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RideAnS [48]
3 years ago
13

Muscles use force to move objects. When using a ramp to do work, would your muscles use more force or less force?

Physics
2 answers:
White raven [17]3 years ago
7 0
Less because the ramp is letting off force but i does depend on the way you are going on the ramp

Mademuasel [1]3 years ago
4 0
If it's up then more if it's down then less
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Answer: A crest is the highest point the medium rises to and a trough is the lowest point the medium sinks to. It is also a point on the wave where the displacement of the medium is at a maximum.

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The major contribution of the French philosopher Rene Descartes was in the realm of
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You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley
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An evacuated long tube contains a coin and a feather. If both objects fall together starting from the top of the tube, it is exp
Natasha_Volkova [10]

Answer:

gexp = 3.65 m/s²

Explanation:

The value of acceleration due to gravity changes with the altitude. The following formula gives the value of acceleration due to gravity at some altitude from the sea level:

gexp = g(1 - 2h/Re)

where,

gexp = expected value of g at altitude = ?

g = acceleration due to gravity at sea level = 9.8 m/s²

h = altitude = 2000 km = 2 x 10⁶ m

Re = Radius of Earth = 6.37 x 10⁶ m

Therefore,

gexp = (9.8 m/s²)(1 - 2*2 x 10⁶ m/6.37 x 10⁶ m)

<u>gexp = 3.65 m/s²</u>

5 0
3 years ago
A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge
Kisachek [45]

Answer:

\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r

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\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

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