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mihalych1998 [28]
3 years ago
10

A person in the back of a pick-up moving at 20 m/s relative to the Earth, throws a baseball with a speed of 15 m/s in a directio

n opposite the motion of the truck (consider the direction of the truck’s velocity to be positive). What is the velocity of the ball relative to the Earth as it leaves the thrower’s hand?
Physics
1 answer:
Bumek [7]3 years ago
3 0

Answer:

This is known as a Galilean transformation where

V' = V - U

Where the primed frame is the Earth frame and the unprimed frame is the frame moving with respect to the moving frame

V - speed of object in the unprimed frame

U - speed of primed frame with respect to the unprimed frame

Here we have:

V = -15 m/s        speed of ball in the moving frame (the truck)

U =  -20 m/s        speed of primed (rest) frame with respect to moving frame

So  V' = -15 - (-20) = 5 m/s

It may help if you draw a vector representing the moving frame and then add

a vector representing the speed of the ball in the moving frame.

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The number of kilograms of water in a human body varies directly as the mass of the body. Upper AA 9393​-kg person contains 6262
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Answer:

54 kg

Explanation:

Mass of person = 93 kg = m_p

Mass of water = 62 kg = m_w

Dividing the above two masses we get

\frac{m_p}{m_w}=\frac{93}{62}\\\Rightarrow \frac{m_p}{m_w}=1.5\\\Rightarrow m_p=1.5m_w

Hence, the mass of the person is 1.5 times the mass of the water in them

Now, Mass of person = 81 kg = m_p

m_p=1.5m_w\\\Rightarrow m_w=\frac{1}{1.5} m_p\\\Rightarrow m_w=\frac{1}{1.5} \times 81\\\Rightarrow m_w=54\ kg

So, the mass of water in a person that has mass of 81 kg is 54 kg

3 0
3 years ago
If the initial upward speed of the ball in Activity 7 C.2 is 10m/s, and the ball is releasedat a height of 1.5 m above the oor,
Ulleksa [173]

Answer:

6.5 m above the floor and 5 m above Christine's hand when it reaches the maximum height.

Explanation:

Let g = 10 m/s2 be the gravitational deceleration that affects the ball vertical motion so it comes to the maximum height at 0 speed. We can use the following equation of motion to find out the distance traveled by the ball from where it's thrown:

v^2 - v_0^2 = 2g\Delta s

where v = 0 m/s is the final velocity of the ball when it reaches maximum level, v_0 = 10m/s is the initial velocity of the ball when it starts, g = -10 m/s2 is the deceleration, and \Delta s is the distance traveled, which we care looking for:

0^2 - 10^2 = -2*(-10)\Delta s

\Delta s = 100 / (2 * 10) = 5 m

So the ball is 5 m above Christine' hands when it reaches maximum height, and since the hand is 1.5 m above the floor, the ball is 5 + 1.5 = 6.5 m above the floor when it reaches maximum height.

5 0
3 years ago
Read 2 more answers
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