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Leya [2.2K]
2 years ago
14

A 450-N rightward force is used to drag a large box across the floor with a constant velocity of 1.2 m/s. The coefficient of fri

ction between the box and the floor is 0.795. Determine the mass of the box.
Physics
1 answer:
Nikolay [14]2 years ago
8 0

Answer:

the mass of the box is 51.98 kg.

Explanation:

Given;

applied horizontal force, F = 450 N

coefficient of friction, μ = 0.795

constant velocity, v = 1.2 m/s

At constant velocity, the acceleration of the object is zero and the net force will be zero.

F_{Net} = F - F_k\\\\0 = F - F_k\\\\F_k = F\\\\\mu \ N = F\\\\\mu (mg) = F\\\\m = \frac{F}{\mu  g} \\\\m = \frac{405}{0.795 \ \times \ 9.8} \\\\m = 51.98 \ kg

Therefore, the mass of the box is 51.98 kg.

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In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.​
galben [10]

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED

6 0
2 years ago
Ohms law is A.(R=E/W). B.(R=E/1). C.(E/Z). D.none of them
nikklg [1K]

Answer:

D. none of them.

Explanation:

This is because Ohm's law is:

Voltage = Current × Resistance

or,

V = IR

6 0
3 years ago
g A 4-foot spring is elongated167feet long after a mass weighing 16 pounds is attached to it. The medium throughwhich the mass m
marta [7]

Answer: hello question b is incomplete attached below is the missing question

a) attached below

b) V = 0.336 ft/s

Explanation:

Elongation ( Xo)  = 16/ 7 feet

mass attached to 4-foot spring = 16 pounds

medium has 9/2 times instanteous velocity

<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>

The motion is an underdamped motion because the value of β < Wo

Wo = 3.741 s^-1

attached below is a detailed solution of the question

3 0
2 years ago
Identify and compare 4 Fundamental forces (lesson 1.04)
Tom [10]
4 fundamental forses are: strong, electro-magnetic, weak, gravity.
The strong force is the force which can hold nucleus together against enormous forces of repulsion of the protons is strong indeed. In comparasing to electro magnetic force, this force in not an inverse square and it has very short range.
The electro-magnetic force manifests itself as trough the forces between charges(Colubos law) and the magnetic force, both of which are summarized in the Lorentz force law. The electro-magnetic force holds atoms and moleculs together.
The weak force is a force between elementary particles certain processes that take place with low probability, as radio-active beta-decay and collisions between neutrinos and other particles.
The gravity force is the weakest of all 4 fundamental forces. It is the force of  attraction between all masses in the universe, especially the attraction of the earth's mass for bodies near its suface. Newton's law of gravity states that gravitational force between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them.
8 0
3 years ago
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
3 years ago
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