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2 years ago

11

If the force of the bullet’s impact was constant and the bullet took 0.2s to become embedded in the pendulum, what was the force

of impact? bullet .02kg speed 124.5m/s

1 answer:

Anastaziya [24]2 years ago

8 0

The correct answer is 24.9

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Give two alterations in a generator to produce more electromotive force

Mrac [35]

Answer:

Rotating the coil faster

Increasing the number of coils

Using stronger magnets

4 0

3 years ago

Read 2 more answers

A 2000 kg car experiences a constant braking force

Marat540 [252]

In that case, there are three possible scenarios:

-- If the braking force is less than the force delivered by the engine,
then the car will continue to accelerate, and the brakes will eventually
overheat and erupt in flame.

-- If the braking force is exactly equal to the force delivered by the engine,
then the car will continue moving at a constant speed, and the brakes will
eventually overheat and erupt in flame.

-- If the braking force is greater than the force delivered by the engine,
then the car will slow down and eventually stop. If it stops soon enough,
then the absorption of kinetic energy by the brakes will end before the
brakes overheat and erupt in flame. Even if the engine is still delivering
force, the brakes can be kept locked in order to keep the car stopped ...
They do not absorb and dissipate any energy when the car is motionless.

4 0

3 years ago

A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the

MaRussiya [10]

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

$v^{2}=u^{2}+2as$ and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

$u=\sqrt {v^{2}-2gs}$

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

$u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s$

5 0

4 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

3 0

3 years ago

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

Triss [41]

The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

$F = q(v \times B)$

$F = qvB \: sin \: θ$

$F = 1.6 \times 10 ^{ - 19} \times 5.02 \times 10 ^{6} \times 0.180 \times \: sin \: 60°$

$= 1.25 \times 10 ^{ - 13} \: N$

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

brainly.com/question/23096032

#SPJ4

4 0

2 years ago