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2 years ago

9

Rocket-powered sleds are been used to test the responses of humans to acceleration. Starting from rest, one sled can reach a spe

ed of 495 m/s in 1.78 s and can be brought to a stop again in 2.16 s.
Find the acceleration of the sled while accelerating.
Please be very thorough in your answer, I have a very hard time understanding if you don't.
Also, find the acceleration of the sled when braking.

1 answer:

Greeley [361]2 years ago

3 0

Answer:cho v₀ =0s

α=Δv/Δt

Explanation:

$\frac{0-495}{2,16-1,78}$

=-1302,631579

chuyển động chậm dầnđều

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3 years ago

Which of the following is an example of an electromagnetic wave? (2 points)

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3 years ago

Read 2 more answers

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

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$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

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3 years ago