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3 years ago

A technique in which the muscles are stretched by an outside force is called _____.

Physics

1 answer:

geniusboy [140]3 years ago

6 0

A technique in which the muscles are stretched by an outside force is called Passive Stretching

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Burning a canful of petrol or dropping it. which is likely to release most energy?

Neporo4naja [7]

Answer:

Burning of canful of petrol will release more energy

Explanation:

If you like my answer than please mark me brainliest

5 0

3 years ago

A dense metal sphere is dropped from a 10-meter tower, and at the exact same time an identical metal sphere is thrown horizontal

givi [52]

Answer:

A. Both spheres land at the same time.

Explanation:

The horizontal motion doesn't affect the vertical motion. Since the two spheres have the same initial vertical velocity and same initial height, they land at the same time.

7 0

3 years ago

Read 2 more answers

In the above lightwave the property labeled a determines which characteristic of visible light

kap26 [50]

The light must be either very dim or else non-existent.
We can't see the light wave or the label.

7 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

3 years ago

A driver drove for 3 hours at a certain speed. if he drove 12 more miles at the same speed, the distance would be 132 miles. at

Zepler [3.9K]

Answer:

40mph

Explanation:

1st leg DATA:

time = 3 hrs ; speed = r mph ; distance = 3r miles

------------

2nd leg DATA:

speed = r mph ; distance = 12 miles

--------------------------------

3r + 12 = 132

3r = 120

rate = 40 mph

4 0

3 years ago