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stich3 [128]
3 years ago
9

6. If a bicyclist, with initial velocity of zero, steadily gained speed until reaching a final velocity of 39m/s, how far would

she travel
during the race in the same amount of time)?
Physics
1 answer:
Bezzdna [24]3 years ago
7 0

Answer:

  175 m

Explanation:

The average velocity for constant acceleration is the average of the beginning and ending velocities. That is (0+39)/2=19.5 m/s. If the bicyclist rides for 9 seconds, the distance traveled is ...

  (9 s)(19.5 m/s) = 175.5 m

She would travel 175.5 meters in that time.

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An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o
Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

cos(\alpha) = 80/320 = 0.25

\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0 relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h

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Lifting a stone block 146m to the top of the Great Pyramid required 146,000 J of work. How much work was done to lift the block
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3 years ago
The idea that John Marshall, the first Chief Justice of the Supreme Court, singularly established the principle of judicial revi
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Answer:

c. Was an idea created and supported by Congress.

Explanation:

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3 years ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
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Answer:

a). 1.218 m/s

b). R=2.8^{-3}

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m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
3 years ago
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