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vladimir2022 [97]
3 years ago
11

A boy is holding a ball 1 m from the ground with a force of 20 N. He holds it still for 60seconds. How much power in watts is be

ing used in this situation?
Physics
1 answer:
Sergio [31]3 years ago
8 0

Answer:

Power = 0.33 Watts

Explanation:

Given the following data;

Distance = 1m

Force = 20N

First of all, we would solve for the work done by the boy.

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 20*1 = 20J

Now to find power;

Power = workdone/time

Power = 20/60

Power = 0.33 Watts.

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S_A_V [24]

Answer:hchv nhgjj

Explanation:

ggchh

3 0
3 years ago
engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi
Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

where ;

m_1=mass of object 1\\v_1i=initial  velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2

Given in the question that;

m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?

Apply the formula as;

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

4 0
2 years ago
Read 2 more answers
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
If this speed is based on what would be safe in wet weather, estimate the radius of curvature for a curve marked 50 km/h . The c
PtichkaEL [24]

Answer:

Explanation:

v = 50 km / h

= 13.89 m /s

When a vehicle runs on a circular path , it is static friction which prevents it from getting overturned .

static friction = μs mg

centripetal force = m v² / R

m v² / R = μs mg

R = v² / μs x g

= 13.89² / .7 x 9.8

= 28.12 m .

7 0
3 years ago
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Katena32 [7]

Answer:

20 m

Explanation:

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mgh = 1/2 mv²

gh = 1/2 v²

h = v² / (2g)

Given v = 20 m/s and g = 10 m/s²:

h = (20 m/s)² / (2 × 10 m/s²)

h = 20 m

4 0
3 years ago
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