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2 years ago

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A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

t the pitcher with a final speed of 24 m/s. Assume the direction of the initial motion of the baseball to be positive.

1 answer:

Maru [420]2 years ago

4 0

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

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A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte

Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

= 3/2×(5.3)×(8.31451)×(24 + 273)

= 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

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Ocean waves are observed to travel to the right along the water surface during a developing storm. A Coast Guard weather station

Nuetrik [128]

Answer:

The amplitude is 2.3 m

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The frequency is 0.16 Hz

The time period is 6.25 sec

The equation that governs the behavior is $Y=(2.3)sin[(\frac{2\pi}{8.6} )x -(\frac{2\pi}{6.2} )t]$

Explanation:

The explanation is shown on the first uploaded image

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2 years ago

En la etiqueta de un bote de fabada de 350 g, leemos que su aporte energético es de 1630 kj por cada 100 g de producto a) La can

fredd [130]

Answer:

(a) 153.37 g

(b) 5705 kJ

Explanation:

(a) To find the amount of bean needed by a man you first calculate the equivalence in beans to 2500kJ

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Thus, 153.37 g has the energy needed by a man that needs 200kJ per day.

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$E=350g*\frac{1630kJ}{100g}\\\\E=5705\ kJ$

Thus, the energy is 5705kJ

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Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

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now potential is given as

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now final potential energy is given as

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Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

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2 years ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

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The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

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Tension becomes :

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So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

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2 years ago