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muminat
3 years ago
13

Kinetic friction is affected by the weight of the object. True or false?

Physics
1 answer:
nevsk [136]3 years ago
3 0

True Kinetic friction is affected by the weight of the object

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Considering only the earth's rotation, determine how much later the asteroid would have had to arrive to put the explosion above
tangare [24]

Answer:

5.1 hours

Explanation:

The only fact we need to know about such a question is that when gazing down at the north pole, the earth spins longitudinally at 360 degrees / day in the clockwise direction.

The planet would have to spin an additional 77 ° to strike the asteroid at 25° E. If the earth rotates in 24 hours 360 degrees, then it must it rotates in 5.1 h at 77 degrees.

8 0
3 years ago
Which of the following is equal to Velocity change per second​
Mademuasel [1]

Acceleration

Explanation:

Acceleration is a physical quantity that expresses the change in the velocity of a body per unit of time.

  Acceleration = \frac{V - U }{T}

V is the initial velocity

U is the final velocity

T is the time

It is has a unit of m/s²

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

8 0
3 years ago
I just need x isolated
mamaluj [8]

Answer:

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

Explanation:

rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0

Solve using the quadratic formula.

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

7 0
3 years ago
Read 2 more answers
:3333333333333333333333
elixir [45]

Answer:

Cannot see all the answer choices but, when pressure goes up, volume goes down. So your best bet is to choose the graph that has a downward slant.

Explanation:

3 0
3 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
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