Please help me im struggling

In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha

eduard

Average force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the net change in momentum of the hailstones in a time interval of $\Delta t$

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

$p_i =m u sin \theta$ is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

$u=4.5 m/s$ is the initial speed

$\theta=25^{\circ}$ is the angle with the window

The final momentum is

$p_f = mv sin \theta$

where

v = 4.5 m/s is the final speed (the collision is elastic so the speed is equal, while the direction changes)

$\theta=-25^{\circ}$ (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

$\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s$

We are interested only in the magnitude, so

$\Delta p = 0.0266 kg m/s$

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

$\Delta p = 595\cdot 0.0266 = 15.8 kg m/s$

And therefore, the average force on the window is

$F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N$

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3 0

2 years ago

Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

Given the following data:

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

For student 1:

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

For student 2:

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

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7 0

2 years ago

Read 2 more answers

A block of mass, m, sits on the ground. A student pulls up on

kakasveta [241]

Answer a

Explanation: a

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2 years ago

The mass of a giraffe on the moon would be

larisa [96]

Mass is an independent quantity. It can neither be created nor be destroyed. So mass of giraffe will not change on moon.

4 0

2 years ago

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A particle vibrates in Simple Harmonic Motion with amplitude. What will be its displacement in one time-period? A 2A 4A 0

yan [13]

'Displacement' is the distance and direction between the starting point and
ending point, regardless of the path followed to get there.

A particle that's executing simple harmonic motion is always in the same place
where it was one time period ago, and where it will be later after another time
period has passed.

So its displacement during exactly one time period is exactly zero.

3 0

3 years ago

Please help me im struggling​

Please help me im struggling