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2 years ago

5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.

Physics

1 answer:

andreev551 [17]2 years ago

6 0

Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

a = 8/4.2

a = 1.904

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Answer:

2.25 billion years

Explanation:

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Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

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3 years ago

The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?

kipiarov [429]

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e I = $\frac{Q}{t}$ ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

Q = I × t

= 1.5 × 2

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The amount of electrons that flow in the given time is 3.0 C.

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3 years ago

When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is

padilas [110]

Answer:

$P_{C} = 3.2\, atm$

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

$P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T$

Bulb B (3 L, 4 atm) - Before opening:

$P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T$

Bulbs A & B (5 L) - After opening:

$P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T$

After some algebraic manipulation, a formula for final pressure is derived:

$P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}$

And final pressure is obtained:

$P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}$

$P_{C} = 3.2\, atm$

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3 years ago