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polet [3.4K]
3 years ago
10

A car travels from point A to point B in 3.991 hours and returns back to point A in 4.109 hours. Points A and B are 134.2 miles

apart along a straight highway. What was the car's average speed and average velocity over this time period?
Physics
1 answer:
crimeas [40]3 years ago
8 0

Answer:

gjsgjdcjpdcojdcoj gud gud gud gud gud gud gud gud gud gud gud gud gud gud get home h hum h hum hth hai h h g c b c b v g c h x corolla

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The motor of a battery-powered scooter can maintain a speed of 5.3 m/s by providing a force of 280 N. What is the power output o
Bond [772]

Answer:

the power output P = 1484 W

Explanation:

Power P  = Force× speed

Given that motor of a battery-powered scooter can maintain a speed of 5.3 m/s by providing a force of 280 N.

Therefore,the power output of the motor P = 280×5.3 = 1484 W

Hence, the power output P = 1484 W

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What is magnetic poles always come in pair
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What is the first step a scientist usually take to solve a problem?
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Make prediction about what will happen certain circumstances
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4 years ago
Mike roller skates with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles.
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3 years ago
A 10.00 kg mass is attached to a 250N/m spring and set into vertical oscillation. When the mass is 0.50m above the equilibrium i
Paraphin [41]

assuming the reference line to measure the height for gravitational potential energy lying at the equilibrium position

m = mass attached to the spring = 10.00 kg

k = spring constant of the spring = 250 N/m

h = height of the mass above the reference line or equilibrium position = 0.50 m

x = compression of the spring = 0.50 m

v = speed of mass = 2.4 m/s

A = maximum amplitude of the oscillation

v' = speed of mass at the maximum amplitude location = 0 m/s

using conservation of energy between the point where the speed is 2.4 m/s  and the highest point at which displacement is maximum from equilibrium

kinetic energy + spring potential energy + gravitational potential energy = kinetic energy at maximum amplitude + spring potential energy at maximum amplitude  + gravitational potential energy at maximum amplitude

(0.5) m v² + m g h + (0.5) k x² = (0.5) m v'² + m g A + (0.5) k A²

inserting the values

(0.5) (10) (2.4)² + (10) (9.8) (0.50) + (0.5) (250) (0.50)² = (0.5) (10) (0)² + (10) (9.8) A + (0.5) (250) A²

109.05 = (98) A + (125) A²

A = 0.62 m

3 0
3 years ago
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