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fiasKO [112]
3 years ago
11

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air? The softball expe

riences force as a result of Amy’s throw. As the ball moves, it experiences from the air it passes through. It also experiences a downward pull because of . NextReset

Physics
2 answers:
scZoUnD [109]3 years ago
7 0
The force of the throw is an applied force

moving through the air is drag 

and the downward pull would be due to gravity
Komok [63]3 years ago
4 0

Answer:

As the ball moves, it experiences from the air it passes through. It also experiences a downward pull because of gravity

Explanation:

When Amy throws a ball in the air then initially the force is applied by Amy to accelerate the ball in the air.

Now after throwing into the air the ball will have forces as per the FBD

Ball will have a downward force which is due to gravity and its magnitude is always remains the same

Now ball also have another force opposite to the direction of motion due to air drag.

This air drag force depends on the speed of the ball and always opposite to the velocity of the ball.

so correct answer will be

As the ball moves, it experiences from the air it passes through. It also experiences a downward pull because of gravity.

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If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s , determine the constant deceleration of the car
Svetlanka [38]

Question:  Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.

Answer:

1.29 m/s²

Explanation:

From the question,

a = (v-u)/t............................ Equation 1

Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

Given: v = 13 m/s, u = 35 m/s, t = 17 s.

a = (13-35)/17

a = -22/17

a = -1.29 m/s²

Hence the deceleration of the car is 1.29 m/s²

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3 years ago
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andreev551 [17]

Answer:

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6 0
4 years ago
Read 2 more answers
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N i
DaniilM [7]

Answer:

The apparent weight of the person as she pass the highest point is  N  =  458.8 \ N

Explanation:

From the question we are told that

   The radius of the Ferris wheel is r = 5.0 \ m

    The period of revolution is T = 8.0 \ s

     The weight of the person is  W  =  670 \ N

   

Generally the speed of the wheel is mathematically represented as

      v =  \frac{2 \pi r}{T }

substituting values

      v =  \frac{2 * 3.142 *  5}{8 }

       v =  3.9 3 \ m/s

The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as

          N  =  mg  - \frac{mv^2}{r}

Where m is the mass of the person which is mathematically evaluated as

     m =  \frac{W}{g}

substituting values

    m =  \frac{670}{9.8}

    m =  68.37 \ kg

So

    N  =  68.37 * 9.8   - \frac{68.37 * {3.93}^2}{5}

    N  =  458.8 \ N

5 0
3 years ago
Starting with a constant velocity of 45 km/h, a car accelerates for 35 seconds at an acceleration of 0.45 m/s2 . What is the vel
DENIUS [597]

Answer:

28.3 m/s

Explanation:

Vi = 45 Km/h = 12.5 m/s

Vf - Vi = at

Vf -12.5 = 0.45(35)

Vf= 28.3 m/s

5 0
3 years ago
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