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mel-nik [20]
3 years ago
5

Please show steps as to how to solve this problem Thank you!

Physics
1 answer:
bezimeni [28]3 years ago
4 0

Explanation:

Let x = distance of F_1 from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque \tau_{net} about the fulcrum is zero:

\tau_{net} = -F_1x + F_2d_2 = 0

-m_1gx + m_2gd_2 = 0

m_1x = m_2d_2

Solving for <em>x</em>,

x = \dfrac{m_2}{m_1}d_2

\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}

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A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
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The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

8 0
2 years ago
Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109
r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

E=\Delta mc^2\\\Delta m=\frac{E}{c^2}(1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J

Hence, by replacing in the equation (1) you have  (c=3*10^{8}m/s)

\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg

HOPE THIS HELPS!!

3 0
3 years ago
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What is the scientific term for rocks formed from lava?
sergeinik [125]

Answer:

extrusive I'm pretty sure that's right

5 0
4 years ago
What is the reaction force if a girl pulls on a cow?
Papessa [141]
Answer : B) The cow pulls back on the girl.

From newton’s third law we know that every action has a reaction force pushing back. So when the girl pulls on a cow, the cow is pulling back on her.
8 0
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1. Earth releases about 44-46 Tw of heat, in fact heat can be converted into
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Answer:

  even if it all could be used, it wouldn't be enough

Explanation:

The land area of the US is about 5.45% of the world's area, so the amount of released heat over the area of the US is on the order of 2.4 Tw. Current technology for converting geothermal energy to electricity is about 12% efficient, so the available energy might amount to 0.29 Tw if it could all be captured.

Energy consumption in the US in 2019 was on the order of 0.46 Tw. This suggests that even if <em>all</em> of the thermal energy radiated by the Earth from the US could be turned to useful forms of energy, it would meet only about 60% of the US need for energy.

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