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velikii [3]
3 years ago
5

Describe two sources of earth's energy that are not produced

Physics
1 answer:
Anvisha [2.4K]3 years ago
6 0
Here is the answer. Two sources of Earth's energy that are not produced would be Cosmic rays and Tidal Energy. Cosmic rays <span>are high-energy protons and atomic nuclei that come from outside the solar system. Whereas, tidal energy is the energy produced by both the moon (2/3) and the sun (1/3). Hope this answers your question.</span>
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A car started from a rest and accelerated at 9.54 m/s2 for 6.5 seconds. How much distance was covered by the car?
ElenaW [278]

Explanation:

S =ut + 1/2at^2

S = 0×6.5 + (1/2 × 9.54) × 6.5^2

S =0 + 4.77 ×42.25

S=201.5m

5 0
3 years ago
A farmer had 15 sheep, and all but 8 died. How many are left?
Troyanec [42]
So he has 7 sheepleft if i did it correctly
15-8=7
3 0
3 years ago
Read 2 more answers
Mohammad has been saving 8 each week. Today he spent 142 of the savings, and he now has 50 left. For how many weeks has he been
poizon [28]

Answer:

24 weeks

Explanation:

142+50=192

192/8=24

3 0
3 years ago
One atom has 14 neutrons. Another has 13 neutrons. Both have 13 protons. These atoms are...
agasfer [191]
Hello!

These atoms are isotopes

Isotopes
are atoms with the same number of protons (which determine the atomic number) and a different number of neutrons. These two atoms have the same number of protons (13), but a different number of neutrons, so they are isotopes. 

According to their number of protons, these are Aluminum isotopes: ²⁷Al (stable), and ²⁶Al (radioactive), respectively. 

Have a nice day!

7 0
3 years ago
Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector
garri49 [273]

Answer:

5.2\ \text{m/s}

70^{\circ} south of east

Explanation:

v_a = 3 m/s

\theta_a = 20^{\circ} north of east

v_b = 6 m/s

\theta_b = 40^{\circ} south of east = 360-40=320^{\circ} north of east

x and y component of v_a

v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

Magnitude

|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}

Direction

\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

6 0
3 years ago
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