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ankoles [38]
4 years ago
5

How are moving pulleys different from fixed-position pulleys? (Points : 3) Moving pulleys have more force than fixed-position pu

lleys. Moving pulleys are easier to use than fixed-position pulleys. Moving pulleys are attached to the object and move with the object. Fixed-position pulleys are connected to a fixed point and do not move. Moving pulleys are attached to the object and are stationary. Fixed-position pulleys are connected to a fixed point and move freely.
Physics
2 answers:
Lubov Fominskaja [6]4 years ago
8 0
"Fixed-position pulleys are connected to a fixed point and do not move" is the one among the following choices given in the question that explains how the moving<span> pulleys are different from fixed-position pulleys. The correct option among all the options that are given in the question is the fourth option.</span>
meriva4 years ago
4 0

Answer : Option D) Fixed-position pulleys are connected to a fixed point and do not move.

Explanation : As the name suggests the moving pulleys move freely whereas the fixed pulleys as per their name is attached to a fixed position and at a fixed point and they can not move freely.

Movable pulleys are ones which has one part of the rope attached to a fixed object, like a bar or a beam. Whereas when both the parts of a rope are attached to a fixed object, this is called as a fixed pulley.

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The acceleration due to gravity is lower on the Moon than on Earth. Which one of the following statements is true about the mass
Helga [31]
mass is the same, weight is less because weight =mass ×gravity
4 0
4 years ago
Read 2 more answers
Laskar, J.: 1990, The chaotic motion of the solar system. A numerical estimate of the size of the chaotic zones, Icarus, 88, 266
balandron [24]

The chaotic nature of the Solar System excluding Pluto was established by the numerical computation of the maximum Lyapunov exponent of its secular system over 200 myr.

<h3>What is chaotic motion of the solar system ?</h3>

There has been an increase in awareness of chaotic dynamics in the solar system over the past 20 years. The orbits of tiny objects in the solar system, such as asteroids, comets, and interplanetary dust, are now known to be chaotic and to experience significant variations across geological time periods.

  • a completely unpredictable orbit, or one where significant changes in the orbit can result from even small changes in the position and/or velocity of the orbiting entity.

Learn more about Chaotic motion here:

brainly.com/question/13717859

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7 0
1 year ago
The magnitude​ R, measured on the Richter​ scale, of an earthquake of intensity I is defined as Requalslog StartFraction Upper I
lapo4ka [179]

Answer:

R = 6.8

Explanation:

Given data:

Richter scaleR = log(\frac{I}{I_o})

where R - magnitude of earthquake of Richter scale

I - quake's intensity =  10^{6.8} \times I_o

I_o - minimum intensity earthquake

Plugging all information in the equation to get Richter's scale

R = log(\frac{10^{6.8} \times I_o}{I_o})

R = log(10^{6.8})

R = 6.8

6 0
3 years ago
Two bar magnets are placed side by side.
Jobisdone [24]

The question is wrongly written

Explanation:

There isn't a question in this.

But hope this helps

Magnets have two poles on positive and one negative

Positive and negative attract each other

3 0
4 years ago
A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electr
Nata [24]

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

E=\frac{1}{2}CV^2

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

6 0
3 years ago
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