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2 years ago

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A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° wit

h the vertical. In such a condition of static equilibrium, what is the net force on the new vine?

1 answer:

natka813 [3]2 years ago

8 0

To solve this problem we will apply the considerations of equilibrium on a body, we will decompose the tensions into their respective components and apply the sum of forces and solving equations for the given system.

According to the graph, the sum of the horizontal components would be

$\sum T_x = 0$

$TSin35\° -TSin35\° = 0$

$0 = 0$

The sum of the vertical components is

$\sum T_y = 0$

$2TCos35\° - mg = 0$

$2TCos35\° = 600N$

$T = \frac{600N}{2Cos35\°}$

$T = 366.2N$

Therefore the net force on the new vine is 366.2N

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In a simple harmonic motion, when the displacement from the equilibrium position is at its maximum,

Keith_Richards [23]

Answer:

option B

Explanation:

The correct answer is option B

When the displacement in the harmonic motion is maximum then kinetic energy at the maximum point is minimum and the potential energy is maximum at that point.

So, when the displacement is maximum, spring force magnitude is also maximum because the force is proportional to the displacement and also the magnitude of the acceleration is maximum so, the net force is also maximum.

7 0

2 years ago

Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D

erma4kov [3.2K]

Answer:

The car C has KE = 100, PE = 0

Explanation:

The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.

The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.

Kinetic energy is the energy that a body possesses due to its motion.

Potential energy is the energy a body possesses due to its position.

From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.

In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.

Therefore, the car C has KE = 100, PE = 0

6 0

2 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

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2 years ago

A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t

nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

$\epsilon = 1 - \dfrac{T_C}{T_H}$

$0.25 = 1 - \dfrac{T_C}{464}$

$T_C= (1 - 0.25) \times 464$

$T_C= 0.75 \times 464$

T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

$T_C= (1 - 0.42) \times 464$

$T_C= 0.58 \times 464$

$T_C= 269.12\ K$

8 0

2 years ago

What’s the difference between atoms and molecules in a substance?

Goryan [66]

Answer:

Atoms are single neutral particles.

example: Ne, O

Molecules are neutral particles made of two or more atoms bonded together.

example:O2,HCl

5 0

2 years ago

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