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velikii [3]
3 years ago
14

O'Malley is riding on a bus which is moving at 10 m/s, and he throws a ball which he observes to be moving at 10 m/s relative to

him in the direction of motion of the bus. What speed will Ophelia observe the ball to have?
Physics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

<em>20 m/s in the same direction of the bus.</em>

Explanation:

<u>Relative Motion </u>

Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.

If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.

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VikaD [51]

Answer:

R_3 < R_1 < R_2

Explanation:

The resistance of a wire is given by:

R=\frac{\rho L}{A}

where

\rho is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

1) The first wire has length L and cross-sectional area A. So, its resistance is:

R_1=\frac{\rho L}{A}

2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

R_2=\frac{2\rho L}{A}

3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

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By comparing the three expressions, we find

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5 0
3 years ago
A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai
shutvik [7]

Answer:

d) None of the above

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θ = angle of launch = 10 deg

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v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

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using the equation

y=v_{oy} t+(0.5)a_{y} t^{2}

- 20 = (0.695) t + (0.5) (- 9.8) t²

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consider the motion along the horizontal direction

x = horizontal displacement

v_{ox} = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

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Using the kinematics equation

x =v_{ox} t+(0.5)a_{x} t^{2}

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y_{o} =initial vertical position at the time of launch = 20 m  

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using the equation

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0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)

y = 20.02 m

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h = y - y_{o}

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h = 0.02 m

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Answer:

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