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balandron
1 month ago
5

What’s the force of a pitching machine on a baseball?

Physics
2 answers:
Ostrovityanka [3.2K]1 month ago
7 0

Answer:

Force, F = 9 N

Explanation:

The initial velocity of the baseball, u = 0 m/s

Final velocity of the baseball, v = 30 m/s

Time taken, t = 0.5 s

Mass of the baseball, m = 0.15 kg

The force applied by a pitching machine on a baseball can be calculated using the second law of motion. The formula is given by:

F=ma

F=m\times \dfrac{v-u}{t}

F=0.15\times \dfrac{30}{0.5}

F = 9 N

Thus, the pitching machine exerts a force of 9 N on the baseball. This is the required solution.

Ostrovityanka [3.2K]1 month ago
6 0

Answer:

F = 9 N

Explanation:

Acceleration of the ball can be defined as the rate at which the velocity changes

and here we commence with

a = \frac{dv}{dt}

noting that

v_i = 0

v_f = 30 m/s

\Delta t = 0.5 s

now we will find

a = \frac{30 - 0}{0.5} = 60 m/s^2

to determine the force, we will apply Newton's second law as follows

F = ma

from the prior formula we derive

F = (0.15)(60) = 9 N

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On the assumption that copper– nickel alloys are random mixtures of copper and nickel atoms, calculate the mass of copper which,
inna [3103]

Response:

65.14 g

Clarification:

The entropy change resulting from mixing two metals is

\delta S= R[(n_{cu}+n_{Ni})ln(n_{cu}+n_{Ni})-n_{cu}lnn_{cu}-n_{Ni}lnn_{Ni}]

Here, R is the real gas constant, n_{cu} denotes the moles of copper, while n_{Ni} signifies the moles of nickel.

Therefore, determining the moles of Nickel yields

n_{Ni}=\frac{m_{Ni}}{M_{Ni}}

n_{Ni}=\frac{100}{58.69} =1.7 moles

Then substituting n_{Ni} = 1.7 into the equation produces ΔS= 15,

with R= 8.314, and solving gives

15= R[(n_{cu}+1.7)ln(n_{cu}+1.7)-n_{cu}lnn_{cu}-1.7ln1.7]

Upon resolving, the quantity of moles of copper in the mixture equals

n_cu= 1.025

Thus, the copper mass is calculated as n_cu×M_cu = 1.025×63.55 = 65.14 g

Consequently, the required mass is found to be = 65.14 g

3 0
2 months ago
HELP !! Maura is deciding which hose to use to water her outdoor plants. Maura noticed that the water coming out of her garden h
Yuliya22 [3333]
THE GREEN HOSE: Define the (x,y) coordinates at a height of 4 feet, which corresponds to where Majra holds the green hose. This indicates the equation for the green hose takes the form y = a(x - h)² + 4. Water from the hose lands on the ground 10 feet away from Majra, thus y(10) = -4. Given that the curve passes through (0,0), this leads to ah² + 4 = 0; therefore, ah² = -4. To satisfy the previous equation, we find a(10 - h)² + 4 = -4, simplifying to a(10 - h)² = -8. Dividing (3) by (4) gives a ratio of h²/(10-h)² = 1/2, leading to 2h² = (10 - h)² = 100 - 20h + h², and resolving yields h² + 20h - 100 = 0. Applying the quadratic formula, we get x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142. We discard the negative solution. The vertex locates at (4.142, 4). From (3), we deduce a = -4/4.142² = -0.2332, leading to the equation for the green hose: y = 0.2332(x - 4.142)² + 4. THE RED HOSE: The vertex of the red hose is positioned at (3,7), represented by the equation y = -(x-3)² + 7. A graph depicting y(x) for both hoses is included in the attached figure. Answers: a. The red hose throws water higher. b. The green hose's equation is y = -0.2332(x - 4.124)² + 4, starting at a height of 4 feet. c. The feasible domain for the green hose is between 0 ≤ x ≤ 10 feet, with the corresponding range being -4 ≤ y ≤ 4 feet.
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1 month ago
In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the
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The city evaluates the continuous increase of carbon monoxide from different origins each year. According to calculations, in the year "C: 2019"<span> (rounded to the closest whole number), the concentration of CO will surpass the allowed threshold.

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8 0
2 months ago
Read 2 more answers
A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
Maru [3345]

The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:

F = mg

Where,

m = mass

g = gravitational acceleration

F = 5*9.8

F = 49N

Consequently, the answer is E.

5 0
2 months ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Yuliya22 [3333]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

It is known that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2..............1

let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}.............3

divide equation (2) by equation (3)

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

6 0
1 month ago
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