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Sever21 [200]
2 years ago
7

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

ry itself just below the surface.
Physics
1 answer:
MArishka [77]2 years ago
6 0

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

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A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36
AlexFokin [52]

Answer:

0.60 m/s

Explanation:

The average velocity from t = a to t = b is:

v_avg = (x(b) − x(a)) / (b − a)

Given that x(t) = 0.36t² − 1.20t, and the time is from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed is 0.60 m/s.

5 0
2 years ago
The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p
zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

Explanation:

Given that,

Average pressure P=1.01\times10^{5}\ Pa

Radius of earth R_{E}=6.38\times10^{6}\ m

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

P=\dfrac{F}{A}  

F=PA

F=P\times 4\pi\times R_{E}^2

Where, P = pressure

A = area

Put the value into the formula

F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2

F=516.6\times10^{17}\ N

Hence, The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

8 0
2 years ago
Read 2 more answers
A microwave oven operates at 2.60 ghz . what is the wavelength of the radiation produced by this appliance?
cestrela7 [59]
<span>10.3 cm The wavelength will be the distance that light travels in 1 second divided by the frequency of the radiation. Since the over operates at 2.60 ghz, the frequence is 2.6 billion times per second, or 2.60 x 10^9. The speed of light is defined as 299792458 m/s exactly. So 299792458 m/s / 2.60 x 10^9 1/s = 0.10337671 m = 10.337671 cm Since we only have 3 significant digits, the answer rounds to 10.3 cm</span>
8 0
2 years ago
Two sources emit beams of light of wavelength 550 nm. The light from source A has an intensity of 10 μW/m2, and the light from s
alexira [117]

Answer:

<em>A) Beam B carries twice as many photons per second as beam A.</em>

Explanation:

If we have two waves with the same wavelength, then their intensity is proportional to their power, or the energy per unit time.

We also know that the amount of photon present in an electromagnetic beam is proportional to the energy of the beam, hence the amount of beam per second is proportional to the power.

With these two facts, we can say that the intensity is a measure of the amount of photon per second in an electromagnetic beam. So we can say that <em>beam B carries twice as more power than beam A, or Beam B carries twice as many photons per second as beam A.</em>

3 0
2 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
2 years ago
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