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vaieri [72.5K]
3 years ago
14

Consider a small child sliding down two differently shaped slides that have the same coefficient of kinetic friction. How must t

he second slide be shaped to provide the same change in gravitational potential energy of the child-Earth system but more change in kinetic energy for the child while they slide down
Physics
1 answer:
timama [110]3 years ago
6 0

Answer:

Having a bigger angle above the horizontal

Explanation:

Applying the energy conservation theorem:

K_1+U_1+W_{ext}=K_2+U_2

The kinetic energy is reduced because of the work done by the friction force.

The friction force is given by:

F_f=F_N*\µ

so the friction force depends on the Normal force, because the slide has an angle the normal force is given by:

F_N=m.g*cos(\theta)

So when the angle of the slide is bigger, the friction force decreases, for example:

for 45 degrees:

F_N=m.g*cos(45)\\F_N=0.70(m.g)\\

for 75 degrees:

F_N=m.g*cos(75)\\F_N=0.26(m.g)\\

as you can see if the angle is bigger above the horizontal, the friction force is reduced and so the work done by that force. We didn't have to change the height of the slide, so the potential gravitational energy remains the same.

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You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation
Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

\omega = 2\pi \times f

\omega = 2\pi \times 4

\omega = 25.13 rad/s

5 0
3 years ago
In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0
3 years ago
A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5
pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

v_w=v_s+v=1+1.5=2.5m/s

Distance=35 m

Time=\frac{distance}{speed}

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving=\frac{35}{v_w}=\frac{35}{2.5}=14s

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground=v_w=v_s+v=1-1.5=-0.5m/s

Time=\frac{-35}{-0.5}=70 s

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

3 0
3 years ago
If John travels 2,387 miles in 2 days, what's his average speed per hour?
klasskru [66]
In 2 days there are 48 hours
to find the average speed per hour, divide 2,387 by 48
Which gets you the answer 49.72
Which rounds up to 50

The average speed is 50mph
3 0
3 years ago
Read 2 more answers
Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. A
Finger [1]

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

r=(100\cos2\theta)\ m

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

v=r\omega

\omega=\dfrac{v}{r}

Where, v = velocity

r = radius

Put the value into the formula

\omega=\dfrac{29.0576}{100\cos30}

\omega=0.335\ rad/s

Hence, The rate at which radar must rotate is 0.335 rad/s.

3 0
3 years ago
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