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Paladinen [302]
3 years ago
12

According to the _______ the amount of energy in the universe doesn't change.

Physics
2 answers:
balandron [24]3 years ago
6 0
The answer is B, Law of Kinetic Energy
Gnesinka [82]3 years ago
6 0
C.) the law of conservation of energy.<span>
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Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit
cestrela7 [59]

Answer:

v=1667.9km/h

a_{cp}=436.6km/h^2

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth C=2\pi R, where R=6371km is the radius of the Earth.

We have then:

v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h

And then we use the centripetal acceleration formula:

a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2

6 0
3 years ago
An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution
givi [52]
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π 

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
   = (15π rad/s)*(1 rev/2π rad)
   = 7.5 rev/s

Answer: 7.5  revolutions per second.

7 0
3 years ago
a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

#SPJ4

4 0
2 years ago
Which of the following is strenuous?
Arada [10]

Answer:

I think it c

Explanation:

7 0
3 years ago
Read 2 more answers
If the light wave has a wavelength of 10m what would be its velocity
s2008m [1.1K]

If this case could ever happen, the speed would follow from this formula:

v = f \cdot \lambda

with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."

v = 600THz\cdot 10m = 6\cdot 10^{14} \frac{1}{s}\cdot 10 m = 6\cdot10^{15}\frac{m}{s}

The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)

7 0
3 years ago
Read 2 more answers
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