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Fed [463]
3 years ago
10

What are two ways in which friction can be useful?

Physics
2 answers:
pishuonlain [190]3 years ago
6 0
If something is going down a hill it can help slow it down

it can stop you from flying off a rollercoaster
Katena32 [7]3 years ago
5 0
Friction slows an object down. the brakes in your car are a good example of friction.
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All waves have wavelength, frequency, rest point, and speed. <br> True or False?
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Answer:

it is true

Explanation:

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Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi
Mamont248 [21]

Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

\Delta X=X_{k}-X_{C}=0.021miles

4 0
3 years ago
Due Ma<br> duart<br> ded<br> out<br> 25 N<br> 35 N<br> 1-03
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Answer:

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3 years ago
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

3 0
3 years ago
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The wave function of a particle is exp(i(kx-omegat)), where x is distance, t is time, and k and co are positive real numbers. Th
Step2247 [10]

Answer:

Momentum, p=\hbar k

Explanation:

The wave function of a particle is given by :

y=exp[i(kx-\omega t)]...............(1)

Where

x is the distance travelled

t is the time taken

k is the propagation constant

\omega is the angular frequency

The relation between the momentum and wavelength is given by :

p=\dfrac{h}{\lambda}............(2)

From equation (1),

k=\dfrac{2\pi}{\lambda}

\lambda=\dfrac{2\pi}{k}

Use above equation in equation (2) as :

p=\dfrac{h k}{2\pi }

Since, \dfrac{h}{2\pi}=\hbar

p=\hbar k

So, the x-component of the momentum of the particle is \hbar k. Hence, this is the required solution.

8 0
3 years ago
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