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uysha [10]
3 years ago
8

Question 5 (1 point)

Physics
1 answer:
katovenus [111]3 years ago
8 0

Answer:

The horizontal distance the ball travels is approximately 4.055 meters

Explanation:

The given parameters are;

The height from which the child kicks the ball = 3.5 m

The horizontal speed of the ball = 4.8 m/s

Therefore, we have;

The time it takes the ball to hit the ground is given by the relation;

h = u·t + 1/2·g·t²

Where;

u = The initial vertical velocity of the ball = 0 m/s

t = The time it takes the ball to hit the ground

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the ball = 3.5 m

3.5 = 0 × t + 1/2 × 9.81 × t²

3.5 = 1/2 × 9.81 × t²

∴ t² = 3.5/(1/2 × 9.81)

∴ t = √(3.5/(1/2 × 9.81) = 0.8447 s

t ≈ 0.8447 s

The time the ball takes in flight = t ≈ 0.8447 s

Therefore;

The horizontal distance the ball travels = The horizontal velocity × The time of flight

∴ The horizontal distance the ball travels = 4.8 × 0.8447 ≈ 4.055

The horizontal distance the ball travels ≈ 4.055 meters.

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Would a bond between potassium and iodine most likely be covalent or ionic? Explain your answer
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The speed of a tennis ball that is served is 73.14 m/s. During a serve, the ball typically starts from rest and is in contact wi
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Answer: Third option

F = 250w

Explanation:

The impulse can be written as the product of force for the time interval in which it is applied.

I = F (t_2-t_1)

You can also write impulse I as the change of the linear momentum of the ball

I = mv_2 -mv_1

So:

F (t_2-t_1) = mv_2 -mv_1

We want to find the force applied to the ball. We know that

(t_2-t_1) = 30 milliseconds = 0.03 seconds

The initial velocity v_1 is zero.

The final speed v_2 = 73.14\ m / s

So

F * 0.03 = 73.14m

F * 0.03 = 73.14m\\\\F=\frac{73.14m}{0.03}\\\\F=2438m

We must express the result of the force in terms of the weight of the ball.

We divide the expression between the acceleration of gravity

g = 9.8\ m / s ^ 2

F=\frac{2438m*g}{g},\ \ m*g=w\\\\g=9.8\ m/s^2\\\\F=\frac{2438w}{9.8}\\\\F=249w

The answer is the third option

3 0
3 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
3 years ago
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