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Gre4nikov [31]
3 years ago
5

Help please? I dont understand what this means.

Physics
1 answer:
Alex Ar [27]3 years ago
5 0
The answer is c for real
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At present most of the world's energy needs are supplied by what kind of energy
antiseptic1488 [7]

Fossil fuels . . . coal, oil, natural gas

Among primitive cultures, wood is an important source.
4 0
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Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration
exis [7]

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it's acceleration is 0

Explanation:

since it is travelling at a constant speed it is not accelerating so its acceleration is 0

3 0
3 years ago
Use Snell's Law to solve the following:
azamat

Answer:

1.171

Explanation:

if n₁sinΘ₁=n₂sinΘ₂, then n₂=n₁sinΘ₁ / sinΘ₂;

n_2=\frac{1.5*sin45}{sin65}=\frac{1.5*0.707}{0.906} =1.1705

3 0
3 years ago
A 12.8-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 156N and break
Varvara68 [4.7K]

Answer:

2.4 m/s^2

Downward

Explanation:

We are given that

Mass of monkey=12.8 kg

Tension=156 N

We have to find the magnitude of the elevator's minimum acceleration.

T=m(g+a)

Where g=9.8 m/s^2

Substitute the values

156=12.8(a+9.8)

9.8+a=\frac{156}{12.8}=12.2

a=12.2-9.8=2.4 m/s^2

Hence, the acceleration =a=2.4 m/s^2

Direction of the elevator's minimum acceleration is downward because the elevator moves downwards.

8 0
3 years ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
vlada-n [284]

Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

3 0
3 years ago
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