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3 years ago

2. A person began running due east and covered 15 kilometers in 2.0 hours. What is the average velocity of the person?

Physics

2 answers:

kicyunya [14]3 years ago

8 0

Answer: the answer is 7200s

Explanation: i just now it

Please brainliest

Angelina_Jolie [31]3 years ago

4 0

Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

$v=\frac{d}{t}$

where

d is the displacement

t is the time taken

In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

$v=\frac{15 km}{2.0 h}=7.5 km/h$

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

$v=\frac{15,000 m}{7200 s}=2.1 m/s$

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A 27.5-g object moving to the right at 19.0 cm/s overtakes and collides elastically with a 11.5-g object moving in the same dire

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Answer:

After collision, velocity of massive object 27.5 g has reduced from 19 cm/s to 16.64 cm/s while that of lighter object 11.5 has increased from 15 cm/s to 20.64 cm/s because of elastic collision.

Explanation:

As, this question has two parts: a) We have to find the velocity after the collision with the mass of the object 27.5 g.

b) We have to find the velocity with 11.5 g mass of the object.

Now,

In this case the massive body collides with the lighter body. As we know;

Mass of massive object = m1 = 27.5 g

Velocity of m1 before collision = v1 = 19 cm/s

mass of lighter object = m2 = 11.5 g

velocity of m2 before collision = v2 = 15 cm/s

We have to find the velocity of m1 and m2 after the collision

Velocity of 27.5 g object after collision = v1' = ?

v1' = (m1-m2)/(m1+m2)×v1 + 2×m2/(m1+m2)×v2

v1' = (27.5-11.5)/27.5+11.5)×(19) + 2×(11.5)/(27.5+11.5)×(15)

v1' = 7.795 + 8.846

v1' = 16.64 cm/s.

Velocity of 11.5 g object after collision = v2' = ?

v2' = 2×m1/(m1+m2)×v1 + (m2-m1)/(m1+m2)×v2

v2' = 2×(27.5)/(27.5+11.5)×(19) + (11.5-27.5)/(27.5+11.5)×(15)

v2' = 26.795 + (-6.154)

v2' = 20.641 cm/s.

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3 years ago

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Angelina_Jolie [31]

Answer:

631.7 Hz

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In particular, the new apparent frequency can be calculated as:

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