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olasank [31]
1 year ago
7

A tennis ball is hit into the air with a racket. when is the ball’s kinetic energy the greatest? ignore air resistance.

Physics
1 answer:
krok68 [10]1 year ago
4 0

When the player strikes the ball, kinetic energy is at its greatest.

Kinetic energy = 1/2*m*v^{2}

where m denotes mass and v denotes velocity.

Therefore, kinetic energy is inversely correlated with square of velocity.

When a player strikes the ball, velocity is at its highest.

Thus, when the player strikes the ball, kinetic energy is at its highest.

<h3>What is kinetic energy?</h3>

It is the energy that a body has because it is moving.

<h3>Which 5 forms of kinetic energy are there?</h3>

Kinetic energy comes in five different forms:

  • radiant
  • thermal
  • sound
  • electrical
  • mechanical.

Kinetic energy is measured in Joules.

Importance of kinetic energy:

  • The capacity to perform work is kinetic energy's most significant quality.
  • Force acting on an object while it is moving is referred to as work.
  • Energy and work are interchangeable because of their tight relationship.

To learn more about kinetic energy visit:

brainly.com/question/12669551

#SPJ4

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A distressed car is rolling backward, downhill at 3.0 m/s when its driver finally manages to
konstantin123 [22]

Answer:

Explanation:

Acceleration is equal to the change in velocity over the change in time, or

a=\frac{v_f-v_i}{t} where the change in velocity is final velocity minus initial velocity. Filling in:

3.0=\frac{v_f-(-3.0)}{6.0} Note that I made the backward velocity negative so the forward velocity in our answer will be positive.

Simplifying that gives us:

3.0=\frac{v_f+3.0}{6.0} and then isolating the final velocity, our unknown:

3.0(6.0) = v + 3.0 and

3.0(6.0) - 3.0 = v and

18 - 3.0 = v so

15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.

5 0
3 years ago
Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
Where does matter come from?
Pie
The Big Bang theory is matter and energy in the universe exploded out from one point. As the explosion occurred, energy and matter spread outward and formed the universe. The matter from the Big Bang formed clouds of gas.
8 0
2 years ago
How does the angle of launch affect the kinetic energy of a rubber band?​
Lady_Fox [76]

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

3 0
3 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
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