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storchak [24]
3 years ago
15

A ball falling through the air has--------

Physics
1 answer:
padilas [110]3 years ago
5 0

Answer:

hey hello thanks

Explanation:

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The melting of a glacier is an example of the interactions among which of Earth’s spheres?
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Atmosphere, geosphere, cryosphere would be the correct answer I think
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How can a high school student have more momentum than a truck
morpeh [17]

Answer:

When truck is at rest while student is under motion

Explanation:

Since it is obvious that the mass of a truck is more than that of a student, we know that momentum is a product of mass and velocity

P=mv where m represent mass, v is velocity. When the student has more speed than that of truck, he exerts more momentum. The only way a student can exert more momentum is by having more speed while the truck is at rest. In such case, the momentum of truck will be zero while momentum of student will have a value

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As shown in the diagram, an inflated balloon released from rest moves horizontally with velocity "v". The velocity of the balloo
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Answer:

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Explanation:

8 0
3 years ago
Read 2 more answers
A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound
Strike441 [17]

Answer:

5.997m/s

Explanation:

We were told to calculate the speed of the ball,

Given speed of sound as 340 m

And we know that the sound of the ball hitting the pins is at 2.80 s after the ball is released from his hands.

Speed of ball = distance traveled/(time of hearing - time the sound travels).

Speed= S/t

Where S= distance traveled

t= time of hearing - time the sound travels

time=time for ball to roll+timefor sound to come back.

time of sound=16.5/340

=0.048529secs

solving for speedof ball

Then,Speed of ball = distance traveled/(time of hearing - time the sound travels).

=16.5/(2.80-0.048529) m/s = 5.997m/s

Therefore, the speed of the ball is

5.997m/s

4 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
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