Answer:
D
Explanation:
Gravity is the force of attraction between two objects.
Each object creates a gravitational field in wich every other object is affected by it.
A stone is thrown vertically upward with a speed of 21.0 m/s . How fast is it moving when it is at a height of 11.0 m ?
Answer:
15.01 m/s
Explanation:
Applying newtons equation of motion,
v² = u²+2gs.................. Equation 1
Where v = final velocity, u = initial velocity, s = distance, g = acceleration due to gravity.
Given: u = 21 m/s, s = 11 m
Constant: g = -9.8 m/s² ( g is negative because the stone against the force of gravity).
Substitute these values into equation 1
v² = 21²-2(11)(9.8)
v² = 441-215.6
v² = 225.4
v = √225.4
v = 15.01 m/s
Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. The amount of charge that passes per unit time is called
The amount of charge that passes per unit time is called electric current .
Current has dimensions of [Charge] / [Time] .
It's measured and described in units of ' Ampere ' .
1 Ampere means 1 Coulomb of charge passing a point every second.
Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 232 and 194 newtons, forming an angle of 67
Answer:
408N at 89.89°
Explanation:
This problem requires that we resolve the force vectors into
x- and y
-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.
Finally, we will convert the resultant force into standard form and find the equilibrant.
Resolve into components:
F1x =F1cos 180°= 232(−1)=−232N
F1y=F1sin180°=0N
F2x=F2cos(−140°)=194(−0.766)=−148.6N
F1y=F1sin(−140°)=232(−0.643)=−149.17N
Note the change of the angle used to give the direction of
F2. Standard angles (rotation from thex
-axis; counterclockwise is +) should be used to avoid sign errors in the results.
Now, we add the components:
Fx=F1x+F2x=−380.6N
Fy=F1y+F1y=−148.17N
Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:
F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N
θ=tan−1(−148.17−380.6)
=89.89°
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum
Answer:
a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.
Explanation:
a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:
[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]
Where:
[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.
[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.
The impact experimented by the ball due to collision is:
[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]
By using the definition of momentum, the expression is therefore expanded:
[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]
Where:
[tex]m[/tex] - Mass of the ball, measured in kilograms.
[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.
If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:
[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]
The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.
b) The magnitudes of initial and final momentums of the ball are, respectively:
[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is:
[tex]\Delta p = p_{f}-p_{o}[/tex]
[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.
c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.
In a nutshell, the right choice is option D.
A drum rotates around its central axis at an angular velocity of 19.4 rad/s. If the drum then slows at a constant rate of 8.57 rad/s2, (a) how much time does it take and (b) through what angle does it rotate in coming to rest
Answer:
Explanation:
Using equations of motion:
(a)
v=u+at
∴0=19.4−8.57t
∴t=19.4/8.57
=2.3s
B. Using s= ut + 1/2 at²
19.4(2.3)-1/28.57(2.3)²
= 21.92rad
An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the electron is 1.48 107 m/s, determine the following.
(a) the radius of the circular path ............ cm
(b) the time interval required to complete one revolution ............ s
Answer:
(a) 3.9cm
(b) 1.66 x 10⁻⁸s
Explanation:
Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,
F = m x a --------------(i)
Where;
m = mass of the particle
a = acceleration of the mass
The centripetal acceleration is given by;
a = v² / r [v = linear velocity of particle, r = radius of circular path]
Therefore, equation (i) becomes;
F = m v²/ r --------------------(ii)
The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;
F = qvBsinθ -------------(iii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = angle between the velocity and the magnetic field
Combine equations (ii) and (iii) as follows;
m (v² / r) = qvBsinθ [divide both side by v]
m v / r = qBsinθ [make r subject of the formula]
r = (m v) / (qBsinθ) ---------(iv)
(a) From the question;
v = 1.48 x 10⁷m/s
B = 2.14mT = 2.14 x 10⁻³T
θ = 90° [since the direction of velocity is perpendicular to magnetic field]
m = mass of electron = 9.11 x 10⁻³¹kg
q = charge of electron = 1.6 x 10⁻¹⁹C
Substitute these values into equation (iv) as follows;
r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)
r = 3.9 x 10⁻²m
r = 3.9cm
Therefore, the radius of the circular path is 3.9cm
(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by
T = d / v --------------(*)
Where;
d = distance traveled in the circular path in one complete turn = 2πr
v = velocity of the motion = 1.48 x 10⁷m/s
d = 2 π (3.9 x 10⁻²) [Take π = 22/7 = 3.142]
d = 2(3.142)(3.9 x 10⁻²) = 0.245m
Substitute the values of d and v into equation (*) as follows;
T = 0.245 / 1.48 x 10⁷
T = 0.166 x 10⁻⁷s
T = 1.66 x 10⁻⁸s
Therefore, the time interval is 1.66 x 10⁻⁸s
A system loses 510 J of potential energy. In the process, it does 430 J of work on the environment and the thermal energy increases by 100 J.
