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Tanzania [10]
4 years ago
15

You hold a barbell in a horizontal position. The barbell’s center of mass is exactly mid-way between your two hands. A friend wa

lks up and attaches a weight on the end near your left hand. How does the torque on the bar exerted by your left hand compare to the torque on the bar from your right hand if the bar remains horizontal and at rest?

Physics
1 answer:
Harrizon [31]4 years ago
6 0

Explanation:

First consider that each hand works as a fulcrum: a pivot point where the barbell can rotate.

Now consider only the left hand. If the center of mass of the barbell is between hands (in the middle) it is displaced respect the fulcrum, therefore the weight which is pushing the bar downwards becomes a rotational force. The same thing happens to the other hand. Now, if more weight is added to the left hand the center of mass is displaced towards the left hand and depending how much weight is added, the center of mass will change its position and therefore the torque each hand experiences changes.

If the center of mass is still between hands: The torque remains almost the same changing only the magnitudes but not the direction.

If the center of mass is on the hand: there is no torque for the left hand because there is no leaver.

If the center of mass is to the left: now the torque changes direction and both hands need to stop it in the same direction.

(see diagram below)

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Recall that force is a change in momentum over a change in time, the force due to radiation pressure reflected off of a solar sa
Zarrin [17]

Answer:

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= \frac{2I}{c}

Explanation:

The momentum of a photon is:

p = E/c

E = the photon energy

c = the speed of light.

take the time derivative (gives the force)

F = dp/dt = (dE/dt)/c

F = 2(dE/dt)/c (is doubled for complete reflection of the light)

Intensity has the units of energy per unit time per unit area

=  I

then,

Force/unit area = 2I/c

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= \frac{2I}{c}

4 0
3 years ago
A charge of 5.4 C experiences a force of 25.0 in an electric field. What is the strength of electric field at that point ? If th
Airida [17]

The strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

<h3></h3><h3>What is the electric field strength?</h3>

The electric field strength is defined as the ratio of electric force to charge.

Given data;

q₁ = 5.4 C

F₁ is the electric force in case1

E is the electric field =?

F₂ is the electric force in case 2

q₂ is the charge 2

The strength of the electric field at that point is;

F₁=Eq₁

E₁=F/q₁

E₁=25.0 N / 5.4 C

E₁=4.587 N/C

The force would this charge experience at that point when the charge is 2.7 C;

F₂=Eq₂

F₂=4.587 N/C × 2.7 C

F₂ = 12.38 N

Hence the strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

To learn more about the electric field strength, refer to the link;

brainly.com/question/4264413

#SPJ1

8 0
2 years ago
3. What is the best way to help with<br> mining problem in the future?
marusya05 [52]

Answer:

1) Mining companies have an impressive track record for delivering continuous improvements in safety and risk governance standards. We have no doubt that the professionalism and expertise present within the industry will ensure that any new and emerging risk challenges are dealt with in an equally determined fashion.

2) Insurers have recognized the approach and achievements of mining companies in identifying, mitigating and retaining their risks. Many mining companies have close and longstanding relationships with their insurers built through regular dialogue with mining company executives and visits to mining sites and processing facilities.

3) This document should be of value to any organization involved in this ever-evolving industry.

4) The very nature of mining natural resources means that many businesses will have operations in some of the most remote and inhospitable areas in the world and very often coupled with a high susceptibility to natural catastrophe.

5) In addition to the traditional risk factors, the mining industry now faces an even wider range of challenges. Factors such as climate change, new technologies, economic uncertainties and secure supply of key consumables like electricity, water, gas and other fuels are all difficult to predict and bring additional complications to securing appropriate balance sheet protection.

4 0
3 years ago
What’s the answer for this problem?
pickupchik [31]
The answer is always true a



4 0
3 years ago
Nicolo` works on weekends at a Slow Food Parlor. He fills a pitcher full of Cola, places it on the counter top and gives the 2.6
Slav-nsk [51]

Answer:4.22 J

Explanation:

Given

mass of pitcher m=2.6\ kg

Force applied  F=8.8\ N

distance moved s=48\ cm

Applying work-Energy theorem which states that work done by all the forces is equal to the change in kinetic energy of the object

Work done by force W=F\cdot s

W=8.8\times 0.48 J

change in kinetic Energy =\frac{1}{2}mv^2-0

8.8\times 0.48=\Delta K.E.

K.E.-0=4.22

K.E.=4.22\ J

8 0
4 years ago
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