Answer:
Radiation is transferred through electromagnetic waves so D.
Explanation:
Answer:
D. Waves
Explanation:
a and b don't make much sense, conduction is transfer of energy through touch
A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. There exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision.
Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself.
Required:
a. If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?
b. If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes?
Answer:
a) f₁ = 3.50 m , b) f₂ = 0.84 m
Explanation:
For this exercise we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image
a) the distance where the image should be placed is q = 3.50 m and the object is located at infinity p = ∞
1 / f₁ = 1 /∞ + 1 / 3.50
f₁ = 3.50 m
b) in this case the image is at q = -0.600 m and the object p = 0.350 m
1 / f₂ = 1 / 0.350 -1 / 0.600
the negative sign, is because the image is in front of the object
1 / f₂ = 1,1905
f₂ = 1 / 1,1905
f₂ = 0.84 m
An object on a rope is lowered steadily decreasing speed. Which is true?
A) The tope tensions is greater than the objects weight
B) the rope tension equals the objects weight
C)the rope tension is less than the objects weight
D) the rope tension can’t be compared to the objects weight
Answer:
C) the rope tension is less than the objects weight
Explanation:
According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.
In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.
Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:
C) the rope tension is less than the objects weight
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, what condition would he have?
A. Nearsightedness
B. Farsightedness
C. Neither nearsightedness nor farsightedness
Answer:
A. NearsightednessExplanation:
A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.
At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.
Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what method is each of the spheres charged?
Answer:
If they are metallic spheres they are connected to earth and a charged body approaches
non- metallic (insulating) spheres in this case are charged by rubbing
Explanation:
For fillers, there are two fundamental methods, depending on the type of material.
If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.
If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.
Recall that the voltages VL(t) and VC(t) across the inductor and capacitor are not in phase with the respective currents IL(t) and IC(t). In particular, which of the following statements is true for a sinusoidal current driver?
VL(t) and VC(t):
a) both lag their respective current
b) both lead their respective currents
c) VL(t) lags IL(t) and VC(t) leads IC(t)
d) VL(t) leads IL(t) and VC(t) lags IC(t)
Answer:
D) VL(t) leads IL(t) and VC(t) lags IC(t)
Explanation:
This is because The phase angle between voltage and current for inductors and capacitors is 90 degrees, or radians, also, this means that no power is dissipated in either the inductor or the capacitor, since the time average of current times voltage,
( I(t), V(t)), is zero.
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interferometer. The hydrogen discharge tube provides light of several different wavelengths (colors) in the visible range. The red light in the hydrogen spectrum has a wavelength of 656.3 nm and the blue light has a wavelength of 434.0 nm. When using the discharge tube and the red filter as the light source, you view a bright red spot in the viewing area of the interferometer. You now move the movable mirror away from the beam splitter and observe 158 bright spots. You replace the red filter with the blue filter and observe a bright blue spot in the interferometer. You now move the movable mirror towards the beam splitter and observe 114 bright spots. Determine the final displacement (include sign) of the moveable mirror. (Assume the positive direction is away from the beam splitter.)
Answer:
final displacement = +24484.5 nm
Explanation:
The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;
Δr = 2d2 - 2d1 = 150λ1
So, 2d2 - 2d1 = 150λ1
Dividing both sides by 2 to get;
d2 - d1 = 75λ1 - - - - eq1
Where;
d1 = distance between the fixed mirror and the beam splitter
d2 = position of moveable mirror from splitter when 158 bright spots are observed
Now, the path difference between the two waves when 114 bright spots were observed is;
Δr = 2d'2 - 2d1 = 114λ1
2d'2 - 2d1 = 114λ1
Divide both sides by 2 to get;
d'2 - d1 = 57λ1
Where;
d'2 is the new position of the movable mirror from the splitter
Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.
(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2
d2 - d1 - d'2 + d1 = 75λ1 - 57λ2
d2 - d'2 = 75λ1 - 57λ2
We are given;
(λ1 = 656.3 nm) and λ2 = 434.0 nm.
