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Alenkasestr [34]
3 years ago
8

how much gravitational potential energy does a 5 kg rock have if is sitting on the edge of a cliff that is 10m high

Physics
1 answer:
docker41 [41]3 years ago
8 0
Gpe =5 x 9.8 x 10 = .... Joules
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What is the acceleration of a 10 kg mass pushed by a 5 N force?
yanalaym [24]
g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}=  \\ g=0,5 \frac{N}{kg}
8 0
3 years ago
Read 2 more answers
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag
Readme [11.4K]

Answer:

(a)0.0675  J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675  J

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0
3 years ago
A loudspeaker diaphragm is vibrating in simple harmonic motion with a frequency of 760 Hz and a maximum displacement of 0.85 mm.
Alchen [17]

Answer:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²

Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m

(a) Angular frequency w= 2πf

w= 2π × 760 = 4775.2Hz

(b) Maximum speed v is given as the product of angular frequency and maximum displacement

V=wx

V= 4775.2 × 0.00085

V= 4.06m/s

(c) The maximum acceleration a

= w²x

= (4775.2)² × (0.00085)

a= 19382.15m/s².

5 0
3 years ago
A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0
3 years ago
How do land and sea breeze work?
Studentka2010 [4]
 by the wind and air flow in the wind

7 0
3 years ago
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