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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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There is a high concentration of this and other metals in the mesosphere
WARRIOR [948]
Hey there! 

The answer to your question is: Iron

There is a very high concentration of iron and other metals in the mesosphere. There is a further buildup of these metals when meteorites burn up in the atmosphere and release such metal atoms. 

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7 0
4 years ago
Three resistors connected in series have potential differences across them labeled /\V1 , /\V2 , and /\V3. What expresses the po
Brrunno [24]

Answer:

\Delta V=\Delta V_1+\Delta V_2+\Delta V_3

Explanation:

We are given that three resistors R1, R2 and R3 are connected in series.

Let

Potential difference across R_1=\Delta V_1

Potential difference across R_2=\Delta V_2

Potential difference across R_3=\Delta V_3

We know that in series  combination

Potential difference ,V=V_1+V_2+V_3

Using the formula

\Delta V=\Delta V_1+\Delta V_2+\Delta V_3

Hence, this is required expression for potential difference.

3 0
4 years ago
Newton’s heliocentric view of the universe meant that the sun was the center of the universe. True False
Anuta_ua [19.1K]
The answer to this question is true.
7 0
4 years ago
Read 2 more answers
A mixture of N2, H2 and He have mole fractions of 0.25, 0.65, and 0.10, respectively. What is the partial pressure of N2 if the
matrenka [14]

Answer: The partial pressure of N_2 if the total pressure of the mixture is 3.9 atm is 0.975 atm

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

p_A=x_A\times P_T

where, x = mole fraction of nitrogen in solution = 0.25

p_A = partial pressure  of nitrogen = ?

P_T = Total pressure = 3.9 atm

Putting in the values :

p_A=0.25\times 3.9=0.975atm

The partial pressure of N_2 is 0.975 atm

3 0
4 years ago
Read 2 more answers
A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.60 m/s. How
kirill115 [55]

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

E_{pot}-E_{friction}=E_{kin}

where:

E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

(56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ]

4 0
3 years ago
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