]\frac{1}{2}mv^{2}[/tex]
(600)1.6×10⁻¹⁹ = ¹/₂ × 9.1×10⁻³¹ × v²
v² ≈ 2.11 × 10¹⁴
v = 1.45 × 10⁷m/s
According to the energy conservation law, the total energy of an isolated system is always constant.
The energy of an isolated system can neither be created nor be destroyed, it can only convert one form to another form.
∴ the maximum electric field
E = ΔV/d
E = 600/d
where d is the distance between the two points
where d = 0.06m
E = 600/0.06
E = 10,000V/m
Note: the electric field due to the potential difference between to points depends upon the potential difference V and the distance between both points d.
a) The speed of the electron as it reaches the inner conductor is closest to: v = 1.45 × 10⁷m/s
b) The electric field magnitude between the cylinders is, E = 10,000V/m
Given:
Inner radius of the cylinder r₁ = 20mm = 0.02m
Outer radius of the cylinder r₂ = 80mm = 0.08m
Potential difference V= 600V
Mass of electron = [tex]9.1*10^{-31}kg[/tex]
Charge on electron = 1.6×10⁻¹⁹C
A)
Calculation for Work Done:
[tex]W=1/2mv^2[/tex]............(1)
Also.
[tex]V=\frac{W}{q}[/tex]
Thus, [tex]W=\triangle V*q[/tex]...........(2)
On equating 1 and 2:
[tex]\triangle V*q=1/2mv^2\\\\(600)1.6*10^{-19} = 1/2 * 9.1*10^{-31}* v^2\\\\v^2 =2.11 * 10^{14}\\\\v = 1.45 * 1067m/s[/tex]
B)
Law of conservation of Energy:The energy of an isolated system can neither be created nor be destroyed, it can only convert one form to another form.
Thus, the maximum electric field
[tex]E = \triangle V/d\\\\E = 600/d[/tex]
where d is the distance between the two points
d = 0.06m
[tex]E = 600/0.06\\\\E = 10,000V/m[/tex]
Thus, the maximum electric field is 10,000V/m.
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Gun was fired with a muzzle velocity of 350m/s, mounted at an angle of 45’ above the ground. Neglecting air resistance, compute for the following;
*Maximum height reached
*Range of the projectile
*Total time of flight
Answer:
Maximum height, h = 3062.5m
Total time of flight, T = 49.49secs or 50secs approx.
Range, R = 12250m
Explanation:
Given data:
U = 350m/s
Angle = 45°
Assume g = 10m/s
At the greatest height, v = 0
Therefore,
V^2 = U^2 sin^2 × angle - 2×g×h
Substituting values:
0^2 = 350^2 sin^2 (45) - 2 × 10 × h
Let h = maximum height reached
Rearranging gives:
350^2 sin^2(45) = 2 x 10 x h
h = 350^2 sin^2(45)/2×10
h = 122500 x 0.5/20
h = 61250/20
h = 3062.5m
2)Total time of flight, T
T = 2U sin(angle)/g
= 2x350 sin(45)/10
= 494.9747/10
= 49.49secs or 50sec approx.
3) Range of projectile, R
R = U^2 sin2(angle)
= 350^2 sin2 (45)
= 122500 x 1/10
= 12250m
To compensate for acidosis, the kidneys will
Answer:
Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.
In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.
A light bulb is completely immersed in water. Light travels out in all directions from the bulb, but only some light escapes the water surface. What happens to the fraction (f) of light that escapes the water's surface as the bulb is moved deeper into the water?
Answer:
The fraction of light that escapes the water surface as the water moves deeper into the water will decrease.
Explanation:
The speed of light in water is small compared to the speed of light in air, and a larger part of the light energy is absorbed in water than in air. When the bulb is immersed in water, some of the light energy is absorbed by the mass of water. When the light bulb is further moved deeper into the water, the fraction of light that escapes decreases, because more mass of water is made available to absorb more of the light energy from the bulb.
A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is _______.
