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VARVARA [1.3K]
3 years ago
7

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]3 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 <em>"opposite direction."</em> The<u> force exerted by the first object</u> is known as "action force" while the <u>force exerted by the second object </u>is known as "reaction force."

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

Anuta_ua [19.1K]3 years ago
5 0
Answer: H


Explanation: It’s returning the same force back to the bat
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6 0
3 years ago
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
How would it be possible for an object to be traveling with a constant speed and still accelerating?
lys-0071 [83]
Yeah!! It's possible for an object by changing it's direction....
4 0
4 years ago
Can I get a direct answer please??
OleMash [197]
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7 0
3 years ago
The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen
Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer:  A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life =  5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 =  5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

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So that expressing   N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0
3 years ago
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