20) If there was no gravity, what would happen to a moving object? Which law is this? 21) What factors are included in gravity

1 answer:

3 0

20) a moving object would still move but it would also float since no pressure is holding it down.
21) The amount of mass determines the gravitational pull, the more the mass the greater the pull.

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A company is creating a new mosquito repellent, and needs to figure out how much is needed to stop mosquito bites. They find vol

Jlenok [28]

The independent variable is the different amount of repellent each person is given. This is because it does not depend on any variable. The dependent variable is what they are measuring, which would be the number of bites on each volunteer.
I hope this helps! :)

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1 year ago

Since all objects are ‘weightless’ for an astronaut in orbit, is it possible for astronauts to tell whether an object is heavy o

ivanzaharov [21]

W=gm
where g - gravitation
m - mass
w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light

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3 years ago

Read 2 more answers

Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a

Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

Force applied on camel feet = 4000 N
Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

${\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}$

Substituting all the given values in the formula to find the pressure exerted by camel feet.

$\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\ \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}} \\ \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\ \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\ \\\star \: \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}$

Hence, the pressure exerted by camel feet is 2000 N/m².

$\rule{300}{2.5}$

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2 years ago

Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations

lapo4ka [179]

Answer: To increase the rigidity of the system you could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

Explanation:

When a rule is displaced from its vertical position, it oscillates back and forth because of the restoring force opposing the displacement. That is, when the rule is on the left there is a force to the right.

By holding a ruler with one hand and deforming it with the other a force is generated in the opposite direction which is known as the restoring force. The restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. The momentum gained causes the ruler to move to the right leading to opposite deformation. This moves the ruler again to the left. The whole process is repeated until dissipative forces reduce the motion causing the ruler to come to rest.

The relationship between restoring force and displacement was described by Hooke's law. This states that displacement or deformation is directly proportional to the deforming force applied.

F= -kx, where,

F= restoring force

x= displacement or deformation

k= constant related to the rigidity of the system.

Therefore, the larger the force constant, the greater the restoring force, and the stiffer the system.

5 0

3 years ago

A man wishes to row the shortest possible distance from north to south across a river which is flowing at 2 km/hr from the east.

sweet-ann [11.9K]

From the diagram we have that

$sin\theta = \frac{2}{4}$