All categories

2 years ago

The condition required to work to be done

Physics

1 answer:

Dvinal [7]2 years ago

6 0

Answer:

work=f(costheta)

Explanation:

work is done when a force acts on a body and displaces it on the direction of force

You might be interested in

Mason notices that his boat sinks lower into the water in a freshwater lake than in the ocean. What could explain this?

zloy xaker [14]

Answer:

Salt water is more dense than freshwater.

8 0

2 years ago

Read 2 more answers

Which statement bet explains the relationship between the electric force between two charged objects and the distance between th

Nataliya [291]

Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

4 0

2 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers

A spring whose spring constant is 260 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required

Readme [11.4K]

Answer:

A spring whose spring constant is 200 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required to compress it another 1 inch.

Step 1 of 4

The force at any point during the deflection of the spring is given by,

where is the initial force

and x is the deflection as measured from the point where the initial force occurred.

The work required to compress the spring is

Therefore work required to compress the spring is

The work required to compress the spring in Btu is calculated by

Where 1Btu =778

The work required to compress the spring,

eman Asked on February 19, 2018 in thermal fluid Sciences 4th solutions.

Explanation:

7 0

2 years ago

What it mean head on collision?

murzikaleks [220]

I don’t understand the question

6 0

2 years ago

Read 2 more answers

The condition required to work to be done​

The condition required to work to be done