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kifflom [539]
3 years ago
10

How much force does it take to stop a 1,000 kg car traveling at 20 m/s in 10 seconds?

Physics
1 answer:
Doss [256]3 years ago
5 0
The initial momentum of the car is

                (mass) x (speed)  =  20,000 kg-m/s

"Impulse" is (force exerted) x (time the force lasts)
and it's equal to the change in momentum.

              (Force) x (10 sec)  =  20,000 kg-m/s

Divide each side by  (10 sec)  and you have

                      Force  =  (20,000 kg-m/s) / (10 sec)

                                 =          2,000 kg-m/s²  =  2,000 Newtons

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horrorfan [7]

Answer:c

Explanation:ganglandd

4 0
3 years ago
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
The first electric generators used direct current, so they needed to be reversed manually on a regular basis in order to work pr
djyliett [7]
Could I see the questions
4 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
3 years ago
You are getting bored as you are stuck at home, and want to find out the effective 'spring constant' of random objects. You find
Luda [366]

Given Information:  

Mass of sock = 0.23 kg

Stretched length of sock = x = 2.54 cm = 0.0254 m

Required Information:  

Spring constant = k = ?

Answer:  

Spring constant = k = 88.82 N/m

Explanation:  

We know from the Hook's law that

F = kx

Where k is spring constant, F is the applied force and x is length of sock being stretched.

k = F/x

Where F is given by

F = mg

F = 0.23*9.81

F = 2.256 N

So the spring constant is

k = 2.256/0.0254

k = 88.82 N/m

Therefore, the spring constant of the sock is 88.82 N/m

4 0
3 years ago
Read 2 more answers
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