Answer:
10 kgm/s
Explanation:
Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.
From the question,
ΔM = m(v-u)...................... Equation 1
Where ΔM = change in momentum, u = initial velocity, v = final velocity.
Note: Let upward direction be negative, and downward direction be positive.
Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s
Substitute into equation 1
ΔM = 0.2(-20-30)
ΔM = 0.2(-50)
ΔM = -10 kgm/s.
The negative sign shows that the change in momentum is Upward
The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
Given data:
The mass of rubber ball is, m = 0.2 kg.
The initial speed of ball is, u = 30 m/s.
The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)
The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,
[tex]p= m ( v-u)[/tex]
Solving as,
[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]
Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
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If not already selected: Select ‘Electric Field’. How does the brightness of the arrow relate to the strength of the field? What happens when you check/uncheck ‘Direction only’? Which way do the arrows point for a positive charge?
The arrows point away from a positive charge and towards a negative charge.
The electric field is a vector field, which means that it has both a magnitude and a direction. The magnitude of the electric field is a measure of how strong the field is, and the direction of the electric field is a measure of the direction in which the force would be exerted on a positive charge.
The brightness of the arrow in an electric field simulation is a representation of the magnitude of the electric field. The brighter the arrow, the stronger the electric field. When you check the "Direction only" box, the arrows will only show the direction of the electric field. This is because the "Direction only" box only shows the direction of the vector field, not the magnitude.
When you uncheck the box, the arrows will show both the direction and the strength of the electric field. This is because the "Direction only" box is unchecked, so the arrows will show the full vector field. The arrows point away from a positive charge and towards a negative charge because positive charges repel each other and negative charges attract each other.
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A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant for this circuit
Answer:
The time constant is [tex]\tau = 1.265 s[/tex]
Explanation:
From the question we are told that
the time take to charge is [tex]t = 2.4 \ s[/tex]
The mathematically representation for voltage potential of a capacitor at different time is
[tex]V = V_o - e^{-\frac{t}{\tau} }[/tex]
Where [tex]\tau[/tex] is the time constant
[tex]V_o[/tex] is the potential of the capacitor when it is full
So the capacitor potential will be 100% when it is full thus [tex]V_o =[/tex]100% = 1
and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so
V = 0.85
Hence
[tex]0.85 = 1 - e^{-\frac{2.4}{\tau } }[/tex]
[tex]- {\frac{2.4}{\tau } } = ln0.15[/tex]
[tex]\tau = 1.265 s[/tex]
The movement of a car on a road is represented in this figure. Between t = 0 and t = 0.6 hrs, what is the displacement made by the car?
1). 4.0 km.
2). 0.0 km.
3). -4.0 km. 4
). 8.0 km.
Answer:
0
Explanation:
On a graph, if there is plot of time vs velocity, then area of velocity plot gives the displacement.
also we can see area of plot is velocity* time which is equal to formula of displacement.
area for this plot is velocity * time
Thus,
from
t =0 to t = 0.2
v = 20
t = 0.2 - 0 = 0.2
thus, displacement till 0.2 seconds = 20*0.2 = 4 Km
____________________________________________________
from
t =0 to t = 0.4
v = 0
t = 0.4 - 0.2 = 0.2
thus, displacement from 0.2 seconds to 0.4 seconds = 0*0.2 = 0 Km
____________________________________________________
from
t =0.4 to t = 0.6
v = -20
t = 0.6 - 0.4 = 0.2
thus, displacement from 0.4 seconds to 0.6 seconds = -20*0.2 = -4 Km
Thus, total displacement = 4+0 -4 = 0
Thus, net displacement made by car is 0.
Two electric force vectors act on a particle. Their x-components are 13.5 N and −7.40 N and their y-components are −12.0 N and −4.70 N, respectively. For the resultant electric force, find the following.
