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2 years ago

What is the IMA of the following pulley system?

Physics

2 answers:

Dmitry [639]2 years ago

6 0

Answer : The IMA of the given pulley system is $\dfrac{F_r}{F_e}$ .

Explanation :

The full form of IMA is an ideal mechanical advantage. For the pulley system, it is defined as the ratio of the output force (resistance force) and the input force (effort force).

In this case, the input force is $F_e$

and the output force is $F_r$

Mathematically, the IMA for the pulley system it can be written as:

$IMA=\dfrac{F_r}{F_e}$

Hence, this is the required solution.

Furkat [3]2 years ago

5 0

Answer: $IMA= \frac {F_r}{F_e}$

Explanation:

IMA stands for Ideal Mechanical advantage.

The IMA of pulley system can be defined as the ratio of output force to input force.

From the given pulley system,

The input force = $F_e$

The output force = $F_r$

Hence, IMA of the given pulley system,

$IMA= \frac {F_r}{F_e}$

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Alik [6]

Answer:

Resultant = 3.05N

Explanation:

$r = \sqrt{{5}^{2} + 5^{2} - 2(5)(5) \cos(120) }$

$r = \sqrt{25 + 25 - 50(0.8142)}$

$r = \sqrt{50 - 40.71} = \sqrt{9.29}$

$r = 3.05$

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2 years ago

Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a

Novosadov [1.4K]

Answer:

A) correct answer is C, B) correct answer is b and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

1 / f = 1 / p + 1 / q

where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

1 / q = 1 / f 1 / p

let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

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2 years ago

Sonja [21]

Answer

2) 1.5×10-2 m

Explanation

The potential difference is related to the electric field by:

$\Delta V=Ed$ (1)

where

$\Delta V$ is the potential difference

E is the electric field

d is the distance

We want to know the distance the detectors have to be placed in order to achieve an electric field of

$E=1 V/cm=100 V/m$

when connected to a battery with potential difference

$\Delta V=1.5 V$

Solving the equation (1) for d, we find

$d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m$

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2 years ago

Micah knows that a car had a change in velocity of 15 m/s. What does micah need to determine acceleration.

katrin2010 [14]

The time component is needed. The acceleration is the change of velocity divided by the time in when this change of velocity happens.

4 0

2 years ago

Read 2 more answers

A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they

Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁ + m₂u₂

P₁ = (1000 x 25) + (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(1000 x 25) + (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

= 27.727(1000 + 1200)

= 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0

2 years ago