Thus;
d2 - d'2 = 75(656.3) - 57(434)
d2 - d'2 = +24484.5 nm
The maximum amount of pulling force a truck can apply when driving on
concrete is 10,560 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
Answer:
Explanation:
Let the weight of the truck be W . reaction force R = W
Maximum frictional force = μ R
= .8 x W
So for movement of truck
Pulling force = frictional force
10560 = .8W
W = 13200 N
weight of heaviest truck required = 13200 N .
What is the average flow rate in cm3 /s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L
Answer:
2.78 cm³/s
Explanation:
From the question,
Q = v/A'.................... Equation 1
Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.
Given: v = 100 km/h, A' = 10 km/L
Substitute this value into equation 1
Q = 100/10
Q = 10 L/h.
Now, we convert L/h to cm³/s.
Since,
1 L = 1000 cm³, and
1 h = 3600 s
Therefore,
Q = 10(1000/3600) cm³/s
Q = 2.78 cm³/s
13. A base is also known as
A) A proton donor
B) An electron donor
C) proton acceptor
D)An electron acceptor
Answer:
C) proton acceptor
Explanation:
A proton acceptor is another name for a base,
A pickup truck has a width of 79.0 in. If it is traveling north at 42 m/s through a magnetic field with vertical component of
40 ut, what magnitude emf is induced between the driver and passenger sides of the truck?
Answer:
The magnitude of the induced Emf is [tex]0.003371V[/tex]
Explanation:
The width of the truck is given as 79inch but we need to convert to meter for consistency, then
The width= 79inch × 0.0254=2.0066 metres.
Now we can calculate the induced Emf using expresion below;
Then the [tex]induced EMF= B L v[/tex]
Where B= magnetic field component
L= width
V= velocity
=(40*10^-6) × (42) × (2.0066)
=0.003371V
Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V
A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface.
Required:
a What is the speed of the block after the collision?
b. What percentage of the ball's initial energy is "lost"?
Answer:
(a) The speed of the block after the collision is 0.0476v0.
(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)
Explanation:
Given;
mass of ball of clay, m₁ = 50 g = 0.05 kg
mass of brick, m₂ = 1 kg
initial velocity of the ball of clay, u₁ = v0
initial velocity of the brick, u₂ = 0
Since the clay ball sticks with the brick after collision, it is inelastic collision.
Therefore, let their final velocity = v
(a) What is the speed of the block after the collision?
Apply the principle of conservation linear momentum
m₁u₁ + m₂u₂ = v (m₁ + m₂)
0.05v₀ + 1(0) = v( 0.05 + 1)
0.05v₀ = 1.05v
v = 0.05v₀ / 1.05
v = 0.0476v₀
Thus, the speed of the block after the collision is 0.0476 of its initial velocity.
(b). What percentage of the ball's initial energy is "lost"?
Initial kinetic energy of the ball = ¹/₂mv₀²
= ¹/₂ x 0.05 x v₀²
= 0.025v₀²
Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²
= ¹/₂ x 1.05 x 0.0476v₀²
= 0.025v₀²
Change in kinetic energy = 0.025v₀² - 0.025v₀²
= 0
percentage change in the initial kinetic energy of the ball;
= (0 / 0.025v₀²) x 100%
= 0 x 100%
= 0 %
Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)
. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the voltage across the capacitor 0.16 seconds after the circuit is connected to the battery
Answer:
The voltage is [tex]V_c = 9.92 \ V[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V_b = 24 \ V[/tex]
The capacitance of the capacitor is [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]
The resistance of the resistor is [tex]R = 100\ \Omega[/tex]
The time taken is [tex]t = 0.16 \ s[/tex]
Generally the voltage of a charging charging capacitor after time t is mathematically represented as
[tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]
Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so
[tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]
[tex]V_c = 9.92 \ V[/tex]
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.24 x 10-3 rad/s2 for 2.80 x 103 s. For the next 1.57 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.01 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.99 rad/s. Find the total angular displacement of the propeller.
