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3 years ago

A truck of mass 1800kg is moving with a speed 54km/h. When brakes are applied, it

Physics

1 answer:

EleoNora [17]3 years ago

6 0

Force = 3200 N

Work done = 640, 000 Nm

Explanation:

We begin by calculating the deceleration of the truck, using the velocity and distance;

a = (v² – u²)/2s

whereby;

a = acceleration

v = initial velocity

u = initial velocity

s = distance

We begin by changing the speed from km/h into m/s;

54km/hr = 15m/s

Then acceleration;

a = (0² – 15²) / 2 * 200

a = -225 / 400

a = - 0.5625 m/ s²

To calculate force;

F = ma

Whereby;

F = force

M = mass (in kgs)

a = acceleration

F = 1800 / 0.5625

F = 3200 N

Work done = Force * displacement

Work done = 3200 * 200

= 640, 000 Nm

Learn More:

For more on force and work done check out;

brainly.com/question/8662583

brainly.com/question/1268612

brainly.com/question/11870590

brainly.com/question/9125094

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Which of the following is an example of technology influencing science?

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3 years ago

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An airplane accelerates from a velocity of 22 m/s to 40 m/s with an acceleration of 2 m/s2. How long does it

andrey2020 [161]

The time it takes the plane to change its velocity is 9s.

Time can be defined the measured or measurable period during which an action, process, or condition exists or continues.

To calculate the time it takes the airplane to change its velocity, we use the formula below.

Formula:

t = (v-u)/a.......... Equation 1

Where:

a = Acceleration
v = Final velocity
u = Initial velocity
t = time

From the question,

v = 40 m/s
u = 22 m/s
a = 2 m/s²

Substitute these values into equation 1

t = (40-22)/2
t = 18/2
t = 9s

Hence, the time it takes the plane to change its velocity is 9s.

Learn more about time here: brainly.com/question/2854969

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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,

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