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faust18 [17]
3 years ago
8

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

ce against the wheels?
Physics
1 answer:
Artemon [7]3 years ago
6 0
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)
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For a block to move down an inclined plane what force has to be the greatest?
Hatshy [7]

Answer:

D) True. This is what creates the body weight

Explanation:

Let's write Newton's second law for this case. For inclined planes the reference system takes one axis parallel to the plane (x axis) and the other perpendicular to the plane (y axis)

X axis

          Wx -fr = ma

Y Axis

          N - Wy = 0

With trigonometry we can find the components of weight

          sin θ = Wₓ / W

         cos θ = W_{y} / W

         Wₓ = W sin θ

          W_{y} = W cos θ

        W  sin θ - fr = ma

From this expression as it indicates that the body is descending the force greater is the gravity that create the weight of the body

Let's examine the answers

A False This force does not apply because it is not a spring

B) False. It is balanced at all times with the component (Wy) of the weight

C) False. For there to be a rope, if it exists you should be less than the weight component for the block to lower

D) True. This is what creates the body weight

E) False. The cutting force occurs for force applied at a single point and gravity is applied at all points

5 0
3 years ago
The ampere is a unit of which physical quantity?
hram777 [196]
Electrical current is measured using the ampere.
7 0
3 years ago
Read 2 more answers
A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a
Elena L [17]

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

F=ma      Eqn. (1)

where

F : is the Net Force in Newtons ( N )

m : is the mass ( kg )

a  : is the acceleration ( m/s^2 )

We also know that the acceleration is denoted by the velocity ( v ) of an object as a function of time ( t ) with

a=\frac{v}{t}         Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

F=m\frac{v}{t}\\ \\Ft=mv     Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. Ft is in fact the Impulse  J of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ 22N and t=4 sec (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

J=Ft\\J=(22N)(4sec)\\J=88 kg m/s

So the impulse of the cart is J=88kg m/s

Then, we know that the cart is moving at v=10 m/s. Plugging in the values in Eqn. (3) we have:

(22N)(4sec)=(10m/s)m\\\\88=10m\\\\m=\frac{88}{10}\\ \\m=8.8kg

So the mass of the cart is m=8.8kg.

8 0
3 years ago
A student wish to measure the gravitational acceleration g. She does it by releasing a small lead ball from rest and measures th
Goryan [66]

Answer:

(9.64 +- 0.86) m/s^2

Explanation:

The generic motion equation for constant acceleration is

x = X0 + v0 * t + \frac{1}{2}*a * t^2

Where

X0: initial position

v0: initial speed

a: acceleration

t: time

If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:

x = \frac{1}{2}*a * t^2

And the acceleration can be obtained as:

a = 2*\frac{x}{t^2}

Where x is the distance fallen and a = g.

So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate

g = 2*\frac{100}{144^2} = 9.64e-3 \frac{mm}{ms^2} = 9.64 \frac{m}{s^2}

For the uncertainty we have to calculate the relative uncertainties first

For the distance (100 * 0.3)/100 = 0.3%

For the time (100 * 3)/144 = 2.08%

For multiplications or divisions the relative uncertainties are added

0.3% + 2.08% + 2.08% = 4.46%

We convert this into absolute uncertainty:

(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)

Finally, this is multiplied by a constant scalar, so:

2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)

We convert the units

0.86 m/(s^2)

And the measurement is (9.64 +- 0.86) m/s^2

A better method is putting the ball in a ramp instead of a free fall, that way the fall is longer and the effect of time measuring uncertainty is reduced.

5 0
3 years ago
Suppose that Lisa walks her dog around the block for a little exercise. The block is 1 mile around. If she walks around the bloc
Alina [70]

Displacement is defined as the net distance traveled through the entire trip. That is to say, if you compared where you are at the end of the walk to where you were at the start, the displacement is the distance between those two points.


Distance doesn't compare the two end points. It just counts the total distance you've walked throughout your trip.


Lisa walks once around the block. She starts and ends at the same place. Her displacement is therefore 0 m.


But she has walked 1 mile throughout the trip. Her distance walked is therefore 1 mile.

4 0
2 years ago
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