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3 years ago

13

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

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