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Elis [28]
2 years ago
8

How much force would be required for a ten strand pulley system to lift 1000 newtons?

Physics
1 answer:
Pepsi [2]2 years ago
8 0

Answer: 250

Explanation:

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A wave with a height of 6 would have greater amplitude than a wave with a height of 5 true or false? HELP
BabaBlast [244]

Answer:

it could be either or because it doesnt just depend on the height but it also depends on the pressure but then again the question didnt ask anything about the pressure so the answer should be true

Explanation:

3 0
2 years ago
An ice skater weighs 500 [N]. He is coasting to the right at a constant velocity of 2 [m/s]. Assume
luda_lava [24]

Answer:

The net force on the skater is zero. (F_{net} = 0\,N)

Explanation:

According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. (F_{net} = 0\,N)

4 0
3 years ago
What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 11.0 m and the downward accel
8090 [49]
<span>When an object moves in a circle, the acceleration points toward the center of the circle. This acceleration is called centripetal acceleration. We can use a simple equation to find centripetal acceleration. a = v^2 / r We can use this same equation to find the speed of the car. v^2 = a * r v = sqrt { a * r } v = sqrt{ (1.50)(9.80 m/s^2)(11.0 m) } v = 12.7 m/s The speed of the roller coaster is 12.7 m/s</span>
3 0
3 years ago
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc
sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

\overrightarrow{v}_{s,w}=6.5 \widehat{j}

velocity of water with respect to earth = 1.5 m/s at 40° north of east

\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}

\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j}  \right )

\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}

The magnitude of the velocity of ship relative to earth is \sqrt{1.15^{2}+5.54^{2}} = 5.66 m/s

5 0
3 years ago
20 POINTS!
CaHeK987 [17]
Potential energy is the answer
6 0
3 years ago
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