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statuscvo [17]
3 years ago
15

A 0.53 kg arrow leaves a bowstring at a velocity of 63 m/s. If the arrow was initially at rest and then the string applied a for

ce on it for 7 x 10-3 seconds, what was the approximate average force in Newtons that acted on the arrow during this time?
Physics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

4770 N

Explanation:

Momentum is the product of mass and velocity and force is the change in momentum divided by change in time.

Given from the question;

Mass of arrow= 0.53 kg

Velocity of arrow = 63 m/s

Initial velocity of arrow = 0 m/s

Change in time = 0.007 s

Finding momentum after the arrow is released as;

p=m*v

p= 0.53 * 63

p= 33.39 kg*m/s

Force is the change in momentum divided by change in time;

F= 33.39 / 0.007

F= 4770 N

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A car of mass 1000 kg is travelling at 10 m/s. What is its kinetic energy?
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Answer:

50,000

Explanation:

kinetic energy=1\2×mv^2

=1\2×1000kg×10×10

=50000

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The state of matter that has particles that slide by one another
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A cylindrical tank has a radius of 53 cm and a height of 1200 mm. It weighs 9.6 kN
xxTIMURxx [149]

Answer:

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Explanation:

assuming that

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Read 2 more answers
Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the
Rzqust [24]

Wow !  This is not simple.  At first, it looks like there's not enough information, because we don't know the mass of the cars.  But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

                                  PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

                                 KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down.  So now, here comes the big jump.  Put a comment under
my answer if you don't see where I got this equation:

                                   KE = 0.9  PE

        (1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)     

Divide each side by (mass): 

               (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass.  As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

               (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

               Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

                          =  (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

                          =  (24,500 / 88.2)  (m²/s²) / (m/s²)

                          =        277-7/9    meters
                                  (about 911 feet)
3 0
3 years ago
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