Required:
Find the change in kinetic energy.
Answer:
20J
Explanation:
The computation of the change in kinetic energy is shown below:
As we know that
Work was done on system = change in potential energy + change in kinetic energy + change in thermal energy
-430J = -510J + change in kinetic energy + 100J
-430J = -410J + change in kinetic energy
So, the change in kinetic energy is 20J
We simply applied the above formula to find out the change in kinetic energy
radiation transfers energy through___. a metal. b liquid. c touch. d waves.
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
The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm. If a cell of e.m.f. 1.2 volts is balanced by it, the balancing length will be:
A) 90 cm
B) 60 cm
C) 50 cm
D) 75 cm
Answer:
Potential gradient = 1.6 v/m
E.M.F. = potential gradient × balancing length 1.2 = 1.6 × l
l=1.2\1.6
=3\4
3\4=0.75m
convert m into cm 0.75x10
0.75m=75 cm
ans=75 cm
Explanation:
A 4.7 kg solid sphere made of metal, whose density is 4000 kg /m3, is suspended by a cord. The density of water is 1000 kg/m3.When the sphere is immersed in water, the tension in the cord is closest to:
Answer:
34.58 N
Explanation:
From the question,
Tension in the cord = Weight of the solid metal- Upthrust
T = W-U................... Equation 1
But,
W = mg
Where m = mass of solid, g = acceleration due to gravity.
Given: m = 4.7 kg, g = 9.8 m/s²
W = 4.7(9.8) = 46.1 N.
U = (Density of water×Volume of water displaced.)gravity
U = D×V×g.
But,
V = Mass of solid/density of solid
V = 4.7/4000
V = 1.175×10⁻³ m³.
Therefore,
U = 1.175×10⁻³(1000)(9.8)
U = 11.52 N
Therefore,
T = 46.1-11.52
T = 34.58 N
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The moment about the origin O of the force F is Mo
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
As more energy from fossil fuels and other fuels is released on Earth, the overall temperature of Earth tends to rise. Discuss how temperature equilibrium explains why Earth’s temperature cannot rise indefinitely.
Answer:
processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Explanation:
The temperature of planet Earth is due to two main types of process, internal and external.
Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect
External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.
These are the two terms that heat the Earth
In addition there are several processes so the planet loses energy,
* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission
* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system
* liquid (water) system that absorbs large amounts of heat to change state and temperature.
These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value
Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed.
In this final state
a) spheres #1 and #2 are still uncharged.
b) sphere #1 carries negative charge and #2 carries positive charge.
c) spheres #1 and #2 both carry positive charge.
d) spheres #1 and #2 both carry negative charge.
e) sphere #1 carries positive charge and #2 carries negative charge
Answer:
sphere #1 carries positive charge and #2 carries negative charge
This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged
A system consists of two particles. The first particle has mass m1 = 6.60 kg and a velocity of (4.20i + 2.00j) m/s, and the second particle has mass m2 = 2.00 kg and a velocity of (2.00i − 3.60j) m/s. (Express your answers in vector form.)
Required:
a. What is the velocity of the center of mass of this system?
b. What is the total momentum of this system?
Answer:
a. 8.465 m/s
b.22.3659
If d is 10% smaller than D, how do the pressure drops experienced by the fluid across region I and region II compare?
