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3 years ago

Galilee said that if you rolled a ball along a level surface it would be

Physics

1 answer:

lana [24]3 years ago

3 0

The answer would be stay because the surface is flat so it will stay!

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A virtual image formed by convex lens is always

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2. erect and magnified

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What will be the weight of a man of mass 90kg when he is on a planet with acceleration due to gravity of 15ms-2?

Feliz [49]

Answer:

1350N

Explanation:

Weight = Mass x Acceleration Due to Gravity

W=mg

W=90*15=1350N

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4 years ago

The diagram shows four paths from point A to point B.

iVinArrow [24]

Path 2.

Displacement is the direction and magnitude of an object from its starting point, so path 2 is the direct route you would need to take to find direction and magnitude.

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2 years ago

If a block of wood has a density of 0.6g/cm3 and a mass of 120g what is the volume

Luda [366]

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It would be 39 :))

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3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago