Can I get a direct answer please??

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

a)

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

b)

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

4 0

2 years ago

A wind-tunnel experimentis performed on a 1/25scale model of a supersonic aircraft. The prototype aircraft flies at 450 m/s in c

KengaRu [80]

Answer: V = 504m/a

F = 4N

Explanation: please find the attached file for the solution

4 0

3 years ago

You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

7 0

3 years ago

24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

il63 [147K]

Answer:

...do

Explanation:

24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

cm. What is the length of the book in mm?

25. Explain the modifications

3 0

3 years ago

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

elena-s [515]

Answer:

The change in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

Explanation:

From the question we are told that

The magnitude of the uniform electric field is $E = 950 \ N/C$

The distance traveled by the electron is $x = 2.50 \ m$

Generally the force on this electron is mathematically represented as

$F = qE$

Where F is the force and q is the charge on the electron which is a constant value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally the work energy theorem can be mathematically represented as

$W = \Delta KE$

Where W is the workdone on the electron by the Electric field and $\Delta KE$ is the change in kinetic energy

Also workdone on the electron can also be represented as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ considering that the movement of the electron is along the x-axis

So

$\Delta KE = F * x cos (0)$

substituting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

Now From the law of energy conservation

$\Delta PE = - \Delta KE$

Where $\Delta PE$ is the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

7 0

3 years ago