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DanielleElmas [232]
3 years ago
13

A 50.0-kg girl stands on a 9.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wag

on at a speed of 5.0 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time
Physics
1 answer:
Kobotan [32]3 years ago
5 0
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What is energy saving..Write any three methods of saving energy..​
Basile [38]

Answer:

Energy conservation is the effort made to reduce the consumption of energy by using less of an energy service. This can be achieved either by using energy more efficiently (using less energy for a constant service) or by reducing the amount of service used (for example, by driving less).

Explanation:

8 0
3 years ago
Mass is constant, but weight can change with location. Explain.
disa [49]
Mass is always constant. You are 50 kgs no matter how far above the surface of the earth you are, imagine standing on a high building you'll still be 50 kgs. But weigh depends on th gravitational acceleration at the point ie as you move up the surface of the earth the gravitational acceleration fluctuates and is not constant. But at school level it's 9.8 m.s-2
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3 years ago
Ano -ano ang uri ng terorismo​
kogti [31]

Answer:

nagaganap sa loob ng bansa sa pagitan ng mga pinuno

6 0
3 years ago
Jim strikes a 0.058 kg golf ball with a force of 272 N, giving it a velocity of 62.0 m/s. How long was the ball in contact with
Mekhanik [1.2K]

Answer: 0.013 seconds

Explanation:

Given that

Mass of golf ball = 0.058 kg

Force = 272 N

Velocity = 62.0 m/s

Time taken = ?

Recall that force is the rate of change of momentum per unit time

i.e Force = Change in momentum / Time

i.e Force = (Mass x velocity) / Time

272N = (0.058 kg x 62.0 m/s) / Time

272N = 3.6kgm/s / Time

Time = (3.6kgm/s / 272N)

Time = 0.013 seconds

Thus, the ball was in contact with the club for 0.013 seconds

8 0
4 years ago
In a race, Usain Bolt accelerates at
jeka94

Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

3 0
3 years ago
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