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2 years ago

How is work and energy inter-related?? (in simple words)

Physics

1 answer:

matrenka [14]2 years ago

4 0

Answer:

The work and energy is inter- related because if we are energetic then only we are able to work.

People who are weak and not energetic cannot work continuously.

Explanation:

hope this helps you....

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Does a rigid object in uniform rotation about a fixed axis satisfy the first and second conditions for equilibrium?

Vladimir [108]

Answer:

Explanation:

a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .

First condition of equilibrium is that net force on the body should be zero.

or F net = 0

A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0

So first condition of equilibrium is not satisfied.

Second condition is that , net torque acting on the body must be zero.

In case of a rigid object in uniform rotation , centripetal force is applied towards the centre ie towards the line joining the body under rotation with the axis .

F is along r

torque = r x F

= r F sinθ

θ = 0 degree

torque = 0

Hence 2nd condition is fulfilled.

7 0

3 years ago

Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec

STatiana [176]

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

Acceleration= $\frac{v-u}{t}$

Using the formula

Acceleration= $a=\frac{0-4.5}{0.5}$

deceleration=a= $-9m/s^2$

We know that

Force =ma

Using the formula and suppose mass of my body=m=40 kg

The force exerted by wall on you

Force= $40\times (9)=360N$

3 0

2 years ago

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of th

sattari [20]

Answer:

48 m

Explanation:

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.

The eastbound train

First convert km/h to m/s

(97 × 1000)/3600

97000/3600

26.944444 m/s

As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion

V^2 = U^2 - 2as

O = 26.944^2 - 2 × 3.5 S

726 = 7S

S = 726/7

S1 = 103.7 m

The westbound train

Convert km/h to m/s

(127×1000)/3600

127000/3600

35.2778 m/s

Using third equation of motion

V^2 = U^2 - 2as

0 = 35.2778^2 - 2 × 4.2 × S

1244.52 = 8.4S

S = 1244.52/8.4

S2 = 148.2 m

S1 + S2 = 103.7 + 148.2 = 251.86

The distance between them once they stop will be

300 - 251.86 = 48.14 m

Therefore, the distance between them once they stop is 48 metres approximately.

5 0

3 years ago

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, produ

aalyn [17]

Answer:

324795 C

252.637820565 N/C

$2.235844712\times 10^{-9}\ C/m^2$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

R = Radius of Mars = $3.4\times 10^6\ m$

A = Area = $4\pi R^2$

$\phi$ = Electric flux = $3.67\times 10^{16}\ Nm^2/C$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=3.67\times 10^{16}\times 8.85\times 10^{-12}\\\Rightarrow q=324795\ C$

The charge is 324795 C

Electric field is given by

$E=\dfrac{\phi}{A}\\\Rightarrow E=\dfrac{3.67\times 10^{16}}{4\pi (3.4\times 10^6)^2}\\\Rightarrow E=252.637820565\ N/C$

The electric field is 252.637820565 N/C

Surface charge density is given by

$\sigma=\dfrac{q}{4\pi R^2}\\\Rightarrow \sigma=\dfrac{324795}{4\pi (3.4\times 10^6)^2}\\\Rightarrow \sigma=2.235844712\times 10^{-9}\ C/m^2$

The surface charge density is $2.235844712\times 10^{-9}\ C/m^2$

6 0

3 years ago

Calculate work when the force exerted to push an object is 75 n and the distance is 1.2 kilometers.

grigory [225]

The magnitude of the work done to push the object is 90,000 J

Work is done on a body when the direction of the force applied is parallel to the direction of the displacement of the body.

The following parameters are given;

Force F = 75 N

Distance S = 1.2 Km

Work W = ?

First convert the distance in kilometer to meter by multiplying it by 1000

S = 1.2 x 1000 = 1200 m

From the definition of work,

W = F x S

Substitute the parameters into the formula

W = 75 x 1200

W = 90,000 J

Therefore, the magnitude of the work done to push the object is 90,000 J

Learn more about Work here: brainly.com/question/25573309

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7 0

1 year ago

How is work and energy inter-related?? (in simple words)​

How is work and energy inter-related?? (in simple words)