Answer:
θ = 26.19 x 10³ radians
Explanation:
FOR ACCELERATED MOTION:
we use 2nd equation of motion for accelerated motion:
θ₁ = ωi t + (1/2)αt²
Where,
θ₁ = Angular Displacement covered during accelerated motion = ?
ωi = Initial Angular Speed = 0 rad/s
t = Time Taken = 2.8 x 10³ s
α = Angular Acceleration = 2.24 x 10⁻³ rad/s²
Therefore,
θ₁ = (0 rad/s)(2.8 x 10³ s) + (1/2)(2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)²
θ₁ = 8.78 x 10³ radians
Now we find final angular velocity (ωf) by using 1st equation of motion:
ωf = ωi + αt
ωf = 0 rad/s + (2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)
ωf = 6.272 rad/s
FOR UNIFORM ANGULAR SPEED:
For uniform angular speed we use following equation:
θ₂ = ωt
where,
θ₂ = Angular Displacement during uniform motion = ?
ω = Uniform Angular Speed = ωf = 6.272 rad/s
t = Time Taken = 1.57 x 10³ s
Therefore,
θ₂ = (6.272 rad/s)(1.57 x 10³ s)
θ₂ = 9.85 x 10³ radians
FOR DECELERATED MOTION:
Now, we use 3rd equation of motion for decelerated motion:
2αθ₃ = ωf² - ωi²
where,
α = Angular deceleration = - 2.01 x 10⁻³ rad/s²
θ₃ = Angular Displacement during decelerated motion = ?
ωf = Final Angular Speed = 2.99 rad/s
ωi = Initial Angular Speed = 6.272 rad/s
Therefore,
2(-2.01 x 10⁻³ rad/s²)θ₃ = (2.99 rad/s)² - (6.272 rad/s)²
θ₃ = (- 30.4 rad²/s²)/(-4.02 x 10⁻³ rad/s²)
θ₃ = 7.56 x 10³ radians
FOR TOTAL ANGULAR DISPLACEMENT:
Total Angular Displacement = θ = θ₁ + θ₂ + θ₃
θ = 8.78 x 10³ radians + 9.85 x 10³ radians + 7.56 x 10³ radians
θ = 26.19 x 10³ radians
A metal rod of length 2.0 m is moved at 6.0 m/s in a direction perpendicular to its length. A 5.0 mT magnetic field is perpendicular to both the rod and its velocity. What is the potential difference between the ends of the rod? 30 mV 15 mV 0 mV 12 mV 60 mV
Answer:
The potential difference is 60mVExplanation:
This problem bothers on application of the expression for motion emf which is expressed as
[tex]E= Blv[/tex]
where B= magnetic field in Tesla
l= length of the conductor
v= speed of conductor
Given data
l= 2 meters
v= 6 m/s
B= 5 Tesla
Applying the formula we have
[tex]E=5*2*6= 60mV[/tex]
What would the Hall voltage be if a 2.00-T field is applied across a 10-gauge copper wire (2.588 mm in diameter) carrying a 20.0-A current
Answer:
The hall voltage is [tex]\epsilon =1.45 *10^{-6} \ V[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B = 2.00 \ T[/tex]
The diameter is [tex]d = 2.588 \ mm = 2.588 *10^{-3} \ m[/tex]
The current is [tex]I = 20 \ A[/tex]
The radius can be evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{2.588 * 10^{-3}}{2}[/tex]
[tex]r = 1.294 *10^{-3} \ m[/tex]
The hall voltage is mathematically represented as
[tex]\episilon = B * d * v_d[/tex]
where[tex]v_d[/tex] is the drift velocity of the electrons on the current carrying conductor which is mathematically evaluated as
[tex]v_d = \frac{I}{n * A * q }[/tex]
Where n is the number of electron per cubic meter which for copper is
[tex]n = 8.5*10^{28} \ electrons[/tex]
