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2 years ago

Altuve hits a ball with a bat. The action force is the impact of the bat pushing against the ball. What is the

Physics

2 answers:

prisoha [69]2 years ago

8 0

Answer:

H. The ball pushing on the bat

Explanation:

The question above is related to "Newton's Third Law of Motion." This law states that if an object exerts a force on another object, the second object will exert the same amount of force on the first object but in an "opposite direction." The force exerted by the first object is known as "action force" while the force exerted by the second object is known as "reaction force."

In the situation above, the ball is exerting an "action force on the bat" by pushing the bat. The bat then reacts to its force by pushing it back with an equal force (reaction force) that is opposing it.

Anuta_ua [19.1K]2 years ago

5 0

Answer: H

Explanation: It’s returning the same force back to the bat

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Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to

fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

6 0

3 years ago

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

2 years ago

Which of the following could be the source of resistance in a household electric circuit?

Tpy6a [65]

D. all of these

all of these use electricity

Hope I helped!

8 0

3 years ago

Read 2 more answers

An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna

Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

$I=\frac{V}{R}$

Where R is the equivalent resistance of the resistors in series

$R=0.0510+0.0090=0.0600[ohm]$

$I=\frac{12.0}{0.0600}=200A$

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

$V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V$

The power dissipated supplied to the motor is given by:

$P=I^2*R_m\\P=(200)^2*0.0510=2.04kW$

now solving adding a 0.0900 ohm resistor:

$I=\frac{12.0}{0.15}=80A$

$V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V$

$P=I^2*R_m\\P=(200)^2*0.0510=0.326kW$

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2 years ago

To check if a stovetop is hot, you place you hand near the top of the stove and feel that it is warm without touching it. You ca

My name is Ann [436]

B.Radiation, Because the heat is Radiating of of the burner. xD

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2 years ago