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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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What uhhhhhhhhhhhhhhhhhhhhh

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3 years ago
outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp
sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s

So, the speed of the box is 0.489 m/s.

3 0
3 years ago
Convert 8.1 kilograms to grams
kirill115 [55]

Answer:

8100 g

Explanation:

8.1 kg × 1000

= 8100 g

6 0
3 years ago
Read 2 more answers
A charged particle is placed in an external magnetic field and it is moving in a circular path of radius 26m.The magnetic force
ElenaW [278]

Answer:

208 Joules

Explanation:

The radius of the circular path the charge moves, r = 26 m

The magnetic force acting on the charge particle, F = 16 N

Centripetal force, F_c = m·v²/r

Kinetic energy, K.E. = (1/2)·m·v²

Where;

m = The mass of the charged particle

v = The velocity of the charged particle

r = The radius of the path of the charged particle

Whereby the magnetic force acting on the charge particle = The centripetal force, we have;

F = F_c = m·v²/r = 16 N

(1/2) × r × F_c = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.

∴ (1/2) × r × F_c = (1/2) × 26 m × 16 N =  = (1/2)·m·v² = K.E.

∴ 208 Joules = K.E.

The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.

4 0
3 years ago
A solenoid that is 78.8 cm long has a cross-sectional area of 15.9 cm2. There are 914 turns of wire carrying a current of 8.25 A
Harlamova29_29 [7]

Answer:

(a) Energy density will be equal to 57.31J/m^3

(b) Total energy will be equal to 0.0718 J

Explanation:

It is given that length of solenoid l = 78.8 cm = 0.788 m

Cross sectional area A=15.9cm^2=15.9\times 10^{-4}m^2

Number of turns of the wire N = 914

Current in the solenoid i = 8.25 A

Inductance of the wire is equal to L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H

(b) Total energy stored in magnetic field U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J

(a) Energy density will be equal to

U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3

7 0
3 years ago
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