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Mkey [24]
3 years ago
14

Define Mechanical advantage

Physics
1 answer:
Kisachek [45]3 years ago
5 0

Answer:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

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With the help of diagram explain reflection from smooth and rough surface​
aleksley [76]

For a smooth surface, reflected light rays travel in the same direction. This is called specular reflection. For a rough surface, reflected light rays scatter in all directions. This is called diffuse reflection. Diffuse reflection is when light hits an object and reflects in lots of different directions.

8 0
3 years ago
A car accelerates uniformly from rest to a speed of 23.7m/s in 6.5 seconds. Find the distance the
frez [133]

The car's average speed during that time is (23.7/2) = 11.85 m/s .

     Distance = (11.85 m/s) x (6.5 sec) = 77.025 meters .

5 0
4 years ago
you push a 51 kg box with a force of 485 N. the friction force on the box is 232 N. calculate the acceleration of the crate.
vichka [17]

Answer:

a = 4.96 m/s²

Explanation:

Given,

The mass of the box, m = 51 Kg

The magnitude of the applied force, Fₐ = 485 N

The friction force on the box, Fₓ = 232 N

The net force acting on the box is,

                                 F = Fₐ - Fₓ

Substituting the given values in the above equation

                                  F = 485 - 232

                                    = 253 N

The acceleration of the crate is given by

                                   a = F/m

                                      = 253 / 51

                                      = 4.96 m/s²

Hence, the acceleration of the crate is, a = 4.96 m/s²

3 0
3 years ago
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is
padilas [110]

Answer:

P_{C} = 3.2\, atm

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T

Bulb B (3 L, 4 atm) - Before opening:

P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T

Bulbs A & B (5 L) - After opening:

P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T

After some algebraic manipulation, a formula for final pressure is derived:

P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}

And final pressure is obtained:

P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}

P_{C} = 3.2\, atm

5 0
3 years ago
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Gnesinka [82]

Answer:

Q1 = 7.25*10^(-16) C 

Explanation:

We are given;

electric field strength = (1 x 10^5 N/C

drag force (F) = 7.25 x 10^(-11) N

The question says it's moving with constant velocity. This means that he particle is in equilibrium and not accelerating.

Columbs law force of attraction or repulsion between two charges is given as;

F=(KQ1Q2)/r²

Now, electric field strength is given as the formula;(K*Q2)/r², thus plugging the relevant values gives us;

7.25 x 10^(-11) N= (1 x 10^(5) N/C)Q1 Q1 = 7.25 x 10^(-11) /(1 x 10^(5))

Q1 = 7.25*10^(-16) C 

5 0
3 years ago
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