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3 years ago

Which of the following provides evidence that geologic activity occurred in the past on Earth's Moon?

Physics

1 answer:

Bingel [31]3 years ago

6 0

Answer: d) the presence of solidified lava flows on the Moon

Explanation:

A geological activity means an occurrence of event such as volcanic eruption, earthquake, sedimentation, erosion etc. The revolution of the Moon around the Earth , the axial tilt of the Moon or the phases of the Moon are not surface features. hence, these events cannot provide the evidence of geological activity in the past of Moon.

The surface features of moon such as Mares, Craters, mountains, Rays and rills are the proof of some geological activity on the Moon. Mares are the dark patches on the moon's surface formed of solidified lava. Due to negligible atmosphere on the moon, the meteors strike its surface and cause craters to form. Thus, the correct answer is d.

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In an extrasolar planetary system containing a single planet, the parent star is measured to move about its center of mass every

Mademuasel [1]

Answer:

Orbital Time Period is 24 years

Explanation:

This can be explained by the definition of time period.

Time period can be defined as the time taken by an object to complete one cycle, here, time taken to complete one revolution.

Also, we know that an extra solar planet which is also called as an exo planet is that planet which is outside our solar system and orbits any star other than our sun. The system in consideration is extra solar system with a single planet.

Therefore, the time taken by the parent star to move about its mass center is the orbital time period that is 24 years.

4 0

3 years ago

Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone

anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0

3 years ago

The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer

zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

Bag filled with = 20 kg

Lifted acceleration = $5.0\ m/s^2$

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

$F = ma$

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in $m/s^2$

Now, we substitute m = 20 kg and a = $5.0\ m/s^2$ in Newton's second law,

$F = 20\times 5.0$

$F = 100\ m/s^2$

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

3 0

3 years ago

Given that 1 pound is equal to 4.45 newton’s what is the weight of a 500N child in pounds?

Bond [772]

Answer:

112.36 pounds

Explanation:

Since 1 pound = 4.45 Newtons, a 500N child in pounds = 500÷4.45 = 112.36 pounds (approximately).

4 0

3 years ago

A 45-mH inductor is connected in series with a 60-Ω resistor through a 15-V dc power supply and a switch. If the switch is close

egoroff_w [7]

Answer:

The current is 0.248 A

Explanation:

Given that,

Inductor $L= 45\times10^{3}\ H$

Resistance $R= 60\Omega$

Voltage = 15 volt

Time $t =7.0\times10^{-3}\ sec$

We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}(1-e^{\dfrac{-R}{L}}t)$

Where, V = voltage

R = resistance

L = inductance

T = time

Put the value into the formula

$I=\dfrac{15}{60}(1-e^{\dfrac{-60}{45\times10^{3}}}\times7\times10^{-3})$

$I=0.248\ A$

Hence, The current is 0.248 A.

4 0

3 years ago