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3 years ago

A weightlifter presses a 400 n weight 0.5 m over his head in 2 seconds, what is the power of the weight lifter

1 answer:

3 0

There are two ways to solve this problem. First we write the given.

Given: Force F = 400 N; Height h = 0.5 m; Time t = 2 s

Formula: P = W/t; but Work W = Force x distance or W = f x d

Weight is also a Force, therefore: W = mg, solve for Mass m = ?

m = w/g m = 400 N/9.8 m/s² m = 40.82 Kg

P = W/t = F x d/t = mgh/t P = (40.82 Kg)(9.8 m/s²)/2 s

P = 100 J/s or 100 Watts

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What is the mass of a crate if a net force of 12 N gives the crate an acceleration of 0.20 m/s2?

Veseljchak [2.6K]

A :-) for this question , we should apply
F = ma
Given - F = 12 N
a = 0.20 m/s^2
Solution -
F = ma
12 = m x 0.20
m = 12 by 0.20
m = 60 kg

.:. The mass is 60 kg.

8 0

3 years ago

A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant

german

The work done to stretch the spring will be 112 J.

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

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

The work is done to stretch the spring is;

$\rm W= \frac{1}{2} kx^2 \\\\ W=\frac{1}{2} \times 350 \times (0.850-0.050)^2 \\\\ W=0.5 \times 350 \times (0.80)^2 \\\\W=112 \ J$

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ1

3 0

2 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

Veronika [31]

Answer:

Explanation:

Given that,

One fragment is 7 times heavier than the other

Let one fragment mass be M

Let this has a velocity v

And the other 7M

And this a velocity V

Initially the fragment is at rest u = 0

Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

Po = Pf

(M+7M) × 0 = 7M •V − Mv

0 = 7M•V - Mv

Divide both sides by M

0 = 7V -v

v = 7V

Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

Work done by M mass

W(m) = −µk•Mg×d'

Since it is a wordone by friction, that is why we have a negative sign.

Using conservation of energy

Work done by 7M mass is equal to work done by M mass

W(7m) = W(m)

−47.6 µk•Mg = −µk•Mg×d

Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

7 0

3 years ago

Read 2 more answers

A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance

Minchanka [31]

Answer:

$F=43570.9N$

Explanation:

We can calculate the acceleration experimented by the passenger using the formula $v_f^2=v_i^2+2ad$ , taking the initial direction of movement as the positive direction and considering it comes to a rest: