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3 years ago

A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?

Physics

1 answer:

Alenkasestr [34]3 years ago

8 0

accn = - 26m/s divided by 13 secs

accn = 2m/s/s

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A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on

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F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N

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Which force is represented by the arrow at

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A is pulling the block straight down toward the center of the Earth, no matter what the slope of the plane may be. A is the force of gravity.

The directions of B and C both depend on the slope of the plane.

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2 years ago

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What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

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3 years ago

choose the correct anwer 9: the range of the gravitional force is given by A: 10_2m B:10_15m C: infinite D: 10_10m

frosja888 [35]

Answer:

The answer is C

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3 years ago

The escape speed from the moon is much smaller than from earth. True or False

Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

$v_{e} = \sqrt{\frac{2GM}{r}}$ (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is $5.972x10^{24}kg$ and its radius is $6371000m$

Then, replacing those values in equation 1 it is gotten.

$v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}$

$v_{e} = 11.18m/s$

For the case of the Moon:

$v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}$

$v_{e} = 2.38m/s$

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

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3 years ago