Answer:
ΔP = - 0.262 ρ v₁²
Explanation:
This is a fluid mechanics problem, let's use the subscript for region I and the subscript for region II. Let's write Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
suppose the pipe is horizontal, therefore
y₁ = y₂
P₁ -P₂ = ½ ρ (v₁² -v₂²) 1
Now let's use the continuity equation
v₁ A₁ = v₂ A₂ 2
The area of a circle is
A = π r²
also the radius is half the diameter r = d /2
A₁ = π d₁² / 4
let's substitute in equation 2
v1 π d₁² / 4 = v2 π d₂² / 4
v₁ d₁² = v₂ d₂²
v₂ = v₁ d₁² / d₂²
let's substitute in equation 1
P₁ - P₂ = ½ ρ (v₁² - v₁² (d₁² / d₂²)² )
Now let's use that the diameter d₂ = d is 10% smaller than the larger diameter d₁ = D, we assume that it is in region 1
d₂ = D -0.1D = 0.9 D
we substitute in the previous equation
P₁ - P₂ = ½ ρ v₁² (1 - (D / 0,9D)⁴)
P₁- P₂ = ½ ρ v₁² (1 - 1 / 0,9⁴ )
P1 - P2 = ½ ρ v12 (- 0.524)
ΔP = - 0.262 ρ v₁²
in this solution we assume that the data in zone I is known
A circular loop of wire has radius of 9.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0295 W/m^2, and the wavelength of the wave is 6.40 m.
Required:
What is the maximum emf induced in the loop?
Answer:
The maximum emf induced in the loop is 0.132 Volts
Explanation:
Given;
radius of the circular loop, r = 9.5 cm
intensity of the wave, I = 0.0295 W/m²
wavelength, λ = 6.40 m
The intensity of the wave is given as;
[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]
where;
B₀ is the amplitude of the field
c is the speed of light = 3 x 10⁸ m/s
μ₀ is permeability of free space = 4π x 10⁻⁵ m/A
[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]
Area of the circular loop;
A = πr²
A = π(0.095)²
A = 0.0284 m²
Frequency of the wave;
f = c / λ
f = (3 x 10⁸) / (6.4)
f = 46875000 Hz
Angular velocity of the wave;
ω = 2πf
ω = 2π(46875000)
ω = 294562500 rad/s
The maximum induced emf is calculated as;
emf = B₀Aω
= (1.572 x 10⁻⁸)(0.0284)(294562500)
= 0.132 Volts
Therefore, the maximum emf induced in the loop is 0.132 Volts
have an electrical charge of +1, while
have an electrical charge of -1.
A. Neutrons, electrons
B. Protons, electrons
C. Electrons, neutrons
D. Electrons, protons
Answer:
B
Explanation:
Protons have a positive electrical charge of +1,
Electrons have a negative charge of -1,
Neutrons have a neutral charge of about 0.
Question 5
A fidget spinner that is 4 inches in diameter is spinning clockwise. The spinner spins at 3000
revolutions per minute.
At t = 0, consider the point A on the outer edge of the spinner that is along the positive horizontal
axis. Let h(t) be the vertical position of A in inches. Suppose t is measured in minutes. Find a
sinusoidal function that models h(t).
h(t) =
Can someone please help me for this question?!!!!! ASAP?!!!!
Answer:
h = 4 sin (314.15 t)
Explanation:
This is a kinematics exercise, as the system is rotating at a constant speed.
w = θ / t
θ = w t
in angular motion all angles are measured in radians, which is defined
θ = s / R
we substitute
s / R = w t
s = w R t
let's reduce the magnitude to the SI system
w = 3000 rev / min (2π rad / 1rev) (1min / 60 s) = 314.16 rad / s
let's calculate
s = 314.16 4 t
s = 1,256.6 t
this is the value of the arc
Let's find the function of this system, let's use trigonometry to find the projection on the x axis
cos θ = x / R
x = R cos θ
x = R cos wt
projection onto the y-axis is
sin θ = y / R
how is a uniform movement
θ = w t
y = R sin wt
In the case y = h
h = R sin wt
h = 4 sin (314.15 t)
which of the following is true about scientific models?
A. models are used to simplify the study of things
B.computer models are the most reliable kind of model
C. models explain past, present and future information
D.a model is accurate if it does not change over time
Answer:
A- to simplify the study of things
Explanation:
a visual reference
a bear has a mass of 500kg and 100,000 J of mechanical kinetic energy. What is the speed of the bear? (velocity) Can someone please answer with the formula included? Step by step pls.
Answer:
Velocity of bear (v) = 20 m/s
Explanation:
Given:
Mass of bear (m) = 500 kg
Mechanical kinetic energy (K.E) = 100,000 J
Find:
Velocity of bear (v) = ?
