All of following are types of kinetic energy EXCEPT for: *

A trapeze artist, with swing, weighs 800 N; he is momentarily heldto one side by his partner using a horizontal force so that th

Tamiku [17]

Answer:

461.88 N

Explanation:

$F_{g}$ = Weight of the swing = 800 N

$T$ = Tension force in the rope

$F$ = Horizontal force being applied by the partner

Using equilibrium of force in vertical direction using the force diagram, we get

$T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N$

Using equilibrium of force in horizontal direction using the force diagram, we get

$F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N$

6 0

2 years ago

The density of mercury is 13.6 g/ml. what is its density in lbs/L

solong [7]

Answer:

can u send a picture to identify it

Explanation:

sorry I just need point:/

4 0

2 years ago

An ideal monatomic gas at temperature T is held in a container. If the gas is compressed isothermally, that is at constant tempe

OlgaM077 [116]

Answer:

a) 0 J

b) W = nRTln(Vf/Vi)

c) ΔQ = nRTln(Vf/Vi)

d) ΔQ = W

Explanation:

a) To find the change in the internal energy you use the 1st law of thermodynamics:

$\Delta U=\Delta Q-W$

Q: heat transfer

W: work done by the gas

The gas is compressed isothermally, then, there is no change in the internal energy and you have

ΔU = 0 J

b) The work is done by the gas, not over the gas.

The work is given by the following formula:

$\\W=nRTln(\frac{V_f}{V_i})$

n: moles

R: ideal gas constant

T: constant temperature

Vf: final volume

Vi: initial volume

Vf < Vi, then W < 0 and the work is done on the gas

c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.

The amount of heat is equal to the work done W

d)

$\Delta U = \Delta Q-W\\\\0=\Delta Q-W\\\\\Delta Q=W=nRTln(\frac{V_f}{V_i})$

5 0

2 years ago

The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo

weeeeeb [17]

Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

<u>Explanation:</u>

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since $L=I \times \omega$ , the turning rate of the two-puck framework must stay unaltered.

4 0

2 years ago

8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer

Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

$M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \ by \ the \ effort}{disatnce \ moved \ by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m$

Therefore, the distance that the object is raised above its initial position is 5.625 m.

3 0

2 years ago