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Ray Of Light [21]
2 years ago
9

What living things began to increase in numbers following this mass extinction?

Physics
1 answer:
RoseWind [281]2 years ago
8 0

pigsExplanation: population

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A scientist is planning an experiment that will include the measurement of temperatures. She wishes to avoid using negative valu
cricket20 [7]

Answer:

C.

Explanation:

Fahrenheit scale is a temperature scale that measures the boiling point of water at 212 degrees F and the freezing point at 32 degrees F. The Fahrenheit scale was developed by the German scientist Daniel Gabriel Fahrenheit in 1724.

In the given scenario, the scientist, who wishes to measure the temperature of her experiment which will expectantly be colder than the point of water, should use the Fahrenheit scale.

So, the correct answer is option C.

8 0
2 years ago
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
3 years ago
The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of
Korvikt [17]

Answer:

The mass of the earth, M=6.023\times 10^{24}\ kg

Explanation:

It is given that,

Time taken by the moon to orbit the earth, T=27.3\ days=2358720\ m

Distance between moon and the earth,r=384000\ km=384\times 10^6\ m

We need to find the mass of the Earth using Kepler's third law of motion as :

T^2=\dfrac{4\pi^2}{GM}r^3

M=\dfrac{4\pi^2r^3}{T^2G}

M=\dfrac{4\pi^2\times (384\times 10^6)^3}{(2358720)^2\times 6.67\times 10^{-11}}

M=6.023\times 10^{24}\ kg

So, the mass of the earth is 6.023\times 10^{24}\ kg. Hence, this is the required solution.

7 0
3 years ago
Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil
tatyana61 [14]

Answer:

The time taken by missile's clock is 4.6\times 10^{6} s

Solution:

As per the question:

Speed of the missile, v_{m = 6.5\times 10^{3}} m/s

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}         (1)

Using binomial theorem in the above eqn:

We know that:

(1 + x)^{a} = 1 + ax

Thus eqn (1) becomes:

1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}

T = \frac{2c^{2}}{v_{m}^{2}}

Now, putting appropriate values in the above eqn:

T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}

T = 4.6\times 10^{6} s

4 0
3 years ago
A rock is rolling down a hill, and it’s halfway down. Would it have both Kintectic engery and Potenial engery? PLEASE HELP
dimaraw [331]

Answer:

Only kinetic.

Explanation:

Potential energy means it has the potential to move. Not something already in motion.

6 0
3 years ago
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