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malfutka [58]
1 year ago
12

A spring with a constant k=400n/m shoots a 1. 00kg ball up a frictionless incline after being compressed 0. 150m. what is the ma

ximum height reached by the ball?
Physics
1 answer:
Zielflug [23.3K]1 year ago
8 0

The maximum height reached by the ball is 0.46m.

To find the answer, we have to know about the potential energy of a spring mass system.

<h3>How to find the maximum height reached by the ball?</h3>
  • It is given that,

                       k=400N/m\\m=1kg\\x=0.150m\\h=?\\

  • We have to find the maximum height reached by the ball.
  • Thus, we have the expression for potential energy of a spring mass system and that of gravitational field as,

                             U=\frac{1}{2}kx^2 \\U=mgh

  • Combining both, we get,

                    h=\frac{kx^2}{2mg} =\frac{400*(0.15)^2}{2*1*9.8} =0.46m

Thus, we can conclude that, the maximum height reached by the ball is 0.46m.

Learn more about the potential energy here:

brainly.com/question/26962934

#SPJ4

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After crossing the finish line, a race car slows down from 47 m/s to 32m/s in 3seconds. What is the car’s acceleration?
Lisa [10]

Answer: -5 m/s^2

Explanation: a = v - u/t

                         = 32 - 47/3

                         = -15/3

                         = -5 m/s^2

7 0
3 years ago
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Which of these metals are magnetic?
yanalaym [24]

Copper is not magnetic

Steel is magnetic

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5 0
3 years ago
Two small plastic spheres are given positive electrical charges. When they are 15.0 cm apart, the repulsive force between them h
Tems11 [23]

Answer:

a) q_1=q_2= 7.42*10^-7 C

b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C

Explanation:

Given:

F_e = 0.220 N

separation between spheres r = 0.15 m

Electrostatic constant k = 8.99*10^9

Find: charge on each sphere

part a

q_1 = q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_1^2 = F_e*r^2/k

q_1=q_2= sqrt (F_e*r^2/k)

Plug in the values and evaluate:

q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)

q_1=q_2= 7.42*10^-7 C

part b

q_1 = 4*q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_2^2 = F_e*r^2/4*k

q_2= sqrt (F_e*r^2/4*k)

Plug in the values and evaluate:

q_2= sqrt (0.22*0.15^2/4*8.99*10^9)

q_2= 3.7102*10^-7 C

q_1 = 14.8*10^-7 C

4 0
3 years ago
Read 2 more answers
Supply the missing force necessary to achieve equilibrium. Show your work.
Mumz [18]

<u>Analysing the Question:</u>

We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it

In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N

<u>Finding the missing force:</u>

Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces

So,  (106 + 106 + 142 + x) = 496

354 + x = 496

x = 496 - 354

x = 142 N

Therefore, the missing force is 142 N

8 0
3 years ago
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