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Reil [10]
3 years ago
10

A ball is thrown straight up in the air. When will its kinetic energy be the least before it is caught? A. at the start of its f

light. . B. halfway to the top of its flight. . C. at the end of its flight. . D. at the top of its flight
Physics
2 answers:
svet-max [94.6K]3 years ago
4 0

The correct answer to the question is D). At the top of its flight.

EXPLANATION:

When a  ball is thrown into air, the only force except air resistance that affects the motion of the ball is the force of gravity which pulls the body in vertically  downward direction.

When the ball was thrown, it has kinetic energy only. As the ball went up, its kinetic energy is converted into potential energy.

Due to the gravity,the velocity of the ball is decreasing each time. At a certain time, the ball will reach at its maximum height where its velocity will be completely zero. At that time whole of its kinetic energy must have converted into potential energy due to the law of conservation of energy.

Hence, the kinetic energy is least at the top of its flight.

vlabodo [156]3 years ago
3 0

Your answer is

D.) At The Top Of Its Flight

because once something gets to the top, or let's say the peak of its journey, it has gathered the most energy


Hope this Helps! :)

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A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
3 years ago
the car starts from a stop to travel 1100 meters in 14 seconds. it is clocked at 65 m/s at point k. find its average speed and a
inysia [295]

Answer:

The average velocity of the car is, V = 74.04 m/s

Explanation:

Given data,

The initial velocity of the car, u = 0 m/s

The displacement of the ca, S = 1100 m

The time period of travel, t = 14 s

The velocity of the car at point k, v = 65 m/s

Using the II equation of motion,

                      S = ut + ½  at²

Substituting the given values,

                      1100 = 0 + ½ x a x 14²

                          a = 11.22 m/s²

Using the III equation of motion

                         v² = u² + 2 as

                          v = √(2as)              (∵ u = 0)

Substituting,

                           v = √(2 x 11.22 x 1100)

                              = 157.11 m/s

The average speed of the car,

                        V=\frac{0+65+157.11}{3}

                        V = 74.04 m/s

Hence, the average velocity of the car is, V = 74.04 m/s

4 0
3 years ago
Which property of potential energy distinguishes it from kinetic energy
Bas_tet [7]

Answer:

Shape and position

Explanation:

Hope this helps! :)

7 0
2 years ago
How many significant digits should the answer to the following problem have? (2.49303 g) * (2.59 g) / (7.492 g) =
Feliz [49]

The number of significant digits to the answer of the following problem is four.

<h3>What are the significant digits?</h3>

The number of digits rounded to the approximate integer values are called the significant digits.

The following problem is

(2.49303 g) * (2.59 g) / (7.492 g) =

On solving we get

= 0.86184566204

The answer is approximated to  0.86185

Thus, the significant digits must be four.

Learn more about significant digits.

brainly.com/question/1658998

#SPJ1

6 0
2 years ago
It took a student 30 minutes to drive from his home to campus on
Gennadij [26K]

Answer:

48 i believe

Explanation:

3 0
3 years ago
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