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IceJOKER [234]
3 years ago
7

Suppose that the most massive car ever built is moving down a 10.00 slope when the driver applies the brakes. The net force acti

ng on the car as it stops is -2.00X104 N. If the coefficient of kinetic friction between the car’s tires and pavement is 0.797, find the car’s mass. What is the normal force that the pavement exerts on the car?
Physics
1 answer:
slava [35]3 years ago
6 0

Hey there!

Use the question and search for a quizlet or science solving program

Here is a link:

www.scienceslovingprogram.com

Or

www.sciencehelp.com

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A transformer is to be used to provide power for a computer disk drive that needs 6.4 V (rms) instead of the 120 V (rms) from th
Amanda [17]

Answer:

The current in the primary is 0.026 A

Explanation:

Using the formula

I1 = (V1/V2)*I2

we have

I1 = (6.4/120)*0.500

I1 = 0.026 A

8 0
3 years ago
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

6 0
1 year ago
For numbers 2 and 3, which one is compression and which one is rarefaction?
DENIUS [597]
Compression is above the equilibrium and rarefaction is below
5 0
3 years ago
A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed
malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here:  brainly.com/question/25573309

#SPJ10

5 0
2 years ago
One knife is longer from the cutting edge to the back of the blade. It's easier to push the longer knife through a pear and more
AlladinOne [14]

By the formula of torque balance we know that

\tau = r\times F

here we know that

r = distance of force

F = applied force

so here we know that as we will take large edge knife then the distance of force will increase

due to this the torque will also increase in it so it is easy to pear it

So here we can say

F_{in}r_1 = F_{out}r_2

F_{out} = \frac{r_1}{r_2} F_{in}

so output force will increase in this case

So here the correct answer would be

<em>b. the longer the knife, the stronger the output force</em>

6 0
3 years ago
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