A is the cross - area of the wire which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [ 1.294 *10^{-3}]^2[/tex]
[tex]A = 5.2611 *10^{-6} \ m^2[/tex]
so the drift velocity is
[tex]v_d = \frac{20 }{ 8.5*10^{28} * 5.26 *10^{-6} * 1.60 *10^{-19} }[/tex]
[tex]v_d = 2.7 *10^{-4 } \ m/s[/tex]
Thus the hall voltage is
[tex]\epsilon = 2.0 * 2.588*10^{-3} * 2.8 *10^{-4}[/tex]
[tex]\epsilon =1.45 *10^{-6} \ V[/tex]
What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C
A screen is placed 43 cm from a single slit which is illuminated with 636 nm light. If the distance from the central maximum to the first minimum of the diffraction pattern is 3.8 mm, how wide is the slit in micrometer
Answer: The width is 1.25692 μm
Explanation:
The data that we have here is:
Distance between the single slit to the screen = L = 43cm
λ = 636 nm
Distance from the central maximum to the first minimum = Z = 3.8mm
We know that the angle for the destructive diffraction is:
θ = pλ/a
where p is the order of the minimum, for the first minimum we have p = 1, and a is the width of the slit,
then we have:
θ = (636nm/a)
And we also know that we can construct a triangle rectangle, where the adjacent cathetus to this angle is the distance between the slit and the screen, and the opposite cathetus is the distance between the first maximum and the first minimum:
Tg(θ) = Z/L
Tan(636nm/a) = 3.8cm/43cm
First, we need to use the same units in the right side:
3.8mm = 0.38cm
Tg(636nm/a) = 0.38cm/43cm
636nm/a = Atg( 0.38cm/43cm ) = 0.506
a = 636nm/0.506 = 1,256.92 nm
1 μm = 1000nm
then:
a = 1,256.92 nm = (1,256.92/1000) μm = 1.25692 μm
(a) Find the speed of waves on a violin string of mass 717 mg and length 24.3 cm if the fundamental frequency is 980 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string? (Take the speed of sound in air to be 343 m/s.)
Answer:
a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c) λ = 0.486 m , d) λ = 0.35 m
Explanation:
a) The speed of a wave on a string is
v = √T /μ
also all the waves fulfill the relationship
v = λ f
they indicate that the fundamental frequency is f = 980 Hz.
The wavelength that is fixed at its ends and has a maximum in the center
L = λ / 2
λ = 2L
we substitute
v = 2 L f
let's calculate
v = 2 0.243 980
v = 476.28 m / s
b) The tension of the rope
T = v² μ
the density of the string is
μ = m / L
T = v² m / L
T = 476.28² 0.717 / 0.243
T = 6.69 10⁵ N
c) λ = 2L
λ = 2 0.243
λ = 0.486 m
d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air
v = λ f
λ= v / f
λ = 343/980
λ = 0.35 m
A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric field in the wide wire is
Answer:
electric field in the wide wire is
E₂ =[tex]\frac{E}{4}[/tex]
Explanation:
given
length of the copper wire = L
radius of the copper wire r₁ = b
length of the second copper wire = L
radius of the second copper wire r₂ = 2b
electric field in the narrow wire = E₁=E
recall
resistance R = ρL/A
where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.
