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serious [3.7K]
2 years ago
6

how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum

on opposite side of the load? calculate mechanical advantage and velocity ratio of the lever​
Physics
1 answer:
dangina [55]2 years ago
6 0

Answer:

206.8965517 n

Explanation:

First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.

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What is the definition of input force?
andrezito [222]

Answer:

The input force (effort) is the amount of effort used to push down on a rod, or pull on a rope in order to move the weight. In this example, the force the little guy is using to pull the elephant is the input force.

Explanation:

5 0
2 years ago
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To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
2 years ago
A proton is moved from a position where the electric potential is 125 V to a position where the electric potential is 275 V. The
kakasveta [241]
We determine the electric potential energy of the proton by multiplying the net electric potential to the charge of the proton. The net electric potential is the difference of the final state to the that of the initial state. So, it would be 275 - 125 = 150 V.

electric potential energy = 150 (<span>1.602 × 10-19) = 2.4x10^-17 J</span>
7 0
3 years ago
[10 POINTS ❗️❗️ and brainlist :)]
ludmilkaskok [199]
The answer to this is D
6 0
2 years ago
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A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3
Inga [223]
m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.

The momentul of the system preserves:

m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with E.

Now, we write the energy conservation law:

\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E

From the above equation, you find E,  and then conclude that the sound energy can certainly not be greater than this.
8 0
2 years ago
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