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3 years ago

7

A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of

the average force on the wall if the duration of the collision is 0.18 s?

1 answer:

Alex73 [517]3 years ago

8 0

Answer:

4.7 N

Explanation:

130 g = 0.13 kg

The momentum of the snowball when it's thrown at the wall is

$p = mv = 0.13*6.5 = 0.845 kgm/s$

Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s

$p = F\Delta t$

$F*0.18 = 0.845$

$F = 0.845 / 0.18 = 4.7 N$

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MatroZZZ [7]

Answer:

600km/h as u are on a platform moving at the speed of 600 km/h where u are moving in relativity to the plane it's self.

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Lila is a track and field athlete.She has to complete four laps around the track, which is 400 meters. The race took her 6 minut

hoa [83]

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V=?
d=400m(4)
=1600m
t=6 min.
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3 years ago

If a metal ball suspended by a rod is at rest, which force is responsible for balancing the force due to gravity?

Viktor [21]

There are different kinds of forces; applied force, force of gravity, friction force, normal force, tension force. We will focus on the common forces, applied force and force of gravity. An applied force is a force that is applied to an object by another object. The force of gravity is the force with which massively large objects such as the earth attracts another object towards itself. All objects of the earth exert a gravity that is directed towards the center of the earth. Therefore, the force of gravity of the earth is equal to the weight of the object.

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3 years ago

natulia [17]

Answer:

$\% Error = 2.6\%$

Explanation:

Given

$x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45$

Required

Determine the percentage error

First, we calculate the mean

$\bar x = \frac{\sum x}{n}$

This gives:

$\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}$

$\bar x = \frac{9.06}{6}$

$\bar x = 1.51$

Next, calculate the mean absolute error (E)

$|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}$

This gives:

$|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}$

$|E| = \sqrt{\frac{1}{6}*0.0132}$

$|E| = \sqrt{0.0022}$

$|E| = 0.04$

Next, calculate the relative error (R)

$R = \frac{|E|}{\bar x}$

$R = \frac{0.04}{1.51}$

$R = 0.026$

Lastly, the percentage error is calculated as:

$\% Error = R * 100\%$

$\% Error = 0.026 * 100\%$

$\% Error = 2.6\%$

4 0

3 years ago

What does the term electron orbital describe? What does the term electron orbital describe? An electron orbital describes a thre

Yuliya22 [10]

Answer:

An electron orbital describes a three-dimensional space where an electron can be found 90% of the time.

Explanation:

According to Heisenberg's theory we cannot observe the position and velocity of an electron in an orbit, but if they were around the nucleus (in orbit), it would be possible to know its velocity and position, which would be contrary to the principle of Heisenberg So we can say that no electron revolves around a certain orbit around the nucleus, so we can only predict if the electron will be in the right position at the right time.

From there we find two definitions for electron orbital let's see:

Orbital is considered the region of space, where each electron spends most of its time.

Orbital is considered the region of space that is most likely to find an electron.

3 0

3 years ago