All categories

3 years ago

A 3900 kg truck is moving at 6.0 m/s what is the kinetic energy

Physics

2 answers:

Stolb23 [73]3 years ago

3 0

Answer:

70200J

Explanation:

k.E = 1/2mv^2

K.E = 1/2(3900)(6)^2

finlep [7]3 years ago

3 0

Here you go, hope this helped

You might be interested in

Which of the following sets of characteristics describe what we know about the outer planets? (2 points)

masha68 [24]

Answer: D

Explanation: Gaseous composition, larger size and many moons

6 0

3 years ago

A 62.0-kg skier is moving at 6.30 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.90

Klio2033 [76]

Here are the missing questions:
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?
Part A
The initial kinetic energy of the skier is:
$E_{k0}=m\frac{v_0^2}{2}$
Part of this energy is then used to do work against the force of friction. Force of friction on the horizontal surface can be calculated using following formula:
$F_f=mg\mu$
The work is simply the force times the length:
$W_f=F_f\cdot L=mg\mu L$
So when the skier passes over the rough patch its energy is:
$E=E_{k0}-W_f$
When the skier is going down the skill gravitational potential energy is transformed into the kinetic energy:
$E_p=E_{k1}\\ mgh=E_{k1}$
So the final energy of the skier is:
$E_f=E_{k0}-W_f+E_{k1}\\ E_f=m\frac{v_0^2}{2}-mg\mu L+mgh=1856.86$J$
This energy is the kinetic energy of the skier:
$E_f=m\frac{v_f^2}{2}\\ v_f=\sqrt{\frac{2E_f}{m}}=7.74\frac{m}{s}$
Part B
We know that skier lost some of its kinetic energy when crossing over the rough patch. This energy is equal to the work done by the skier against the force of friction.
$E_{int}=W_f\\
E_{int}=mg\mu L=894.1$J$

4 0

4 years ago

Newton has Three Laws of Motion. Seat belts are used to help keep the passengers remaining in their seats during an accident. If

kherson [118]

Newton’s first law of inertia
Less force, same work

5 0

3 years ago

Derive the following equations of motion

xz_007 [3.2K]

Answer:

___________________________________

<h3>a. Let us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so that the final velocity 'v' after a time interval 't'. Now, By the definition of acceleration, we have:</h3>

$a = \frac{v - u}{t} \\ or \: at = v - u \\ v = u + at \:$

It is first equation of motion.

___________________________________

<h3>b. Let us assume a body moving with an initial velocity 'u'. Let it's final body 'v' after a time interval 't' and the distance travelled by the body becomes 's' then we already have,</h3>

$v = u + at...........(i) \\ s = \frac{u + v}{2} \times t.........(ii)$

Putting the value of v from the equation (i) in equation (ii), we have,

$s= \frac{u + (u + at)}{2} \times t \: \: \\ or \: s = \frac{(2u + at)t}{2} \\ or \: s = \frac{2ut + a {t}^{2} }{2} \\ s = ut + \frac{1}{2} a {t}^{2}$

It is third equation of motion.

________________________________

<h3>c. Let us assume a body moving with an initial velocity 'u'. Let it's final velocity be 'v' after a time and the distance travelled by the body be 's'. We already have,</h3>

$v = u + at.....(i) \\ s = \frac{u + v}{2} \times t......(ii) \\$

$v = u + at \\ or \: at = v - u \\ t = \frac{v - u}{a}$

Putting the value of t from (i) in the equation (ii)

$s = \frac{u + v}{2} \times \frac{v - u}{a} \\ or \: s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ or \: 2as = {v}^{2} - {u}^{2} \\ {v}^{2} = {u}^{2} + 2as$

It is forth equation of motion.

________________________________

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

Why is the moons surface much more heavily cratered than earths surface

wlad13 [49]

The earth has less creators because there is a protective atmosphere which makes the asteroids burn up before they hit the surface of earth. Which means by the moon not having an atmosphere small rocks are easily colliding with the moons surface.

4 0

3 years ago

A 3900 kg truck is moving at 6.0 m/s what is the kinetic energy￼

A 3900 kg truck is moving at 6.0 m/s what is the kinetic energy