(a) the x-component N
(b) the y-component N
(c) the magnitude of the resultant electric force N
(d) the direction of the resultant electric force, measured counterclockwise from the positive x-axis ° counterclockwise from the +x-axis
Answer:
Explanation:
Given two vectors as follows
E₁ = 13.5 i -12 j
E₂ = -7.4 i - 4.7 j
Resultant E = E₁ + E₂
= 13.5 i -12 j -7.4 i - 4.7 j
E = 6.1 i - 16.7 j
a ) X component of resultant = 6.1 N
b ) y component of resultant = -16.7 N
Magnitude of resultant = √ ( 6.1² + 16.7² )
= 17.75 N
d ) If θ be the required angle
tanθ = 16.7 / 6.1 = 2.73
θ = 70° .
counterclockwise = 360 - 70 = 290°
By working with the vector forces, we will get:
a) The x-component is 1.5 Nb) The y-component is -12.2 Nc) The magnitude is 12.9 Nd) The direction is 277.01°.How to find the resultant force?
Remember that we can directly add vector forces, so if our two forces are:
F₁ = <13.5 N, -7.5 N>
F₂ = < -12 N, -4.70 N>
Then the resultant force is:
F = F₁ + F₂ = <13.5 N + (-12 N), -7.5 N + ( -4.70 N) >
F = < 1.5 N, -12.2 N>
so we have:
a) The x-component is 1.5 N
b) The y-component is -12.2 N
c) The magnitude will be:
|F| = √( (1.5 N)^2 + (-12.2 N)^2) = 12.29 N
d) The direction of a vector <x, y> measured counterclockwise from the positive x-axis is given by:
θ = Atan(y/x)
Where Atan is the inverse tangent function, then here we have:
θ = Atan(-12.2 N/1.5 N) = 277.01°
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Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in its secondary coil, and an input voltage of 120 V. Randomized Variables Δ 33%
Part (a) What is the voltage output Vs, in volts, of the transformer used for to charge the batteries? Grade Summar Deductions Potential sin tan) ( Submissions Attempts remain coso cotan) asin) acos() atan acotan)sinh( cosh)tanhcotanh0 % per attempt detailed view END Degrees Radians DEL CLEAR Submit Hint I give up! Hints:% deduction per hint. Hints remaining:I Feedback: 1% deduction per feedback. - 쇼 33%
Part (b) what input current ,. İn milliamps, is required to produce a 3.2 A output current? 33%
Part (c) What is the power input, in watts?
Answer:
a) 0.72 V
b) 19.2 mA
c) 2.304 Watts
Explanation:
A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.
number of primary turns = [tex]N_{p}[/tex] = 500 turns
input voltage = [tex]V_{p}[/tex] = 120 V
number of secondary turns = [tex]N_{s}[/tex] = 3 turns
output voltage = [tex]V_{s}[/tex] = ?
using the equation for a transformer
[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]
substituting values, we have
[tex]\frac{V_{s} }{120 } = \frac{3 }{500} }[/tex]
[tex]500V_{p} = 120*3\\500V_{p} = 360[/tex]
[tex]V_{p}[/tex] = 360/500 = 0.72 V
b) by law of energy conservation,
[tex]I_{P}V_{p} = I_{s}V_{s}[/tex]
where
[tex]I_{p}[/tex] = input current = ?
[tex]I_{s}[/tex] = output voltage = 3.2 A
[tex]V_{s}[/tex] = output voltage = 0.72 V
[tex]V_{p}[/tex] = input voltage = 120 V
substituting values, we have
120[tex]I_{p}[/tex] = 3.2 x 0.72
120[tex]I_{p}[/tex] = 2.304
[tex]I_{p}[/tex] = 2.304/120 = 0.0192 A
= 19.2 mA
c) power input = [tex]I_{p} V_{p}[/tex]
==> 0.0192 x 120 = 2.304 Watts
Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface
Answer:
(a) 4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface, (b) The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
Explanation:
The complete statement is: "Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface, (b) How far has the rocket traveled when half the total work has occurred?"
(a) The work required to propel the rocket is given by the change in gravitational potential energy, whose expression derives is described below:
[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]
Where:
[tex]U_{g,o}[/tex], [tex]U_{g,f}[/tex] - Initial and final gravitational potential energies, measured in joules.
