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ycow [4]
3 years ago
14

How much work is required to lift a 2 kg mass to a height of 10 meters?

Physics
1 answer:
scoray [572]3 years ago
6 0
Use the equation potential energy =m*g*h
m-mass,
g-gravitational acceleration,
h-height
Potential energy = 2*10*10
=200
kg {m}^{2}   {s}^{ - 2}
This is the unit of the energy
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What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s
Furkat [3]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity v and at a distance of r from the axis of rotation is given as:

a=\frac{v^2}{r}

Here, v=8\ m/s,r=0.70\ m

∴ a=\frac{8}{0.70}=11.43\ m/s^2

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

3 0
4 years ago
What will one see during an annular eclipse?
KATRIN_1 [288]

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7 0
3 years ago
At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0
3 years ago
(ix) A pin hole camera gives a sharp image because;
sdas [7]

Answer:

d) The screen is in a short distance from the hole

8 0
3 years ago
Read 2 more answers
A string exerts a force of 20 N on a box at an angle of 38° from the horizontal. What is the horizontal component of the force o
dem82 [27]

Answer:

  15.8 N

Explanation:

The component in a given direction is the magnitude of the applied force, multiplied by the cosine of the angle between its application and the direction of interest.

The horizontal component is ...

  (20 N)cos(38°) ≈ 15.8 N

7 0
3 years ago
Read 2 more answers
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