Resistance of narrow wire, R₁
R₁ = ρL/A
where A = πb²
R₁ = ρL/πb²---------- eqn 1
Resistance of wide wire, R₂
R₂ = ρL/A
where A = π(2b)²
R₂ = ρL/π(2b)²
R₂ = ρL/4πb²-------------- eqn 2
R₂ = ¹/₄(ρL/πb²)
comparing eqn 1 and 2
R₁ = 4R₂
calculating the current in the wire,
I = E/(R₁ + R₂)
recall
R₁ = 4R₂
∴ I = E/(4R₂ + R₂)
I = E/5R₂
calculating the potential difference across R₁ & R₂
V₁ = IR₁
I = E/5R₂
∴ V₁ = ER₁/5R₂
R₁ = 4R₂
V₁ = 4ER₂/5R₂
∴V₁ = ⁴/₅E
potential difference for R₂
V₂= IR₂
I = E/5R₂
∴ V₂ = ER₂/5R₂
V₂ = ER₂/5R₂
∴V₂ = ¹/₅E
so, electric field E = V/L
for narrow wire E₁ = V₁/L ----------- eqn 3
for wide wire, E₂ = V₂/L------------ eqn 4
compare eqn 3 and 4
E₂/E₁ = V₂/V₁( L is constant)
E₂/E₁ = ¹/₅E/⁴/₅E
E₂ = E₁/4
note E₁ = E
∴E₂ =[tex]\frac{E}{4}[/tex]
A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film
Answer:
3.867 μm
Explanation:
The index of refraction, μ = 1.6
Wavelength of the light, λ = 580 nm
N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have
(N2 - N1) λ = 2L (n2 - n1)
L = [(N2 - N1) * λ] / 2(n2 - n1)
L = (8 * 580) / 2(1.6 - 1.0)
L = 4640 nm / 1.2
L = 3867 nm or 3.867 μm
Therefore we can come to the conclusion that the thickness of the film is 3.867 nm
You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.
Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
Find out more information about voltage here:
https://brainly.com/question/4389563
What happens to a bar of metal when it's heated?
A.It gets longer.
B.The effect depends on the density of the bar.
C.It stays the same length.
D.It gets shorter
Hey There!!
Your answer will be A. It get longer.
Because, The kinetic energy of its atoms increase. They vibrate faster. This means that each atom will take up more space due to its movement so the material will expand. Some metals expand more than others due to differences in the forces between the atoms / molecules. . . .
Hope This Helps <3!!
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
Suppose you sit on a rotating piano stool and hold a 2kg mass in each outstreched hand. If without your arms relative to your body you now drop these masses,
a) your angular velocity remains unchanged
b) your angular velocity increases
c) your angular velocity decreases but your kinetic energy increases.
d) your kinetic energy and angular velocity increases.
Answer:
C is the best answer for this question
A typical home uses approximately 1600 kWh of energy per month. If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home
Answer:
7.68×10^25kg
Explanation:
The formula for energy used per year is calculated as
Energy used per year =12 x Energy used per month
By substituting Energy used per month in the above formula, we get
Energy used per year =12 x 1600kWh
= 19200kWh
Conversion:
From kWh to J:
1 kWh=3.6 x 10^6 J
Therefore, it is converted to J as
19200 kWh =19200 x 3.6 x 10^6 J
= 6.912×10^10 J
Hence, energy used per year is 6.912×10^10 J
To find the mass that is converted to energy per year.
E = MC^2 ............1
E is the energy used per year
C is the speed of light = 3.0× 10^8m/s
Where E= 6.912×10^10 J
Substituting the values into equation 1
6.912×10^10 J = M × 3.0× 10^8m/s
M = 6.912×10^10 J / (3.0× 10^8m/s)^2
M = 6.912×10^10 J/9×10^16
M = 7.68×10^25kg
Hence the mass to be converted is
7.68×10^25kg
9. You are given a number of 10 Ω resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 10 Ω resistance that is capable of dissipating at least 5.0 W?
Answer:
here are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
Explanation:
Let's work carefully this exercise, they indicate that the total resistance 10 ohm and dissipates 5W, so we can use the power equation to find the circuit current
P = Vi = i² R
i = √ P / R
i = √ (5/10)
i = 0.707 A
This is the current that must circulate in the circuit.
Let's build a circuit with three resistors in series and each resistor in series has three resistors in parallel
The equivalent resistance is
1 /[tex]R_{equi}[/tex] = 1/10 + 1/10 + 1/10 = 3/10
Requi = 10/3
Requi = 3.3 Ω
The current in the three series resistors is I = 0.707 A, and this is divided into three equal parts for the parallel resistors
current in each residence in parallel
i_P = 0.707 / 3
I_p = 0.2357 A
now let's look at the power dissipated in each resistor
P = R i²
P = 10 0.2357²
P = 0.56 W
the power dissipated by each resistance is within the range of 1 A, let's see the total power that the 9 resistors dissipate
P_total = 9 P
P = total = 9 0.56
P_total = 5 W
we see that this combination meets the specifications of the problem.
