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2 years ago

If a change in speed “Δv” occurs and the mass starts at rest, would the change in speed also equal the final speed of the mass?

Physics

1 answer:

stealth61 [152]2 years ago

3 0

Yes, if the mass starts at rest, <u>the change in speed will be equal the final speed</u>, because:

Δv = Vf - Vo

How Vo (Initial velocity) is equal zero, we simplificate:

Δv = Vf

Then, the change of the speed, if the mass starts at rest, will be equal to final velocity.

Greetings.

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A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

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x = 524.14 meters

Hence, this is the required solution.

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3 years ago

Juan lives 100 m away from Bill.What is Juan's averege speed if he reaches Bill's home in 50 s?

Tom [10]

Juan lives a hundred miles away from Bill then the average speed that he reaches Bill's Home in 50 seconds means that Juan lives 50 seconds away from bill because 50 + 50 seconds equals 100 seconds so Juansorry my bad Juan lives 50 miles away from Belle but he probably how I must have run to get to bill in 50 seconds .

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3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

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Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$