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vampirchik [111]
3 years ago
14

During a safety test, a car hits a wall and stops in 0.55 s. The net force on the car is 6500 N during the collision. What is th

e magnitude of the change in momentum of the car? Round answer to two significant digits. kg m
Physics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

ΔP = 3600 kg m/s.

Explanation:

Change in momentum (ΔP) of the car is the amount of impulse on it. Impulse (I) can be define as the product of force and the time at which it acts.

i.e I = ΔP

But, I = Ft

Ft = ΔP

ΔP = 6500 x 0.55

     = 3575 Ns

     = 3600 kg m/s^{2} x s

ΔP = 3600 kg m/s

The magnitude of the change in momentum of the car is 3600 kgm/s.

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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of
lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = 20.6\times 10^3\ m

\theta = The angle subtended by moon = 9.6\times 10^{-3}\ rad

Distance traveled is given by

s=r\times\theta

\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}

\Rightarrow s=197.76\ m

The distance traveled by the jet is 197.76 m

8 0
3 years ago
An x-ray beam of wavelength 1.4×10−10m makes an angle of 20° with a set of planes in a crystal(the Bragg angle)causing first ord
Gnom [1K]

Answer:

Second order line appears at 43.33° Bragg angle.

Explanation:

When there is a scattering of x- rays from the crystal lattice and interference occurs, this is known as Bragg's law.

The Bragg's diffraction equation is :

n\lambda=2d\sin\theta      .....(1)

Here n is order of constructive interference, λ is wavelength of x-ray beam, d is the inter spacing distance of lattice and θ is the Bragg's angle or scattering angle.

Given :

Wavelength, λ = 1.4 x 10⁻¹⁰ m

Bragg's angle, θ = 20°

Order of constructive interference, n =1

Substitute these value in equation (1).

1\times1.4\times10^{-10} =2d\sin20

d = 2.04 x 10⁻¹⁰ m

For second order constructive interference, let the Bragg's angle be θ₁.

Substitute 2 for n, 2.04 x 10⁻¹⁰ m for d and 1.4 x 10⁻¹⁰ m for λ in equation (1).

2\times1.4\times10^{-10} =2\times2.04\times10^{-10} \sin\theta_{1}

\sin\theta_{1} =0.68

<em>θ₁ </em>= 43.33°

4 0
3 years ago
Which example best represents translational kenetic energy
Mila [183]

Answer:

an apple falling off a tree

Explanation:

5 0
3 years ago
When the displacement of a mass on a spring is 12 a the half of the amplitude, what fraction of the mechanical energy is kinetic
Sveta_85 [38]
You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).

And you know that the potential energy, PE, is [ 1/2 ] k (x^2)

Then, use x = A, to calculate the PE in the point where ME = PE.

ME = PE = [1/2] k (A)^2.

At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2

=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME

So, if PE is 1/4 of ME, KE is 3/4 of ME.

And the answer is 3/4


7 0
3 years ago
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