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Ray Of Light [21]
3 years ago
10

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. what is the ball's horizont

al acceleration at the top of its trajectory?
Physics
1 answer:
andreyandreev [35.5K]3 years ago
6 0

the horizontal acceleration of the ball at the top of the trajectory is zero.

in case of projectile motion,only the force of gravity acts on the object along the vertical. There are no horizontal forces acting on the object.

now force = m a

m= mass

a= horizontal acceleration

f= horizontal force=0

so a= F/m

a=0/m=0

so the horizontal acceleration at all the points including the top of the trajectory is zero.

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How much force must a locomotive exert on a 12840-kg boxcar to make it accelerate forward at 0.490 m/s2?
ddd [48]

Data given:

m=12840kg

a=0.490m/s²

Formula:

F=ma

Solution:

F= 12840kg×0.490m/s²

F=6291.6N

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Suppose you exert 150 n on your refrigerator and push it across the kitchen floor at constant velocity. what friction force acts
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3 0
3 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

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3 years ago
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That's a not-bad description of a capacitor.
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