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Vanyuwa [196]
2 years ago
9

An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits

the ground. How large was the frictional force acting upon it ?
Physics
1 answer:
faltersainse [42]2 years ago
6 0

The  frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

 a= 1.5m/s²



By Using conservation  of energy ,

30F+1/2mv²=mgh

30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

to learn more about Friction clickhttps://brainly.com/question/17608236

#SPJ9

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What are the three parts of Kinetic Theory
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You set a small eraser at a distance of 0.27 m from the center of a turntable you find in the attic and establish that when the
Licemer1 [7]

Answer with Explanation:

We are given that

Distance,r=0.27 m

Tangential speed=v=0.49 m/s

a.Angular speed ,\omega=\frac{v}{r}

Using the formula

\omega=\frac{0.49}{0.27}=1.8 rad/s

\omega=\frac{2\pi}{T}

T=\frac{2\pi}{\omega}

Time period,T=\frac{2\pi}{1.8}=3.49 s

b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m

c.Maximum speed,V_{max}=0.49 m/s

d.Maximum acceleration=a=r\omega^2=0.27(1.8)^2=0.87 m/s^2

5 0
3 years ago
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two blocks are held together with a compressed spring between them on the surface of a slippery table .one block has three times
Marina CMI [18]

Explanation:

The initial kinetic energy KE_0 for both blocks is zero. Let m_1= m and m_2 =3m. So using the conservation law of linear momentum, we can write

0 = m_1v_1 - m_2v_2

or

v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1

The final kinetic energies for the two masses are

KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2

KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)

Therefore, the ratio of their kinetic energies is

\dfrac{\Delta KE_2}{\Delta KE_1}  = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}

4 0
3 years ago
A spring with a spring constant of 23.6 N/m has a mass attached that exerts a force of 6.28 Newtons. What is the displacement?
Ostrovityanka [42]

Answer:

26.6 cm

Explanation:

We are given that

Spring constant=23.6 N/m

Force exert=6.28 N

We have to find the displacement.

We know that Hooke's law

F=kx

Where k= Spring constant

x=Displacement

Using the formula

Then, we get

6.28=23.6 x

x=\frac{6.28}{23.6}=0.266m

We know that 1 m=100 cm

x=0.266\times 100=26.6 cm

Hence, the displacement =26.6 cm

7 0
3 years ago
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