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uranmaximum [27]
3 years ago
11

A exerted by an object on another is a force

Physics
1 answer:
Gelneren [198K]3 years ago
4 0

hope this helps

Answer:

If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.

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You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling a
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Answer:

L=55.9m

Explanation:

The equation for the period of a simple pendulum is:

T=2\pi\sqrt{\frac{L}{g}}

In our case what we know is the period and the acceleration of gravity, and we need to know the length of the pendulum, so we can write:

L=(\frac{T}{2\pi})^2g

Which for our values is:

L=(\frac{15s}{2\pi})^2(9.81m/s^2)=55.9m

6 0
3 years ago
The volume of a balloon an be approximated by V = 4 3 π r 3 V=43πr3. If air is leaking from the balloon at a rate of 56 cubic ce
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Answer:

The rate  the radius of the balloon shrinking at the moment the radius is 5 centimeters is 0.1783 cm/s

Explanation:

Here we have

dV/dt = 56 cm³/s

\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3}\pi r^3)  = \frac{4}{3}\pi\cdot3r^2 \frac{dr}{dt}

When the radius is 5 cm we have

56 \hspace {0.09cm}cm^3/s= \frac{4}{3}\pi\cdot3\cdot 5^2 \cdot \frac{dr}{dt} = 314.16 \times \frac{dr}{dt}

Therefore,

56 \hspace {0.09cm}cm^3/s= 314.16   \hspace {0.09cm}cm^2\times \frac{dr}{dt}

From which,

\frac{dr}{dt} = 56 \hspace {0.09cm}cm^3/s \div314.16   \hspace {0.09cm}cm^2

\frac{dr}{dt} = 0.1783 \hspace {0.09cm}cm/s

The rate  the radius of the balloon shrinking at the moment the radius is 5 centimeters = 0.1783 cm/s.

6 0
3 years ago
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The green car traveled a shorter distance, but the displacement of the cars was equal.

Explanation:

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What substance can be used to electrolyze water?
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Amalgamating is the coating of zinc plate with mercury.
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