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3 years ago

A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter

Physics

1 answer:

icang [17]3 years ago

7 0

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P = power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

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WHICH TYPES OF ENERGY ARE HAPPENING

Lady bird [3.3K]

Answer:

mecanichal energy

Explanation:

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4 years ago

A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac

Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

kinetic energy at base of track = ½(20)5.0² = 250 J

energy (work) of friction 352.8 - 250 = 102.8 J

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3 years ago

A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha

cricket20 [7]

The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

$k=\dfrac{4\pi ^2m}{T^2}$

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

$k=\dfrac{4 \pi \times 0.2}{(1.2)^2}$

$k=\dfrac{39.48\times 0.2}{1.44}$

$k=\dfrac{7.90}{1.44}$

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

To know more about spring constant follow

brainly.com/question/1968517

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4 0

2 years ago

1 Point

irakobra [83]

Answer:

SECOND LAW OF NEWTON

Explanation:

When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression

I = F t = Δp

As we can see this expression is a form of Newton's second law

F = m a

a = dv / dt

F = m dv / dt

F dt = m dv

p = mv

F dt = dp

Therefore the station moves through the SECOND LAW OF NEWTON

7 0

3 years ago

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

3 0

3 years ago