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Oksi-84 [34.3K]
3 years ago
13

Select ALL of the following statements that provide evidence that there is friction acting on a cart moving along a level track.

[you can pick more than one answer!]
The cart slows down while moving down the track

The cart's speed increases only when the hand is in contact with the cart

The temperature of the track increases as the car moves over it

The wheels are warmer after the cart moves down the track
Physics
1 answer:
Sati [7]3 years ago
6 0

Answer:

1st and 4th one................

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The human ear canal is about 2.8 cm long. If it is regarded as a tube that is open at one end and closed at the eardrum, what is
Diano4ka-milaya [45]

Answer:

f = 3.1 kHz

Explanation:

given,

length of human canal =2.8 cm = 0.028 m

speed of sound = 343 m/s

fundamental frequency  = ?

The fundamental frequency of a tube with one open end and one closed end is,

f = \dfrac{v}{4L}

f = \dfrac{343}{4\times 0.028}

f = \dfrac{343}{0.112}

       f = 3062.5 Hz

       f = 3.1 kHz

hence, the fundamental frequency is equal to f = 3.1 kHz

8 0
3 years ago
A car traveling in a straight line has a velocity of 6m/s at some instant. After 6.32s its velocity is 13.2m/s . What is the ave
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Average acceleration  =  (change in speed) / (time for the change) .

Average acceleration  =  (13.2 - 6) / (6.32) = 7.2 / 6.32 = about  <em>1.139... m/s²</em> .
8 0
3 years ago
The speed of the 2 kg cart after 5 seconds is ? cm/s.
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The answer to this question is 1cm/s
3 0
3 years ago
The Netherlands controlled ______ for centuries <br><br> A. Vietnam<br> B. Malaysia<br> C. Indonesia
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The Netherlands controlled Indonesia for centuries!
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Read 2 more answers
Consider a block on frictionless ice. Starting from rest, the block travels a distance din
sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{F}{m}

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

\displaystyle d = vo.t+\frac{at^2}{2}

If the block starts from rest, vo=0:

\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

\displaystyle d = \frac{\frac{F}{m}t^2}{2}

Simplifying:

\displaystyle d = \frac{Ft^2}{2m}

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

\displaystyle a'=\frac{4F}{m}

And the distance is now:

\displaystyle d' = \frac{4Ft^2}{2m}

Dividing d'/d:

\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}

Simplifying:

\displaystyle \frac{d' }{d}=4

Thus:

d' = 4d

The distance is now 4d

3 0
3 years ago
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