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olga nikolaevna [1]
1 year ago
11

Tennis balls experience a large drag force. A tennis ball is hit so that it goes up and then comes back straight down.

Physics
1 answer:
Effectus [21]1 year ago
8 0

A tennis ball is hit by a large force so that it goes up into the air and then it comes back straight down because of gravity.

<h3>How object move upward and downward?</h3>

We know that objects move upward due to application of force on it while on the other hand, object comes to the ground because of the attraction of earth which we called gravity.

So we can conclude that a tennis ball is hit by a large force so that it goes up into the air and then it comes back straight down because of gravity.

Learn more about force here: brainly.com/question/12970081

#SPJ1

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natulia [17]
A or d would be most likely it
7 0
3 years ago
When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g
mrs_skeptik [129]

Answer:

3°C

Explanation:

We can that heat Q=mc_p dT

Where m is the mass c_p = specific heat capacity

dT = Temperature difference

here we have given m=625 g =.625 kg

specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)

T_1 =25°C

T_2 we have to find

we have also given Q=10.9 KJ

10.9=0.625×0.79×(25-T_2)

25-T_2 =22

T_2=3°C

7 0
3 years ago
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Vera_Pavlovna [14]

Answer: when in doubt go with B

Explanation:

3 0
3 years ago
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From the lens equation calculate the position of the following images produced by a convex lens.
fiasKO [112]

Explanation:

(i)

O is the object and I is the image.

The image formed is enlarged and it is erect. So the magnification will be positive (+) and greater than 1.

Refer above image. 1

(ii)

O is the object and I is the image.

The image formed is diminished and erect. So the magnification will be positive (+) and less than1.

Refer above image. 2

(iii)

The image will be formed as the 2F on the other side of the lens and it will be of same of the object.

5 0
3 years ago
A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti
vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

q(t) = e^{-\frac{t}{(R*C)}}

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}

Ln(\frac{1}{2}) = -\frac{0.01}{RC}

- RC = \frac{0.01}{Ln(1/2)}

RC = 0.014s

We know that RC is equal to the time constant, then

T = RC  = 0.014s = 14ms

Therefore the time constant for the process is about 14ms

5 0
3 years ago
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