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harkovskaia [24]
4 years ago
15

WILL MARK BRAINLIEST!! PLEASE HURRY

Physics
1 answer:
Evgen [1.6K]4 years ago
3 0
The correct answer is Compressions, Hope this helps!
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There are 1,660 megawatts of wind-generated electricity produced globally every year. This amount is equivalent to A. 1,660,000
disa [49]
Let us first write down the known things .

1660 megawatts = 1660 X 10^6 watt
                           = 166000 kilowatt

From the above deduction we can conclude that the correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that you were looking for and the answer has actually come to your desired help.
5 0
3 years ago
Read 2 more answers
If a galaxy has an apparent radial velocity of 2000 km/s and the Hubble constant is 70 km/s/Mpc, how far away is the galaxy
ddd [48]

Answer:

28.57 Mpc

Explanation:

This question is going to be solved by applying Hubble's Law.

This Hubble's Law is actually an observation in physical cosmology. This observation makes it clear that galaxies are moving away from the Earth, and are doing so at speeds proportional to their distance. This essentially means that the farther they are from the Earth, the faster they are moving away from Earth.

It is represented by this formula

v = H(0)D, where

v = speed

H(0) = Constant of proportionality, or otherwise, Hubble's constant.

D = Distance to a galaxy

Applying the given parameters to the formula, we have

v = H(0).D

D = v / H(0)

D = 2000 / 70

D = 28.57 Mpc

3 0
3 years ago
A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change
lord [1]

The mass that must be added is 0.628 kg

Explanation:

The period of a mass-spring system is given by

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

For the initial mass-spring system in the problem, we have

m = 0.500 kg

T = 1.36 s

Solving for k, we find the spring constant:

k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m

In the second part, we want the period of the same system to be

T = 2.04 s

Therefore, the mass on the spring in this case must be

m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg

Therefore, the mass that must be added is

\Delta m = 1.128 - 0.500 = 0.628 kg

Learn more about period:

brainly.com/question/5438962

#LearnwithBrainly

5 0
4 years ago
A sailboat is traveling across the ocean with a speed of 4 m/s when large gusts of wind pick up that starts to accelerate the sa
Ivahew [28]

solution:

Using Cartesian co-ordinate system

Final velocity =v= -4m/s

Initial velocity = u

Acceleration a = m/s²

Time (t).= 60s

By the first kinematical equation

V= u +at

U = v – at

=(-4)-(-3)(60)

176m/s

So, initial velocity was 176m/s


8 0
4 years ago
What is net force?
kramer

Answer:

A. The sum of all the forces acting on an object.

3 0
3 years ago
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