[tex]m[/tex], [tex]M[/tex] - Masses of the rocket and planet Earth, measured in kilograms.
[tex]G[/tex] - Universal gravitation constant, measured in newton-square meters per square kilogram.
[tex]r_{o}[/tex], [tex]r_{f}[/tex] - Initial and final distances of the rocket with respect to the center of the Earth, measured in meters.
The initial distance and rocket mass are converted to meters and kilograms, respectively:
[tex]r_{o} = (4000\,mi)\cdot \left(1609.34\,\frac{m}{mi} \right)[/tex]
[tex]r_{o} = 6,437,360\,m[/tex]
[tex]m = (7\,ton)\cdot \left(1000\,\frac{kg}{ton} \right)[/tex]
[tex]m = 7000\,kg[/tex]
Given that [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]r_{f} \rightarrow +\infty[/tex], the work equation is reduced to this form:
[tex]U_{g,f} - U_{g,o} = \frac{G\cdot m \cdot M}{r_{o}}[/tex]
[tex]U_{g,f} - U_{g,o} = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)}{6,437,360\,m}[/tex]
[tex]U_{g,f} - U_{g,o} = 4.334\times 10^{11}\,J[/tex]
4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface.
(b) The needed change in gravitational potential energy is:
[tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex]
The expression for the change in gravitational potential energy is now modified by clearing the final distance with respect to the center of Earth:
[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]
[tex]\frac{U_{g,o}-U_{g,f}}{G\cdot M \cdot m} = \frac{1}{r_{f}} - \frac{1}{r_{o}}[/tex]
[tex]\frac{1}{r_{f}} = \frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m}[/tex]
[tex]r_{f} = \left(\frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m} \right)^{-1}[/tex]
If [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex], then:
[tex]r_{f} = \left[\frac{1}{6,437,360\,m}-\frac{2.167\times 10^{11}\,J}{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)} \right]^{-1}[/tex]
[tex]r_{f} \approx 12,874,502.49\,m[/tex]
The final distance with respect to the center of the Earth in miles is:
[tex]r_{f} = (12,874,502.49\,m)\cdot \left(\frac{1}{1609.34}\,\frac{mi}{m} \right)[/tex]
[tex]r_{f} = 7999.865\,mi[/tex]
The distance travelled by the rocket is: ([tex]r_{f} = 7999.865\,mi[/tex], [tex]r_{o} = 4000\,mi[/tex])
[tex]\Delta r = r_{f}-r_{o}[/tex]
[tex]\Delta r = 7999.865\,mi - 4000\,mi[/tex]
[tex]\Delta r = 3999.865\,mi[/tex]
The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.
Answer:
a
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
b
[tex]L = 1.183 *10^{-7} \ H[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 6 \ cm = \frac{6}{100} = 0.06 \ m[/tex]
The current it carries is [tex]I = 1.50 \ A[/tex]
The magnetic flux of the coil is mathematically represented as
[tex]\phi = B * A[/tex]
Where B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * I}{2 * r}[/tex]
Where [tex]\mu_o[/tex] is the magnetic field with a constant value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = \frac{4\pi * 10^{-7} * 1.50 }{2 * 0.06}[/tex]
[tex]B = 1.571 *10^{-5} \ T[/tex]
The area A is mathematically evaluated as
[tex]A = \pi r ^2[/tex]
substituting values
[tex]A = 3.142 * (0.06)^2[/tex]
[tex]A = 0.0113 m^2[/tex]
the magnetic flux is mathematically evaluated as
[tex]\phi = 1.571 *10^{-5} * 0.0113[/tex]
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
The self-inductance is evaluated as
[tex]L = \frac{\phi }{I}[/tex]
substituting values
[tex]L = \frac{1.78 *10^{-7} }{1.50 }[/tex]
[tex]L = 1.183 *10^{-7} \ H[/tex]