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Otrada [13]
3 years ago
5

Thomas needs to move an 80 kg rock, but cannot lift it. He decides to use a

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Answer:

4

Explanation:

The weight of the rock is W = mg = (80 kg) (10 m/s²) = 800 N.

The mechanical advantage is therefore 800 N / 200 N = 4.

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Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha
Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N.  Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.  

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation  is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

a_B = a_A-0.5 \frac{m}{s^2}

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0
3 years ago
Read 2 more answers
Car a runs a red light and broadsides car b, which is stopped and waiting to make a left turn. car a has a mass of 1,800kg. car
frez [133]

The law of conservation of momentum tells us that momentum is conserved, therefore total initial momentum should be equal to total final momentum. In this case, we can expressed this mathematically as:

mA vA + mB vB = m v

where, m is the mass in kg, v is the velocity in m/s

since m is the total mass, m = mA + mB, we can write the equation as:

mA vA + mB vB = (mA + mB) v

furthermore, car B was at a stop signal therefore vB = 0, hence

mA vA + 0 = (mA + mB) v

1800 (vA) = (1800 + 1500) (7.1 m/s)

<span>vA = 13.02 m/s</span>

7 0
3 years ago
Suppose a thick nimbostratus cloud contains ice crystals and cloud droplets all about the same size. Which precipitation process
Ronch [10]

Answer:The most important process would be the ice crystal process

Explanation:

Ice crystal process also called Bergeron process requires numerous small water drops that are supercooled, which is a common feature in clouds between about 0° and -20°C or below, along with a small number of ice crystals. Therefore because the collision-coalescence process requires that cloud droplets be of varying size so that drops will fill at different speeds, the most important process would be the ice crystal process.

7 0
3 years ago
A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga
BartSMP [9]

Answer:

Total work done in expansion will be 3.60\times 10^5J

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm =1.01\times 10^5Pa

So 2.10 atm =2.10\times 1.01\times 10^5=2.121\times 10^5Pa

Volume is increases from 3370 liter to 5.40 liter

So initial volume V_1=3.70liter

And final volume V_2=5.40liter

So change in volume dV=5.40-3.70=1.70liter

For isobaric process work done is equal to W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J

So total work done in expansion will be 3.60\times 10^5J

5 0
3 years ago
Take a look at a bicycle with gears. Using what you have just learned, answer the following questions:
Harlamova29_29 [7]

Answer:

answer a: a large front gear with a small back gear

answer b: a small front gear with a large back gear

Explanation:

just simple gearing ratios

4 0
3 years ago
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