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irinina [24]
2 years ago
14

6.A truck is carrying a steel beam of length 13.0 m on a freeway. An accident causes the beam to be dumped off the truck and sli

de horizontally along the ground at a speed of 30.0 m/s. The velocity of the center of mass of the beam is northward while the length of the beam maintains an east–west orientation. The vertical component of the Earth's magnetic field at this location has a magnitude of 30.0 T. What is the magnitude of the induced emf between the ends of the beam?
Physics
2 answers:
nataly862011 [7]2 years ago
7 0
11234533534534334634354
labwork [276]2 years ago
3 0

The magnitude is 13.12 mV.

The steps are attached below.

<h3>How do you find the magnitude of an induced emf?</h3>

The standard SI unit of the magnetic field is the tesla (T). As an end result, we can use these equations and the equation for an induced emf due to changes in magnetic flux, ϵ=−NΔϕΔt ϵ = − N Δ ϕ Δ t, to calculate the importance of a precipitated emf in a solenoid.

The magnitude of the precipitated contemporary depends on the rate of trade of magnetic flux or the fee of reducing the magnetic area strains.

Learn more about the magnitude here: brainly.com/question/18109453

#SPJ2

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If a fish is trying to capture an insect hovering above the surface of the water – how will it jump to catch it? Will it aim abo
Viktor [21]

Answer:

It will have aim at a point "below" the insect.

From the insect's point of view, the fish will appear to be shallower than it actually is because a ray of light from the insect to the fish will be bent "towards" the normal when the ray enters the water

5 0
2 years ago
Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface
nekit [7.7K]

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

sin\theta_c=\frac{n_2}{n_1}

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30

hope this helps!!

6 0
3 years ago
A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn
kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (\vec{v}_{S}), measured in meters per second, is:

\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}

Where:

v_{o,x} - Initial velocity in +x direction, measured in meters per second.

a_{x} - Acceleration in +x direction, measured in meter per square second.

t - Time, measured in seconds.

v_{y} - Velocity in +y direction, measured in meters per second.

If we know that v_{o,x} = 0\,\frac{m}{s}, a_{x} = 0.250\,\frac{m}{s^{2}}, t = 45\,s and v_{y} = 21.4\,\frac{m}{s}, the final velocity of the satellite is:

\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}

\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}

\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }

\tan \alpha = 1.902

\alpha = \tan^{-1}1.902

\alpha \approx 62.266^{\circ}

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0
3 years ago
How many electrons flow through a point in a wire in 7.00 s if there is a constant current of I = 4.35 A?
OLEGan [10]

Answer:

1.90×10²⁰ Electrons

Explanation:

From the question,

Q = It.................... Equation 1

Where Q = charge flowing through the wire, I = current, t = time

Given: I = 4.35 A, t = 7.00 s

Substitute these values into equation 1

Q = 4.35(7.00)

Q = 30.45 C.

But,

1 electron contains 1.6×10⁻¹⁹ C

therefore,

30.45 C = 30.45/1.6×10⁻¹⁹  electrons

= 1.90×10²⁰ Electrons

8 0
3 years ago
A 4-kg object falls vertically a distance of 5 m. its potential energy has changed by approximately how much?
Brums [2.3K]
Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen

Neglect air resistance.

The PE (potential energy) is 
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J

The PE is converted into KE (kinetic energy) after the fall. 
Therefore the PE decreased by 196 J ≈ 200 J

Answer: d. It has decreased by 200 J
7 0
4 years ago
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