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4 years ago

A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?

Physics

1 answer:

iVinArrow [24]4 years ago

8 0

Answer:

Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

Explanation:

Force applied on baseball = 30 times weight of the ball.

Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.

We have force applied is also equal to product of mass and acceleration.

F = ma = 30 x mg

a = 30g

So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

3 years ago

The image shows the electric field lines around two charged particles.

DENIUS [597]

It's either 3 or 4 I know this becuase I have read a book about electricity

8 0

3 years ago

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A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec

Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec$

Velocity of the string wave is equal to $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec$

Power of wave propagation is equal to $P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt$

So power of the wave will be equal to 5.464 watt

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3 years ago

The electrons of an atom store nuclear energy.

Sladkaya [172]

False. The nuclear energy is found within the nucleus. Electrons are located outside the nucleus.

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3 years ago

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On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86

Leto [7]

Answer:

6.575

Explanation:

T1 = 30C = 30 + 273 = 303 K

T2 = - 10 C = - 10 + 273 = 263 K

The coefficient of performance of heat pump

k = T2 / (T1 - T2)

k = 263 / (303 - 263) = 6.575

7 0

4 years ago