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shutvik [7]
3 years ago
6

A 3.0 kg mass is released from rest at point A. The mass slides along the curved surface to point B in 6.0 seconds. Point B is 2

.0 m lower than A and the mass is moving at 2.0 m/s at point B. Find the average power exerted by kinetic friction on the mass during this segment. a) 0 W b) -3 W c) -6 W d) -9 W e) -18 W
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

option d) -9 J

Explanation:

Given:

Mass, m = 3.0 kg

time, t = 6.0 seconds

Velocity of mass, v = 2.0 m/s

height, h = 2 m

Now, using the concept of work-Energy theorem

we have

Net work done = change in kinetic energy

or

Work done by gravity + work done by the friction = Final kinetic energy - Initial kinetic energy

mgh +W_f = \frac{1}{2}mv^2-0

on substituting the values in the above equation, we get

3 × 9.8 × 2 + W_f = \frac{1}{2}\times 3\times2^2

or

58.8 + W_f = 6

or

W_f = -52.8 J

here negative sign depicts that the work is done against the motion of the mass

also,

Power = (Work done)/time

or

Power = -52.8/6 = -8.8 W ≈ 9 J

Hence, option d) -9 J is correct

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Arlecino [84]

Answer:

a) v=0.5405\ m.s^{-1}

b) v_p'=0.1143\ m.s^{-1}

Explanation:

Given:

mass of ball, m_b=4\ kg

initial speed of the ball, v_b=10\ m.s^{-1}

mass of the person, m_p=70\ kg

a)

Using the conservation of linear momentum:

When the person catches the ball, assuming that the person catches it with an impact without absorbing the shock.

m_b.v_b=(m_b+m_p)v

4\times 10=(4+70)\times  v

v=0.5405\ m.s^{-1}

b)

When the ball hits the person and bounces off with the velocity of v_b'=8\ m.s^{-1}.

Using the conservation of linear momentum:

m_b.v_b+m_p.v_p=m_b.v_b'+m_p.v_p'

where:

v_b'= final speed of the ball after collision

v_p'= final speed of the person after collision

v_p= initial velocity of the person = 0

putting the respective values in the above eq.

4\times 10+0=4\times 8+70\times v_p'

v_p'=0.1143\ m.s^{-1}

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3 years ago
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How does speed relate to motion energy?
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the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

4 0
3 years ago
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A projectile has an initial x-velocity of 4 m/s, and an initial y-velocity of 27.7 m/s. What is the range of the projectile
a_sh-v [17]

Answer:

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2 years ago
Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets
PolarNik [594]

 Step by step solution :

standard deviation is given by :

\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}

where, \sigma is standard deviation

\bar{x} is mean of given data

n is number of observations

From the above data, \bar{x}=24.88

Now, if x=24.8, then (x-\bar{x})^2=0.0064

If  x=23.9, then (x-\bar{x})^2=0.9604

if x=26.1, then (x-\bar{x})^2=1.4884

If x=25.1, then (x-\bar{x})^2=0.0484

If x=24.5, then (x-\bar{x})^2=0.1444

so, \sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}

\sum (x-\bar{x})^2 =2.648

\sqrt{\sum \frac{(x-\bar{x})^2}{n}}

\sigma =0.7277

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

6 0
2 years ago
A student performing a double-slit experiment is using a green laser with a wavelength of 550 nm. She is confused when the m = 5
sweet [91]

Answer:

d = 52 μm

Explanation:

given,

wavelength of the light source (λ)= 550 nm

distance to form interference pattern(D) = 1.5 m

y = 1.6 cm = 0.016 m

width of the slits = ?

now, using displacement formula

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for the first maxima, m = 1

 d = \dfrac{1\times \lambda\ D}{y}

 d = \dfrac{1\times  550 \times 10^{-9}\times 1.5}{0.016}

       d = 5.2 x 10⁻⁶ m

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hence, the width of her slits is equal to d = 52 μm

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