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2 years ago

5

a student throws a coin vertically downward frok the top of a building. the coin leaves the throwers hand with a speed of 15.0m/

s. what is its speed after falling freely for 2.00s?

1 answer:

Elina [12.6K]2 years ago

8 0

Answer:

Final speed after 2 seconds = 34.6 m/s

Explanation:

Given:

Initial speed of coin (u) = 15 m/s

Time taken = 2 seconds

Find:

Final speed after 2 seconds

Computation:

Gravitational acceleration of earth = 9.8 m/s²

Using first equation of motion;

v = u + at

or

v = u + gt

where,

v = final velocity

u = initial velocity

g = Gravitational acceleration

t = time taken

v = 15 + 9.8(2)

v = 15 + 19.6

Final speed after 2 seconds = 34.6 m/s

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# A cheetah can start from rest and attain the velocity 72km/h in 2 seconds. Calculate the acceleration of cheetah

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Answer:

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Explanation:

Solution,

When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s

Initial velocity ( u ) = 0 m/s

Final velocity ( v ) = 72 km/h ( Given)

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Final velocity ( v ) = 20 m/s

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Now,

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$a = \frac{v - u}{t}$

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Hope this helps...

Good luck on your assignment..

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3 years ago

1. In terms of momentum conservation, why does a cannon recoil when fired?

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Newtons law of motion for every action there’s an equal and opposite reaction.

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2 years ago

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2 years ago

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14

Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

$\omega_f^2 - \omega_i ^2 =2 \alpha \theta$

where

$\omega_f = 3.14\cdot 10^4 rad/s$ is the final angular speed

$\omega_i = 1.10 \cdot 10^4 rad/s$ is the initial angular speed

$\theta= 2.00 \cdot 10^4 rad$ is the angular distance covered

$\alpha$ is the angular acceleration

Re-arranging the formula, we can find $\alpha$ :

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2$

Now we want to know the time the bit takes starting from rest to reach a speed of $\omega_f=7.85\cdot 10^4 rad/s$ . So, we can use the following equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where:

$\alpha=2.16\cdot 10^4 rad/s^2$ is the angular acceleration

$\omega_f = 7.85\cdot 10^4 rad/s$ is the final speed

$\omega_i = 0$ is the initial speed

t is the time

Re-arranging the equation, we can find the time:

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s$

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3 years ago

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