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Drupady [299]
2 years ago
9

A car driver sees a rabbit on the road. The driver makes an emergency stop after he sees the rabbit. Figure given shows the spee

d of the car from the time the driver sees the rabbit until the car stops. ) Calculate the distance that the car travels in the first 0.5 seconds.
2. A car of mass 1000 kg is moving with a velocity of 10 m/s. If the velocity-time graph for this car is a horizontal line parallel to the time axis, then the velocity of the car at the end of 25 s will be:
Physics
2 answers:
svetoff [14.1K]2 years ago
4 0
The car must be very fast kuh on bro
Fed [463]2 years ago
3 0
It was nothing becauseuibdadvbipduadbcuipdbaduicphdsa
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X=(150^2•sin(2•42))/9.8
amid [387]

Answer:

2283.3410863

Explanation:

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Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and
bazaltina [42]

Answer:

The answer is X: Thermal energy is converted to light energy

Y: A cold spoon placed in hot liquid gets warmer

Explanation:

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3 years ago
A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.
aleksley [76]
The acceleration that the cheeseburger experienced is 20 m/s^2.
6 0
2 years ago
Protons have a positive charge<br><br> A.True<br> B.False
PIT_PIT [208]

Answer:

TRUE

Explanation:

Protons have a positive charge. Electrons have a negative charge. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.

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3 years ago
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If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
fgiga [73]

Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

7 0
3 years ago
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