Therefore there are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite from the planet is 6600 N. What is the kinetic energy of the satellite
Answer:
The kinetic energy is [tex]KE = 7.59 *10^{10} \ J[/tex]
Explanation:
From the question we are told that
The radius of the orbit is [tex]r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m[/tex]
The gravitational force is [tex]F_g = 6600 \ N[/tex]
The kinetic energy of the satellite is mathematically represented as
[tex]KE = \frac{1}{2} * mv^2[/tex]
where v is the speed of the satellite which is mathematically represented as
[tex]v = \sqrt{\frac{G M}{r^2} }[/tex]
=> [tex]v^2 = \frac{GM }{r}[/tex]
substituting this into the equation
[tex]KE = \frac{ 1}{2} *\frac{GMm}{r}[/tex]
Now the gravitational force of the planet is mathematically represented as
[tex]F_g = \frac{GMm}{r^2}[/tex]
Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
[tex]KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r[/tex]
=> [tex]KE = \frac{ 1}{2} *F_g * r[/tex]
substituting values
[tex]KE = \frac{ 1}{2} *6600 * 2.3*10^{7}[/tex]
[tex]KE = 7.59 *10^{10} \ J[/tex]
In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ???? in the wire?
Answer:
The current in the wire is 31.96 A.
Explanation:
The current in the wire can be calculated as follows:
[tex] I = \frac{q}{t} [/tex]
Where:
q: is the electric charge transferred through the surface
t: is the time
The charge, q, is:
[tex] q = n*e [/tex]
Where:
n: is the number of electrons = 7.93x10²⁰
e: is the electron's charge = 1.6x10⁻¹⁹ C
[tex] q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C [/tex]
Hence, the current in the wire is:
[tex] I = \frac{126.88 C}{3.97 s} = 31.96 A [/tex]
Therefore, the current in the wire is 31.96 A.
I hope it helps you!
When a certain gas under a pressure of 4.65 106 Pa at 21.0°C is allowed to expand to 3.00 times its original volume, its final pressure is 1.06 106 Pa. What is its final temperature?
Answer:
-72.0°C
Explanation:
PV = nRT
Since n, number of moles, is constant:
PV / T = PV / T
(4.65×10⁶ Pa) V / (21 + 273.15) K = (1.06×10⁶ Pa) (3V) / T
T = 201.16 K
T = -72.0°C
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attached by ropes to a massless pulley on a crane, and four members of the film's construction crew lift the prop at constant speed by delivering 135 W of power each. If 18.0% of the mechanical energy delivered to the pulley is lost to friction, what is the time interval required to lift the spacecraft to the specified height?"
Answer:
The time interval required to lift the spacecraft to this specified height is 123.94 seconds
Explanation:
Height through which the spacecraft is to be lifted = 32.0 m
Mass of the spacecraft = 260.0 kg
Four crew member each pull with a power of 135 W
18.0% of the mechanical energy is lost to friction.
work done in this situation is proportional to the mechanical energy used to move the spacecraft up
work done = (weight of spacecraft) x (the height through which it is lifted)
but the weight of spacecraft = mg
where m is the mass,
and g is acceleration due to gravity 9.81 m/s
weight of spacecraft = 260 x 9.81 = 2550.6 N
work done on the space craft = weight x height
==> work = 2550.6 x 32 = 81619.2 J
this is equal to the mechanical energy delivered to the system
18.0% of this mechanical energy delivered to the pulley is lost to friction.
this means that
0.18 x 81619.2 = 14691.456 J is lost to friction.
Total useful mechanical energy = 81619.2 J - 14691.456 J = 66927.74 J
Total power delivered by the crew to do this work = 135 x 4 = 540 W
But we know tat power is the rate at which work is done i.e
[tex]P = \frac{w}{t}[/tex]
where p is the power
where w is the useful work done
t is the time taken to do this work
imputing values, we'll have
540 = 66927.74/t
t = 66927.74/540
time taken t = 123.94 seconds