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grandymaker [24]

2 years ago

13

A tennis ball covers a distance of 12 meters in 0.4 seconds what is the velocity of the tennis ball?

2 answers:

Usimov [2.4K]2 years ago

7 0

V=d/t
V=12/0.4
V=30 m/s

Ksenya-84 [330]2 years ago

4 0

Answer:

30m/s

Explanation:

Given parameters:

Distance = 12m

Time taken = 0.4s

Unknown:

Velocity of the tennis ball = ?

Solution:

Velocity of a body is the displacement per unit of time.

Velocity = $\frac{displacement}{time}$

So;

Velocity = $\frac{12}{0.4}$ = 30m/s

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Answer:

The Stefan–Boltzmann constant (also Stefan's constant), a physical constant denoted by the Greek letter σ (sigma), is the constant of proportionality in the Stefan–Boltzmann law: "the total intensity radiated over all wavelengths increases as the temperature increases", of a black body which is proportional to the ...

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There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

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Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

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2 years ago

A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

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Density of water, d = 1000 kg/m^3

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P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

By substituting the values, we get

$\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}$

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

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