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charle [14.2K]
3 years ago
15

A student is bicycling north along Main Street to school. Another student is timing the bicycling student in order to determine

the average velocity. The bicycling student is 450 m away when the timer starts, and is 200 m away when the timer stops. If the student traveled the distance in 60 seconds, what is the students average velocity?
Physics
1 answer:
MatroZZZ [7]3 years ago
8 0

The average velocity is -4.17 m/s

Explanation:

The average velocity of a body is given by:

v=\frac{d}{t}

where

d is the displacement of the body

t is the time elapsed

For the student in this problem, we have:

Initial position: x_i = 450 m

Final position: x_f = 200 m

So the displacement is

d=x_f -x_i = 200 - 450 = -250 m

The time elapsed is

t = 60 s

Therefore, the average velocity is

v=\frac{-250}{60}=-4.17 m/s

Where the negative sign means the student is moving towards the origin.

Learn more about average speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

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2 years ago
The transfer of energy when particles of a fluid move from one place to another is called
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Explanation:

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5 0
2 years ago
A shot is fired at an angle of 60 degree horizontal with Kinetic energy E. If air resistance is ignored, the K.E at the top of t
Lapatulllka [165]
I'm not sure what "60 degree horizontal" means.

I'm going to assume that it means a direction aimed 60 degrees
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Now, I'll answer the question that I have invented.

When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is    S cos(60)  =  0.5 S ,
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-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.

-- 0.25 of its kinetic energy is due to its horizontal velocity.
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-- So at the top of its trajectory, its KE is 0.25 of what it had originally. 

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3 0
2 years ago
After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m>s. Her 5.0-kg cat
Vinil7 [7]

To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_1= mass of the skier

m_2= mass of the cat

v_1 = initial velocity of skier

v_2 = initial velocity of cat

v_f= final velocity of both

Re-arrange to find V_f we have,

V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}

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V_f = 13.55m/s

Once the final velocity is found it is possible to calculate the change in kinetic energy, so

\Delta KE = KE_i-KE_f

\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2

\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2

\Delta KE = 819.1J

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0
2 years ago
A room has a wall with an R value of 18 F sq.ft. hr/BTU. The room is 15 feet long and 11 feet wide with walls that are 9 ft high
seropon [69]

Answer:

The heat transferred through the wall that day is  13728 BTUs

Explanation:

Here, we have the area of the wall given as

Area of wall = 2 × Length × Height + 2 × Width × Height

Length = 15 feet

Width = 11 Feet and

Height = 9 feet

Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²

Temperature difference is given by

Average outside temperature - Wall temperature = 40 - 18 = 22 °F

Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;

468 × 22 × 24/18 = 13728 = 13728 BTUs.

5 0
3 years ago
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