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3 years ago

| A 1.0 kg stone is dropped from a bridge 100 m above a canyon. What will be the kinetic energy of the stone after it

Physics

1 answer:

Mnenie [13.5K]3 years ago

8 0

Answer:

Option D

490 J

Explanation:

When at a height of 100 am above and released, the ball initially posses only potential energy. When it falls, some potential energy is converted to kinetic energy.

Initial potential energy= mgh where m is the mass, g is the acceleration due to gravity and h is height. Substituting 1 Kg for m, 9.81 for g and 100 m for h then

PE initial = 1*9.81*100= 981 J

At 50 m, PE will be 1*9.81*50=490.5 J

Subtracting PE at 50 m from initial PE we get the energy that has been converted to kinetic energy hence

981-490.5= 490.5 J

Approximately, 490 J

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a farmer lifts a bale of hay that is 12m off the ground . If he uses a force of 350 n, how much work does the farmer do?

hjlf

Answer:

4200 Joules

Explanation:

Work done =force x distance

From the question , we’re given f =350N ,

d = 12m

Using the above formula, we have

Workdone = 350 x 12

= 4200 Joules

5 0

3 years ago

Which of the following represents an image that is located in front of a lens?

Lostsunrise [7]

-di represents an image in front of a lens

5 0

3 years ago

Read 2 more answers

A converging lens of focal length 20 cm is placed in contact with a diverging lens of focal length 30 cm. The focal length of th

kramer

Answer:

The focal lenth (F) =+10.0cm

Explanation:

The formular for combined focal length (F) is given as;

$1/f = 1/f_{1} + 1/f_{2}$

In this question,

F1 = 20cm

F2 = -30cm

Plugging the values into the formuar above,

$1/f = 1/20 + 1/(-30)$

$1/f = 0.05 - 0.033[tex]1/f = -0.017f = [tex]1/ -0.017$

f = 58.82cm

i.e. the combination behaves as a converging lens (because of the postive sign) of focal length 58.82cm .

7 0

3 years ago

Convert 5g/cm^3 into kg/m^3

Sliva [168]

Answer:

0.01135624

Explanation:

7 0

2 years ago

In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0

2 years ago