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OverLord2011 [107]

2 years ago

6

If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan

t velocity? a no force applied b a greater force c an equal force d a lesser force

1 answer:

nlexa [21]2 years ago

5 0

Answer:

Explanation:

hi

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Guys help...A spring that obeys Hooke's law, with a spring constant k1, is cut into N identical springs, each with a spring cons

Tems11 [23]

Answer:

K2 = N*K1

Explanation:

The force you apply to each section is the same you apply to the whole spring, but the extension of each section is dX/N (if dX is the extension of the entire spring)

4 0

3 years ago

How is air resistance similar to gravity? give me two ways.

OLga [1]

Answer:

1. they both act on an object in free fall

Explanation:

2. both help determine how fast the object will accelerate

4 0

2 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

numerical question : what is the required heat to raise the temperature of 2 kg parrafin by 10 Celsius if 44000 joules is requir

kirill115 [55]

Answer:

The heat energy required is, E = 2200 J

Explanation:

Given,

The mass of paraffin, m = 2 Kg

The energy required to raise the temperature of the paraffin by 200° C = 44000 J

Then the heat energy required to raise the temperature of the paraffin by 10° C is given by,

Since 44000 J raises temperature by 200° C, then

E = 44000 J / 20

= 2200 J

Hence, the energy required to raise the temperature of the paraffin by 10° C is, E = 2200 J

8 0

2 years ago

A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the ar

kramer

<span>If there isn't any force then the normal contact force will be

N=m*g=7.5*9.81=73.58N

which is 73.58-23=50.58N less

so, there the person must pull at 23 degree upward

break down the tension in two components, vertical and horizontal.

vertical tension= 50.58=T*sin23

T=50.58/sin23=129.45N</span>

7 0

2 years ago

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