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Levart [38]
3 years ago
11

A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy​

Physics
1 answer:
AURORKA [14]3 years ago
8 0

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

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9/10 in percents. Examples please, at least 2.
Len [333]

Answer:

90/100%, 9/10%

Explanation:

These represent 9 of 10 parts.

Hope this answer helps you out.

8 0
3 years ago
Read 2 more answers
Electric power companies measure energy consumption in kilowatt-hours, denoted kWh. One kilowatt-hour is the amount of energy tr
Natali [406]

The amount of energy used in the billing period is 5,400,000,000 joules.

One kWH is the amount of energy transferred in one hour, there 1 kWh is equal to

1 kWh=1*1000*(J/s)*3600 s

=3600000 J

Thus the amount of energy in joules consumed by the user for the billing period is =1500*3600000=5,400,000,000 J.

The amount of energy used in the billing period is 5,400,000,000 joules.

7 0
3 years ago
The height of a helicopter above the ground is given by h = 3.15t3, where h is in meters and t is in seconds. At t = 2.10 s, the
alexandr1967 [171]

Answer:

The mailbag will take 2.44 seconds to reach the ground.

Explanation:

The height of a helicopter above the ground is given by:

h = 3.15\times t^3

Height of helicopter at t = 2.10 seconds

h(2.10 )=3.15\times (2.10 )^3 m=29.17 m

The helicopter releases a small mailbag from the height of 29.17 m.

The initial velocity of mailbag = u = 0 m/s

Duration in which mailbag will reach the ground = T

Acceleration due to gravity = g = 9.8 m/s^2

Using second equation of motion ;

s=ut+\frac{1}{2}gt^2

We have , s = 29.17

u = 0 m/s

t = T

29.17m=0 m/s\times T+\frac{9.8 m/s^2\times T^2}{2}

Solving for T, we gte :

T = 2.44 seconds

The mailbag will take 2.44 seconds to reach the ground.

5 0
3 years ago
With a speed of 75 m sl. Determine
lana [24]

Answer:

120000    kgxm/s

Explanation:

momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000  kgxm/s

6 0
3 years ago
Why are drills performed in drum beats ​
Whitepunk [10]

Answer:

The drum drill is just one option in stride frequency development. Most of the time, the drum drill can be seen as just a rhythm drill that allows an athlete to relax and experiment with the right range of motion and bounce. A solid background in floating drills and developing reactivity should help athletes mold their stride into a balanced motion that maximizes their speed.

I have used frequency drills for years and now understand the nature of stride development mainly from shaping the stride parameters we all have known about for a long time. The drum drill is a special exercise that can make a great change in athletes who are receptive to improving and with a coach who is worth their salt in instruction. The drum drill is just one option for improving an athlete, and it’s more than fine to use any method you see fit that helps improve stride frequency.

Explanation:

6 0
2 years ago
Read 2 more answers
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