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3 years ago

What's the relationship between work done and force applied

1 answer:

3 0

When a force acts on a body along some path, the work done is W=F*s, where W is the work done, F is the force that is doing the work on the body and s is the path. The force doing the work has to be in the same direction, or parallel, as the path. This is called positive work. If the force and the path are anti-parallel, the work is negative. So the relationship between work and force is W=F*s.

You might be interested in

A mother pushes a baby stroller 10 meters by applying 40 newtons of force. how much work was done?

topjm [15]

Based on the situation above the the work done was 400 Joules. Q = FS cos(theta) is the so-called work function. It's important to learn the work physics; you'll see it over and over in science/physics class. Theta is the angle between the force vector F and the distance vector S. In your problem we assume theta = 0, the two vectors were assumed aligned.

7 0

3 years ago

Read 2 more answers

A quantum well is a region of semiconductor or metal film that confines particles to a small region, resulting in quantized ener

murzikaleks [220]

Answer:

226.2 m/sec

Explanation:

We have given $\Delta x=510nm=510\times 106{-9}m$

The plank's constant $h=6.67\times 10^{-34}$

Mass of electron $m=9.11\times 10^{-31}kg$

Now according to Heisenberg uncertainty principle $\Delta v\geq \frac{h}{2\pi m\Delta x}$

So $\Delta v\geq \frac{6.6\times 10^{-34}}{2\times 3.14\times 9.11\times 10^{-31}\times 510\times 10^{-9}}=226.2m/sec$

5 0

2 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

Conservation of linear momentum:

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

Conservation of kinetic energy:

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

2 years ago

ehicle collisions involve huge amounts of energy. How does the crashworthiness of a car affect the transfer and transformations

Angelina_Jolie [31]

When the two-vehicle collides transformation of the energy is done in terms of kinetic energy.

<h3>What is the law of conservation of energy?</h3>

According to the Law of Conservation of Energy, energy can neither be created nor destroyed, but it can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

In the above problem, the Vehicle get collides so that the kinetic energy of the vehicle is converted into the kinetic energy of another vehicle the speed of the vehicle will reduce when they collide.

Momentum also gets conserved when the two vehicles collide.

Hence, the transformation of the energy is done in terms of kinetic energy.

To learn more about the law of conservation of energy, refer to brainly.com/question/2137260.

#SPJ1

8 0

1 year ago

A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou

qwelly [4]

Answer:

-v/2

Explanation:

Given that:

a disc of mass m

Collides with the wall going through a sliding motion on on the plane smooth surface.

Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

We know, kinetic energy is given as:

$KE_i=\frac{1}{2}. m.v^2$

consider this to be the initial kinetic energy of the body.

Now after collision:

$KE_f=\frac{1}{4}\times KE_i$

$KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2$

Considering that the mass of the body remains constant before and after collision.

$KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2$

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

3 0

2 years ago