Computation:
Mechanical kinetic energy (K.E) = 1/2(m)(v)²
100,000 = 1/2(500)(v)²
200,000 = 500(v)²
400 = (v)²
Velocity of bear (v) = 20 m/s
a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all directions. What is the waves intensity 25 km from the antenna
Answer:
[tex]I=2.67\times 10^{-6}\ W/m[/tex]
Explanation:
Given that,
Frequency of a radio antenna is 1 MHz
Power, P = 21 kW
We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :
[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m[/tex]
So, the wave intensity 25 km from the antenna is [tex]2.67\times 10^{-6}\ W/m[/tex].
If a 20kg mass hangs from a spring, whose elastic constant is 1800 N / m, the value of the spring elongation is
Explanation:
F = kx
mg = kx
(20 kg) (10 m/s²) = (1800 N/m) x
x = 0.11 m
4. The Richter scale describes how much energy an earthquake releases. With every increase of 1.0 on the scale, 32 times more energy is released. How many times more energy would be released by a quake measuring 2.0 more units on the Richter scale?
Answer:
64 times
Explanation:
if increase of 1 gives you 32
then increase of 2 will give you its double
64
If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.
What is Earthquake?An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.
The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.
The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.
To know more about Earthquake:
https://brainly.com/question/1296104
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A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.
Answer:
The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 11^o[/tex]
The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as
[tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]
Where [tex]\beta[/tex] is mathematically represented as
[tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]
[tex]\beta = 708.1 \ rad[/tex]
So
[tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Which of the following is true of children with chronic illness? a.) They are all eligible to recievie special education services. b.) Very few such children are enrolled in public schools c.) Their eligibility for soeical education services depends on whether their conditions adversely affect their educational functioning. d.) They represent a large proportion of the children eligible for speical education services.
Answer:
c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.
Explanation:
Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.
The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.
1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible.
Answer:
3.26m
Explanation:
See attached file
A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the ground if the ball is thrown forward
Answer:
The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s
Explanation:
Given;
speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s
speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s
Determine the speed of the ball relative to the ground if the ball is thrown forward.
If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.
[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]
Therefore, the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.
An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. What is the ratio of the masses X/Y of the asteroids
Explanation:
It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.
As they are in equilibrium, the force of gravity due to each other is same. So,
[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]
So, the ratio of masses X/Y is 0.0847
At the scene of an accident, there are skid marks which are 200 feet long showing where a car put on the brakes. The driver swears that he was obeying the speed limit. If we assume that the car braked with a constant deceleration of 16ft/sec^2, we can determine whether or not the driver is telling the truth. how fast was he going when he applied the brakes?
Answer:
He was going at 54.5mph
Explanation:
Measuring distances in feet and time (starting with the brakes being applied) in seconds,
we have a(t) = −16, s(0) = 0, and v(T) = 0, s(T) = 200,
where T is the time of braking.
So Finding antiderivatives, v(t) = −16t + A and so s(t) = −8t²+ At +B. Since s(0) = 0,
we have B = 0. Since v(T) = 0,
we have T = A /16.
Putting this into s() gives
200 = −8(A/ 16)² +A· (A /16) = A²/32.
This gives A2 = 6,400, so A = 80 = v(0). That is, the initial speed was 80 ft/s ≈ 54.5 mph.
An inductor is connected to a 18 kHz oscillator. The peak current is 70 mA when the rms voltage is 5.4 V What is the value of the inductance L
Answer:
The value of the inductance is 0.955 mH
Explanation:
Given;
frequency of the oscillator, f = 18 kHz = 18,000 Hz
the peak current, I₀ = 70 mA = 0.07 A
the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V
The root mean square current is given as;
[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]
[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]
Inductive reactance is given by;
[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]
Inductance is given by;
[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]
L = 0.955 mH
Therefore, the value of the inductance is 0.955 mH
The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.
Given the following data:
Oscillator frequency = 18 kHzPeak current = 70 mARms Voltage = 5.4 VTo determine the value of the inductance (L):
First of all, we would find the root mean square (rms) current by using the formula:
[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]
Next, we would calculate the inductive reactance of the oscillator by using the formula:
[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]
Now, we can solve for the value of the inductance (L):
[tex]L = \frac{X_L}{2\pi f}[/tex]
Where:
L is the inductance.f is the frequency.[tex]X_L[/tex] is the inductive reactance.Substituting the parameters into the formula, we have;
[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]
L = [tex]9.55 \times 10^{-4}[/tex] Henry.
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