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Minchanka [31]
3 years ago
10

Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2

Physics
1 answer:
liubo4ka [24]3 years ago
7 0

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

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Answer:

The appropriate response will be "F_2=\frac{3}{4}F". A further explanation is given below.

Explanation:

According to the question,

⇒ F_1=\frac{G(m_1 m_2)}{r^2} ....(equation 1)

and,

⇒ F_2=\frac{G(m_1 m_2)}{r^2} ....(equation 2)

Now,

On dividing both the equations, we get

⇒ \frac{F_1}{F_2}=\frac{\frac{G(2)(2)}{(1)^2}}{\frac{G(1)(3)}{(1)^2}}

⇒ \frac{F_1}{F_2}= \frac{4}{3}

⇒ F_2=\frac{3}{4}F

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Answer:

Explanation:

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What is so remarkable about the migratory monarch butterflies?
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? Question
Harrizon [31]

No Solution:

4x+10y = 40

2x + 5y = 40

2x - 8y=17

x - 4y = 7

One Solution:

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Infinitely Many Solutions:

x + 2y = 1

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Explanation: You only had 11 equations listed. But I believe these are the ones you want. I took the test and got them right. Hope it helps

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What force must be applied to move a 251 kg rock on a pavement like surface
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(251 kg) x (9.8 m/s²) x (μ) =

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