For a projectile, why is the initial horizontal velocity equal to the final horizontal velocity?

Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an

Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

$W = F\cdot \Delta s$

Where:

$W$ - Work, measured in joules.

$F$ - Force exerted on the object, measured in newtons.

$\Delta s$ - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

$W_{net} = W_{Joel} - W_{Jerry}$

$W_{net} = F\cdot \Delta s - F\cdot \Delta s$

$W_{net} = (F-F)\cdot \Delta s$

$W_{net} = 0\cdot \Delta s$

$W_{net} = 0\,J$

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

$W_{net} = W_{Joel}$

$W_{net} = F\cdot \Delta s$

Hence, the work done by Joel is greater than the work done by Jerry.

7 0

3 years ago

In which of the following situation is light most likely to be a reflection

maxonik [38]

Answer:

when the reflecting surface is plain and without even small hurdles that are not the visible by our naked eyes. Eg : plain mirror

Explanation:

7 0

3 years ago

The length of a 100 mm bar of metal increases by 0.3 mm when subjected to a temperature rise of 100°C. The coefficient of linear

Juli2301 [7.4K]

Answer:

α = 3×10^-5 K^-1

Explanation:

let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.

then, the coefficient of linear expansion is given by:

α = ΔL/(ΔT×L)

= (0.3×10^-3)/(100)(100×10^-3)

= 3×10^-5 K^-1

Therefore, the coefficient of linear expansion is 3×10^-5 K^-1

5 0

4 years ago

Based on the principles of convection, conduction and thermal radiation, which scenario below is most similar to the following s

Ksivusya [100]

The situation (heat going through the ceiling) describes
conduction ... heat going from one place to another by
soaking through some material.

A). This is the one. Heat goes from from the marshmallow
to your hand by soaking through the wire. This is conduction too.

B). No. The heat in the room goes from the floor to the ceiling
because the warm air rises and carries it there. This is convection.

C). No. There's nothing for the heat to soak through between
the sun and the roof, and nothing that can move from the sun
to the roof and bring the heat with it. This is radiation.

D). No. Cold water sinks from the surface to the bottom because
warm water rose from the bottom to the surface, taking heat with it.
This is convection.

5 0

4 years ago

Read 2 more answers

1. A student lifts a box of books that weighs 185 N. The box is

aksik [14]

1) 148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

$W = \Delta U = (mg) \Delta h$

where

mg is the weight of the object

$\Delta h$ is the change in height

For the box in this problem,

mg = 185 N

$\Delta h = 0.800 m$

Substituting into the equation, we find:

$W=(185)(0.800)=148 J$

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

$W=Fd$

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

$W=(825)(35)=28875 J$

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

$W=Fd$

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

$W' = 2(28875)=57750 J$

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

$F = mg$

where

m = 0.180 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation,

$F=(0.180)(9.8)=1.76 N$

Now we find the work done by gravity using the same formula applied before:

$W=Fd$

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

$W=(1.76)(2.5)=4.4 J$

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

$\Delta h = 1.2 m$

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

$W=mg \Delta h$

Solving for m, we find

$m=\frac{W}{g \Delta h}$

And substituting the numerical values, we find the mass of the box:

$m=\frac{7000}{(9.8)(1.2)}=595.2 kg$

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

$W=mg \Delta h$

Where $\Delta h$ is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of $\Delta h$ is the same for both of them, so the work they have done is exactly the same.

5 0

4 years ago