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never [62]
3 years ago
6

A seagull flying horizontally at 8.00m/s carries a clam with a mass of 300g in its beak. Calculate the total mechanical energy o

f the clam when the seagull is 30.0m abouve the ground.
Physics
1 answer:
Stells [14]3 years ago
7 0

Answer:

9.6J+88.2J=97.8J

Explanation:

Here the velocity of the seagull is given,mass is given and its height.

We have to find its mechanical energy my friend.

Mechanical energy=kinetic energy + potential energy.

First we will find kinetic energy.

For calculating kinetic energy we need mass and velocity,which are given here.

So, Ek=

1 \div 2mv {?}^{2}

So by substituting the values we get 9.6J.

Now we find the potential energy which is mgh.

By substituting the values we get 88.2J.

Then we add both of those and get 97.8J

I hope this satisfies you and make sure you contact me if it doesn't

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Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

\frac{mv^{2}}{2} = mgh

h = \frac{v^{2}}{2g}

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0
2 years ago
Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts
podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height =?

y_{0}: is the initial height = 1 m

v_{0_{y}: is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time    

First, we need to find the time by using the following equation:

t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s

Now, the height is:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m      

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}

1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!                                                                                  

4 0
3 years ago
A miner of mass 90kg travels down a slide calculate the potential energy of the miner when he moves15m vertically downwards
Lena [83]
Answer:

Gravitational Potential Energy = mgh

Explanation:

As the miner moves down, the GPE changes because the height changes.

Gravitational Potential Energy = mgh


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Eddi Din [679]

Answer:

I think your bones, muscles, and joints your welcome :)

Explanation:

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