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soldier1979 [14.2K]
3 years ago
14

A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont

al distance of 56 m. The only force acting on the ball is gravity. Which of the following statements best describes the magnitude of the horizontal component of the velocity of the ball? a. It is constant the whole time the ball is in free-fall. b. It is smallest when the height of the ball is maximum. c. It is smallest just before the ball hits the ground.
Physics
1 answer:
Mumz [18]3 years ago
6 0

Answer:

a. It is constant the whole time the ball is in free-fall

Explanation:

If we divide the movement on its vertical and horizontal components, and we concentrate on the vertical component, let's call x-component, and analyze Newton's second's law:

\sum\overrightarrow{F_{x}}=m\overrightarrow{a_{x}}

with \sum\overrightarrow{F_{x}}, a_{x} the acceleration on horizontal direction and m the mass of the ball, because the only force acting on the object is gravity that is always vertical, there're not forces on the horizontal direction that means \sum\overrightarrow{F_{x}}=0 and by (1) that implies a_{x}=0 there's not acceleration on horizontal direction.

Because acceleration is the rate at what velocity changes and there's no acceleration, there's no change in velocity, in other words velocity is constant on horizontal direction.

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A 6 kg mass collides with a body at rest .After collision they travel together with a velocity equal
Mars2501 [29]

Answer:

12 kg

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

After the collision, they have the same velocity, so v₁ = v₂ = v:

m₁ u₁ + m₂ u₂ = m₁ v + m₂ v

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

We know that m₁ = 6 kg, u₂ = 0 m/s, and v = u₁ / 3.

(6 kg) u₁ + m₂ (0 m/s) = (6 kg + m₂) (u₁ / 3)

(6 kg) u₁ = (6 kg + m₂) (u₁ / 3)

6 kg = (6 kg + m₂) (1/3)

18 kg = 6 kg + m₂

m₂ = 12 kg

3 0
4 years ago
How do forces affect the motion of an object?
cricket20 [7]

Explanation:

Unbalanced Forces in Action

Unbalanced forces can change the motion of an object in two ways. ... Second, when unbalanced forces act on a moving object, the velocity of the object will change. Remember that a change in velocity means a change in speed, direction or both speed and direction.

3 0
3 years ago
Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m
gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

Learn more about velocity here-

brainly.com/question/18084516

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7 0
2 years ago
An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
jekas [21]

Answer:

\varphi_1= 796.25 N m^2/C

\varphi_2= 0 N m^2/C

\varphi_3=686.1  N m^2/C

Explanation:

From the question we are told that

Electric field of intensity E= 3.25 kN/C

Rectangle parameter Width W=0.350 m  Length L=0.700 m

Angle to the normal \angle=30.5 \textdegree

Generally the equation for Electric flux at parallel to the yz plane \varphi_1 is mathematically given by

\varphi_1=EA cos theta

\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0

\varphi_1= 796.25 N m^2/C

Generally the equation for Electric flux at parallel to xy  plane \varphi_2 is mathematically given by

\varphi_2=EA cos theta

\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90

\varphi_2= 0 N m^2/C

Generally the equation for Electric flux at angle 30 to x plane \varphi_3 is mathematically given by

\varphi_3=EA cos theta

\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5

\varphi_3=686.072219  N m^2/C

\varphi_3=686.1  N m^2/C

7 0
3 years ago
If the density of iron is 7.87 g/cm3, what is the volume of the 12.00 g piece of iron? use proper significant figures.
Lelechka [254]
1.524777636594663 cm^3
7 0
3 years ago
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