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Flauer [41]
2 years ago
15

Move 13 m west and then 8 m east?HELP!!!

Physics
1 answer:
Hoochie [10]2 years ago
5 0

Answer:

Displacement will be 5m west

Distance would be 21m No direction

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Q9. Inertia is the tendency of an object to resist any change in its motion.

Q10. A moving object's momentum depends on the object's mass and velocity.

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3 years ago
WHOEVER RESPONDS TO THIS FIRST WILL GET BRAINIEST!!!!!!
postnew [5]

Answer:

alright i responded

Explanation:

you didnt say i had to answer

6 0
3 years ago
Read 2 more answers
How much force is needed to accelerate a toy car of mass 8 kg at 2m/s
atroni [7]

Answer:

4N

Explanation:

f = ma

8kg x 2m/s = 4N

4 0
3 years ago
What should a sailboat operator do when approaching a PWC head-on
nekit [7.7K]

<u><em>In accordance with the International Regulation for the prevention of collisions at sea</em></u><u>: </u>

<u>1.- A sailing boat has a passing preference over a motorized boat, </u><u>except when the motor boat is limited by its draft</u><u>. </u>

<u>2.- The sailboat must maintain its course and speed. </u>

<u>3.- </u><em><u>If it is evident that the PWC does not respond</u></em><u>, the sailboat must sound the warning signal, and change its course to starboard. </u>

<u>4.- </u><u><em>All actions must be taken as soon as possible</em></u><u>. </u>

<u>5.- If a sailboat is using its engine, the situation changes, and in that case, both ships must alter to starboard.</u>

8 0
3 years ago
A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is
OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

U = \frac{GMm}{r}

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}

U_2= - 5.1180*10^9J

Point B )

We now use the average radius distance from the earth.

U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}

U_2= -5.8925*10^9J

Then,

\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)

\Delta U = 774.5*10^6

By the law of conservation of energy we know that,

\Delta U = \frac{1}{2}mv^2

clearing v,

v= \sqrt{2 \Delta U/m}

v= \sqrt{2*774.5*10^6 /94.2}

v= 4055.08m/s

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0
3 years ago
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