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Elena L [17]
3 years ago
7

Action-reaction forces are always the same ____ but are in opposite ____​

Physics
1 answer:
aivan3 [116]3 years ago
5 0

Answer:

Magnitude, direction

Explanation:

Magnitude, direction

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The indian ocean is the third largest ocean at 68,556,000 sq km
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As the mass of object increased it is density increased
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Answer:

True.

Explanation:

The density of an object is given by its mass divided by its volume. It can be given as follows :

d=\dfrac{m}{V}

It can be seen that the density of an object is directly proportional to its mass. It means if the mass of an object increase, its density will also increase. Hence, the given statement is true.

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3 years ago
In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
4 years ago
An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t
Fofino [41]

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

P(z)=P_{0}+\rho _wgh

Now the gauge pressure is given by

P(z)-P_{0}=\rho _wgh

Applying values we get

P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa

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