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Ann [662]
4 years ago
11

When the fulcrum is shifted to the right, which side has a mechanical advantage and which side has a mechanical disadvantage?

Physics
1 answer:
o-na [289]4 years ago
8 0

The lever is a movable bar that pivots on a fulcrum attached to or positioned on or across a fixed point. The lever operates by applying forces at different distances from the fulcrum, or pivot.

A lever amplifies an input force to provide a greater output force, which is said to provide leverage. The ratio of the output force to the input force is the mechanical advantage of the lever.

The mechanical advantage of the lever is increased by moving the fulcrum closer to the load i.e., by shifting the fulcrum towards the right. Moving the fulcrum away from the load has mechanical disadvantage as the distance between the load and the fulcrum increases, the input force becomes greater than the output force.

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An echo is sound that returns to you after being reflected from a distant surface (e.g., the side of a cliff). Assuming that the
hoa [83]

Answer:

The correct answer is B

An echo underwater and an echo in the air will return at different times. The echo underwater will return more quickly than the echo in the air.

Explanation:

The physics of this is simple.

Water and air are both made up of particles. The particles for water are more closely or densely arranged that those of the air molecules. Hence sound travels faster in water than in air. When measured, the speed actually differs by as much as 5 times with water being the fastest medium.

Think of it like this. Assume you have two stacks of dominoes, one closely packed than the other but exactly the same amount of dominos, you'd notice that the stack that is more tightly arranged will be the first to topple over because it takes less time for the kinetic energy from the first domino to reach the next and on and on like that until the last domino.

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4 0
3 years ago
Read 2 more answers
Can someone help me with 45a-g and 46?
LUCKY_DIMON [66]

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7 0
3 years ago
The tip of a triangle is held 12.0 cm above the surface of a flat pool of water. A submerged swimmer in the pool sees the tip of
Rufina [12.5K]

Answer: 9cm

Explanation:

Refractive index can also be defined as the ratio of the real depth to the apparent depth.

Given that the

Real depth = 12 m

Refractive index of water = 1.33

Refractive index of air = 1.00

nair/nwater = real depth/apparent depth

Substitute all the parameters into the formula

1.33/1 = 12/ apparent depth

Cross multiply

1.33 Apparent depth = 12

Apparent depth = 12/1.33

Apparent depth = 9.02 cm

Therefore,  A submerged swimmer in the pool sees the tip of the triangle at 9cm approximately distance above the water.

8 0
3 years ago
A 12.6-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial dis
Zinaida [17]

Answer:

a)27.3N

b)150.78N

Explanation:

Having in mind the conservation of energy, as the monkey goes up (gaining potential gravitational energy) the kinetic energy must be reduced, so reducing the velocity of the monkey. So the maximum velocity will be at this lower point with a velocity of 1.36m/s

From this velocity and the radius we can calculate the angular velocity for the monkey center of mass:

w=\frac{1.36m/s}{0.853m} =1.594 /s

with this we can calculate the centripetal force magnitude:

F=m*w^{2} *r=12.6kg*(1.594/s)^{2} *0.853m=27.3N

On the mokey center of mass we have two opposite forces acting, the tension of the arm and the weight, in order for the monkey to continue swinging the resolt of this two forces must be equal to the centripetal force:

T-P=F=>T=F+P

T=27.3N+12.6kg*9.8m/s^{2}=150.78N

8 0
3 years ago
b. A force 100N makes an angle of Ø with the x axis and has a y component of 30 N. Find both the x and y component of the force
horsena [70]

Answer:

Explanation:

The y component of the force is 100 sinØ . But given that y component is 30N

so 100 sinØ = 30

sinØ = 0.3

Ø = 17.5°.

X component of force = 100 cosØ

= 100 cos17.5

= 95.35 N .

Y component of force = 30 N .

Angle Ø = 17.5°.

6 0
3 years ago
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