Ask question

Ask question

All categories

3 years ago

10

A weightlifter lifts a set of weights a vertical distance of 2 meters. If a constant net force of 350 N is exerted on the weight

s, what is the amount of work done on the weights? :
350N multiplied by 2m = 700

**if the weightlifter in the previous question takes a total of 2 seconds to life the weights, how much power is performed

1 answer:

o-na [289]3 years ago

6 0

Answer:

The work done by the weightlifter, W = 700 J

The power of the weightlifter, P = 350 watts

Explanation:

A weightlifter lifts a set of weights a vertical distance, s = 2 m

The force exerted to lift the weight, F = 350 N

The work done by the body is defined as the product of the force applied by the body to the displacement it caused.

W = F x s

= 350 N x 2 m

= 700 J

The work done by the weightlifter, W = 700 J

The time taken by the weightlifter to lift the weight, t = 2 s

The power is defined as the rate of body to do work. It is given by the equation,

P = W / t

= 700 J / 2 s

= 350 watts

Hence, the power of the weightlifter, P = 350 watts

You might be interested in

Earth’s crust continuously melts and solidifies in the upper mantle. Why does oceanic crust solidify faster than continental cru

Softa [21]

The oceanic crust solidifies faster than the continental crust. Because when magma comes into contact with ocean water it cools faster than when it reaches continental rock. Option C is correct.

<h3>What is the crust?</h3>

The outermost layer of a terrestrial planet is referred to as its "crust."

In the upper mantle, the Earth's crust is constantly melting and solidifying. Compared to magma beneath the continental crust, the oceanic crust has a lower temperature.

Magma cools more quickly when it comes into touch with ocean water than when it hits continental rock. causes the continental crust to solidify more slowly than the oceanic crust.

Hence option C is correct.

To learn more about the crust refer;

brainly.com/question/13428623

#SPJ1

3 0

2 years ago

A force unit is defined as the product of a mass unit and a unit of

butalik [34]

Answer:

A force unit is defined as the product of a mass unit and a unit of acceleration.

5 0

3 years ago

Read 2 more answers

A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

Pie

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) $I=0.0085\ A$

Explanation:

Given:

radius if the coil, $r=0.0395\ m$

no. of turns in the coil, $n=520$

variation of the magnetic field in the coil, $B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4$

resistor connected to the coil, $R=560\ \Omega$

(a)

we know, according to Faraday's Law:

$emf=n.\frac{d\phi}{dt}$

where:

$d \phi=$ change in associated magnetic flux

$\phi= B.A$

where:

A= area enclosed by the coil

Here

$A=\pi.r^2$

$A=\pi\times 0.0395^2$

$A=0.0049\ m^2$

$\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049$

So, emf:

$emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]$

$emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]$

$emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]$

$emf= 0.0306+3.516\times 10^{-4}\ t^3$

(b)

Given:

$t_0=5.25\ s$

Now, emf at given time:

$emf=4.7755\times 10^{-2}\ V$

∴Current

$I=\frac{emf}{R}$

$I=\frac{4.7755\times 10^{-2}}{560}$

$I=8.5\times 10^{-5} A$

6 0

4 years ago

A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where

const2013 [10]

Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:

$W = ^{x_{2}}_{x_{1}}\int F_{x} dx$

Where:

x₁: is the lower limit = 0 m

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N

$W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx$

$W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}$

$W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}$

$W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}$

$W = 29.85 J$

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!

6 0

3 years ago

A jar is completely filled with water, and then it is sealed with a lid. If you push down on the lid with a pressure of 1000 Pa,

notsponge [240]

Answer:

The water pressure at the bottom of the jar will increase by 1000 Pa.

Explanation:

The Pascal principle states that:

In a fluid, a change in pressure at any point in the fluid is transmitted equally throughout the fluid, as it is occuring everywhere.

If we apply this principle at the case mentioned in the problem, we can say that:

- An initial pressure of 1000 Pa is applied on the top of the fluid (the water)

- According to Pascal's law, this pressure is transmitted with equal intensity (1000 Pa) to every point of the fluid

- So, the water pressure at the bottom of the jar will also increase by the same amount, 1000 Pa

5 0

3 years ago