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3 years ago

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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Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite

Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass $_{Hank$ × Acceleration $_{Hank$ = Mass $_{Henry$ × Acceleration $_{Henry$

Mass $_{Hank$ / Mass $_{Henry$ = Acceleration $_{Henry$ / Acceleration $_{Hank$

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass $_{Hank$ / Mass $_{Henry$ = 1 / 1.26

Mass $_{Hank$ / Mass $_{Henry$ = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0

3 years ago

You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p

user100 [1]

Answer:

The potential energy at point A is 17.1675 J

Explanation:

The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries

In meteorology it is used to describe the level of saturated soil above the water table

Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are

Point A, elevation = 75 cm capillary potential = -100 cm

Point B, elevation = 25 cm capillary potential = -200 cm

The total potential energy at point A is

Elevation above reference - capillary potential =75-(-100) = 175 cm

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PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J

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3 years ago

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2 years ago

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3 years ago

On a day when the temperature reaches 50°F, the temperature in degrees Celsius is: 20°C

anzhelika [568]

Answer:

10°C

Explanation:

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