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2 years ago

How do i figure this out ? Anyone help me please

Physics

1 answer:

kondor19780726 [428]2 years ago

6 0

Answer:

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Explanation:

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The eagle drops the trout a height of 6.1 m the fish travels 7.9 m horizontaly before hitting the water what is the velocity of

Semenov [28]

The velocity of the eagle is 7.0 m/s

Explanation:

The motion of the fish is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction, where the horizontal velocity of the fish is equal to the horizontal velocity of the eagle

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion, using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where:

s = 6.1 m is the vertical displacement of the fish

u = 0 is the initial vertical velocity of the fish

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t, we find the time of flight of the fish:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(6.1)}{9.8}}=1.12 s$

Now we know that during this time, the fish travels a horizontal distance of

x = 7.9 m

Therefore, the horizontal velocity of the fish is

$v_x = \frac{x}{t}=\frac{7.9}{1.12}=7.0 m/s$

And therefore, this is the initial velocity of the eagle.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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3 years ago

A car that brakes suddenly comes to a screeching halt. Is the sound energy produced in this conversion a useful form of energy?

Nezavi [6.7K]

Answer:

No, the sound produced by screeching brakes is not a useful form of energy, because it does not contribute to stopping the car.

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2 years ago

Soft ball,kick ball,wiffle ball??

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1. No they aren’t because they all belong to different sports and are used differently

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A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, $\theta=33^{\circ}$

Time for which the player's foot is in contact with it, $\Delta t = 5.1\times 10^{-2}\ s$

Part A,

The x component of the soccer ball's change in momentum is given by :

$\Delta p_x=mv\ cos\theta$

$\Delta p_x=0.425\times 15\ cos(33)$

$p_x=5.34\ kg-m/s$

The y component of the soccer ball's change in momentum is given by :

$\Delta p_y=mv\ sin\theta$

$\Delta p_y=0.425\times 15\ sin(33)$

$p_y=3.47\ kg-m/s$

Hence, this is the required solution.

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3 years ago

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