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lukranit [14]
3 years ago
13

A speed-time graph is shown below:

Physics
1 answer:
VladimirAG [237]3 years ago
5 0

Answer: 0.5 m/s^{2}

Explanation:

Average acceleration a_{ave} is the variation of velocity \Delta V over a specified period of time \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

\Delta V=V_{f}-V_{o} being V_{o}=0 cm/s the initial velocity and V_{f}=4 cm/s the final velocity  (according to the information given from the described graph)

\Delta t=8 s

Then:

a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}

a_{ave}=0.5 m/s^{2}

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<h2>Magnetic field strength in that region is 1.2 T</h2>

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3 years ago
- A cannon of 2000 kg fires a shell of 10 kg at
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1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

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After the shot, the momentum is:

p=MV+mv

where

M = 2000 kg is the mass of the cannon

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v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

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2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

v^2-u^2=2as

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v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

a=-0.25 m/s^2 is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m

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4 years ago
If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.
Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

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