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wariber [46]
2 years ago
10

3) A 30 kg child slides freely across a "Slip and Slide" on LEVEL GROUND. While the child slides, the force applied to keep them

sliding is 0 N. The coefficient of kinetic friction is 0.02. What is the acceleration experienced by the child?
Physics
1 answer:
zavuch27 [327]2 years ago
4 0

Answer:

a = 0.1962 m/s^2

Explanation:

The magnitude of kinetic friction exerted is given by

F_k=\mu_kN

Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg

Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2

F_k=0.02×30×9.81 =5.886 N

Now, since, there is no applied force this kinetic friction force will cause acceleration of the child

⇒ ma = F_k

here, a is the acceleration

⇒30a = 5.886

⇒ a = 0.1962 m/s^2

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3 years ago
parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference bet
Rom4ik [11]

Answer:

V = 576 V

Explanation:

Given:

- The area of the two plates A = 0.070 m^2

- The space between the two plates d = 6.3 mm

- Te energy density u = 0.037 J /m^3

Find:

- What must the potential difference between the plates V?

Solution:

- The energy density of the capacitor with capacitance C and potential difference V is given as:

                               u = 0.5*ε*E^2

- Where the Electric field strength E between capacitor plates is given by:

                               E = V / d

Hence,

                               u = 0.5*ε*(V/d)^2

Where, ε = 8.854 * 10^-12

                               V^2 = 2*u*d^2 / ε

                               V = d*sqrt ( 2*u / ε )

Plug in values:

                               V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )

                               V = 576 V

4 0
3 years ago
Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an
krok68 [10]

Answer:

A) 31 kJ

B)  1.92 KJ

C) 40 ,  2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈  0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.013) )

          = 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh  hence  F = mgh / d

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.345) )

          = 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

=  31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

  Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 0
3 years ago
Consider the four masses. Describe the motion of the masses if you used THE SAME amount of force to move each one over THE SAME
Illusion [34]
The answer is c. if itis heavier, u have to push hardier or it to move the same distance. make sense??
4 0
3 years ago
Read 2 more answers
Does latitude has an effect on weight?​
lord [1]

Answer:

I think so but i could be wrong..

Explanation:

5 0
3 years ago
Read 2 more answers
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