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3 years ago

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It turns out that Mercury and Mars have the same gravity as one another – that is, you would weigh the same on the surface of Me

rcury as you would on the surface of Mars, despite the fact that Mercury is smaller. Why is this the case?

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

Explanation:

The value of acceleration due to gravity of mars is same as that the value of acceleration due to gravity of Mercury.

The value of acceleration due to gravity of a planet depends on its mass and the radius.

The formula for the acceleration due to gravity is given by

$g=\frac{GM}{R^{2}}$

Where, M be the mass of the planet, R be the radius of the planet.

As according to the question, the value of acceleration due to gravity for Mercury is same as that of Mars.

$\frac{GM_{mercury}}{R_{mercury}^{2}}=\frac{GM_{mars}}{R_{mars}^{2}}$

$\frac{M_{mercury}}{R_{mercury}^{2}}=\frac{M_{mars}}{R_{mars}^{2}}$

As teh radius of Mercury is small, and acceleration due to gravity is same, it is possible because the mass of Mercury is more than the mass of Mars.

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8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a

Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61 N

work = 62.61 * 2 = 125.22 j

Net Force downplane = force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

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2 years ago

Difference between artificial magnet and natural magnet??

marissa [1.9K]

Answer:

A natural magnet is a magnet that occurs naturally in nature. Artificial magnets are magnets made by people. ... An example of a natural magnet is the lodestone, also called magnetite. Other examples are pyrrhotite, ferrite, and columbite.

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3 years ago

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41. 2072 Set E Q.No. 11 A source of sound produces a note of

Nimfa-mama [501]

A. is the answer for this

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2 years ago

A ball is hurled straight up at a speed of 15 m/s, leaving the hand of the thrower 2.00 m above the ground. Compute the times an

SVETLANKA909090 [29]

Answer:

5.37 m/s

0.98 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 10+15^2}\\\Rightarrow v=5.37\ m/s$

Velocity of the ball when it passes an observer sitting at a window is 5.37 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{5.37-15}{-9.81}\\\Rightarrow t=0.98\ s$

Time taken by the ball to pass the observer sitting at a window is 0.98 seconds

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3 years ago

A factory worker pushes a 32.0 kgkg crate a distance of 4.0 mm along a level floor at constant velocity by pushing horizontally

Luda [366]

Answer:

a) The force that the worker must apply has a magnitude of 75.317 newtons.

b) The external force does a work of 0.301 joules.

c) The friction force does a work of -0.301 joules.

d) Both normal force and gravity have done a work of 0 joules.

e) The total work done on the crate is 0 joules.

Explanation:

a) As the crate is moving at constant velocity, we know that magnitude of the force done on the crate must be equal to the friction force. Hence, we must use the following formula:

$F = \mu\cdot m\cdot g$ (1)

Where:

$F$ - External force, in newtons.

$\mu$ - Coefficient of kinetic friction, no unit.

$m$ - Mass, in kilograms.

$g$ - Gravity acceleration, in meters per square second.

If we know that $\mu = 0.24$ , $m= 32\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the external force is:

$F = (0.24)\cdot (32\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 75.317\,N$

The force that the worker must apply has a magnitude of 75.317 newtons.

b) The direction of the force is parallel to the direction of motion. The work done by this force ( $W_{F}$ ), in joules, is determined by this formula:

$W_{F} = F\cdot \Delta s$ (2)

Where $\Delta s$ is the distance travelled by the crate, in meters.

If we know that $F = 75.317\,N$ and $\Delta s = 0.004\,m$ , then the work done by the force is:

$W_{F} = (75.317\,N)\cdot (0.004\,m)$

$W_{F} = 0.301\,J$

The external force does a work of 0.301 joules.

c) The direction of the friction force is antiparallel to the direction of motion. The work done by this force ( $W_{f}$ ), in joules, is determined by this formula:

$W_{f} = -W_{F}$ (3)

$W_{f} = -0.301\,J$

The friction force does a work of -0.301 joules.

d) The direction of the normal force is perpendicular to the direction of motion. Therefore, no work is done due to normal force.

$W_{N} = 0\,J$

Likewise, no work is done by gravity.

$W_{g} = 0\,J$

Both normal force and gravity have done a work of 0 joules.

e) The total work is the sum of the works done by the external force and the friction force:

$W = W_{F}+W_{f}$

$W = 0.301\,J - 0.301\,J$

$W = 0\,J$

The total work done on the crate is 0 joules.

5 0

2 years ago