A 38.5 kg man is in an elevator

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for

Artemon [7]

Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks.

W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ

Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)

6 0

4 years ago

Tech A says a camshaft position sensor must be positioned correctly to accurately signal the position of the camshaft to the pow

qaws [65]

Answer:

Both Tech A and Tech B

Explanation:

It's important to position sensor accurately to the position of camshaft to the powertrain control in the ignition module as said by Tech A. Equally, to achieve what Tech A says, a special tool can be used for comparing the position of camshaft sensor to crankshaft sensor. Therefore, both technicians are correct.

6 0

3 years ago

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 43.

zmey [24]

To solve this problem, we must take two important steps. First we will convert all the given units, to international system. Later we will define the torque, which is given as the product between the radius of application of the force and the Force acting on the body. Mathematically the latter is,

$\tau = r F$

Here,

r = Radius

F = Force

Now the units,

$r = 1.95cm = 1.95*10^{-2}m$

Replacing,

$\tau = (1.95*10^{-2})(43.5N)$

$\tau = 0.8483N\cdot m$

Therefore the torque that the muscle produces on the wrist is $0.85N\cdot m$

3 0

4 years ago

Satellite A orbits a planet at a distance d from the planet’s center with a centripetal acceleration a0. A second identical sate

leva [86]

To solve this problem it is necessary to use the concepts related to the Gravitational Force and Newton's Second Law, as far as we know:

$F_g = \frac{GMm}{r^2}$

Where,

G = Gravitational constant

M = Mass of earth (in this case)

m = mass of satellite

r = radius

In the other hand we have the second's newton law:

$F = ma$

Where,

m = mass

a = acceleration

Equation both equations we have,

$ma = \frac{GMm}{r^2}$

For the problem we have that,

<em>Satellite A:</em>

$ma_A = \frac{GMm}{r^2}$

<em>Satellite B:</em>

$ma_B = \frac{GMm}{(2r)^2}$

The ratio between the two satellites would be,

$\frac{ma_A}{ma_B}= \frac{\frac{GMm}{r^2}}{\frac{GMm}{(2r)^2}}$

Solving for a_B,

$a_B = \frac{a_A}{4}$

Therefore the centripetal acceleration of $A_B$ is a quarter of $a_A$

7 0

3 years ago

When a force of 15 newtons is applied to a stationary chair, it starts moving. What can you say about the frictional force betwe

harkovskaia [24]

First you need to make a difference between friction while object is stationary and the friction while object is moving. Force required to start moving some object is slightly greater than force required to maintain objects movement. That means that to move a chair you need some force F1 but you can than slightly reduce force and chair will still be moving.

Now to the problem in this question: It can be said that "stationary friction force" is equal to 15 Newtons. Its also good to know that friction force between chair and floor while you are increasing your push is also increasing and is equal to force of your push. Once it reaches 15N which is it "critical value" for that chair, chair starts moving and friction force drops a little bit and now it is called friction force of moving chair.

3 0

3 years ago