When should scientific claims be questioned?

As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l

deff fn [24]

Answer: 62 μT

Explanation:

Given

Length of rod, l = 1.33 m

Velocity of rod, v = 3.19 m/s

Induced emf, e = 0.263*10^-3 V

Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

0.263*10^-3 = b * 4.2427

b = 0.263*10^-3 / 4.2427

b = 0.0000620 T

b = 62 μT

Thus, the strength of the magnetic field is 62 μT

8 0

2 years ago

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What is the refraction index of a medium?

Sati [7]

<span>the ratio of the velocity of light in the medium over the velocity of light in a vacuum</span>

6 0

3 years ago

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find

butalik [34]

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy ..............................1

so it can be written as

work = force × distance ...................2

and

KE is express as

K.E = 0.5 × m × v²

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)² ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² = 919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

5 0

3 years ago

The buoyant force exerted by a liquid is equal to the _____ of the fluid displaced. mass weight volume density

castortr0y [4]

The buoyant force exerted by a liquid is equal to the weight of the fluid <span>displaced.</span>

6 0

3 years ago

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Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di

Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, $u=20m/s$

And the mass of the car is, $m=1000 kg$

The total distance covered by the car before stop, $s=45m$

And the final speed of the car is, $u=0m/s$

Now initial kinetic energy is,

$KE_{i}=\frac{1}{2}mu^{2}$

Substitute the value of u and m in the above equation, we get

$KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J$

Now final kinetic energy is,

$KE_{f}=\frac{1}{2}mv^{2}$

Substitute the value of v and m in the above equation, we get

$KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J$

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

$F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN$

Here, the force is negative because the force and acceleration in the opposite direction.

6 0

2 years ago