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kykrilka [37]
3 years ago
5

Would the phases of the moon be affected if the moon did not make one rotation for each revolution of earth. Explain .

Physics
1 answer:
elixir [45]3 years ago
3 0

Answer:

No. If the moon did not rotate at all we would still see the same visible "fraction" of the moon's surface.

At the full moon  phase we would still the entire surface of  1/2 of the moon regardless of which side of the moon we were observing.

Actually the moon does not make one rotation for each revolution of the earth, but rather, one rotation for each complete revolution of the moon around the the earth. That's why we always see the same side of the moon from earth. The back side of the moon is never visible from earth.

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At which boundary does subduction stop occurring, resulting in a mountain range?
NARA [144]
Subduction occurs at convergent tectonic plate boundaries, resulting in high rates of volcanism, earthquakes and mountain building.
6 0
3 years ago
Read 2 more answers
How to read a micrometer on a clark cm-100 vickers hardness tester
allsm [11]

Answer:

Explanation:

Equipment manufactured by LECO(8 Corporation, St. Joseph, Michigan is warranted free from defect in material

and workmanship for a period of six months from the date of purchase. Equipment not manufactured by LECO is

covered to the extent of warranty provided by the original manufacturer and this warranty does not cover any

equipment, new or used, purchased from anyone other than the LECO Corporation. All replacement parts shall

be covered under warranty for a period of thirt days from date of purchase. LECO MAKES NO OTHER

REPRESENTATION OR WARRANTY OF ANY OTHER KIND, EXPRESSED OR IMPLIED, WITH RESPECT TO

THE GOODS SOLD HEREUNDER, WHETHER AS TO MERCHANTABILITY, FITNESS FOR PURPOSE, OR

OTHERWISE.

Expendable items such as crucibles, combustion tubes, chemicals and items of like nature are not covered by

this warranty.

LECO's sole obligation under this warranty shall be to repair or replace any part or parts which, to our

satisfaction, prove to be defective upon return prepaid to LECO Corporation, St. Joseph, Michigan. This

obligation does not include labor to install replacement parts, nor does it cover any failure due to accident, abuse,

neglect, or use in disregard of instructions furnished by LECO. In no event shall damages for defective goods

exceed the purchase price of the goods, and LECO SHALL NOT BE LIABLE FOR INCIDENTAL OR

CONSEQUENTIAL DAMAGES WHATSOEVER.

All claims in regard to the parts or equipment must be made within ten (10) days after Purchaser learns of the

facts upon which the claim is based. Authorization must be obtained from LECO prior to returning any other

parts. This warranty is voided by failure to comply with these notice requirements.

NOTICE

The warranty on LECO equipment remains valid only when genuine LECO replacernent parts are employed.

Since LECO has no control over the quality or purity of consumable products not manufactured by LECO, the

specifications for accuracy of results using LECO instruments are not guaranteed unless genuine LECO

consumables are employed in conjunction with LECO instruments. If purchaser defaults in making payment for

any parts or equipment, this warranty shall be void and shall not apply to such parts and equipment. No late

payment or cure of default in payment shall extend the warranty period provided herein.

LECO Corporation is not responsible for damage to any associated instruments, equipment or apparatus nor wil

LECO be held liable for loss of profit or other special damages resulting from abuse, neglect, or use in disregard

of instructions. The Buyer, their employees, agents and successors in interest assume all risks and liabilities for

the operation, use and/or misuse of the product(s) described herein and agree to indemnify, hold harmless and

defend the seller from any and all claims and actions arising from any cause whatsoever, including seller's

negligence for personal injury incurred in connection with the use of said product(s) and any and all damages

proximately resulting therefrom.

CAUTION

The instrument should be operated only by technically qualified individuals who have fully read and understand

these instructions. The instrument should be operated only in accordance with these instructions.

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5 0
3 years ago
What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT
sergiy2304 [10]

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

8 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
A 5.10 kgkg watermelon is dropped from rest from the roof of a 18.5 mm-tall building and feels no appreciable air resistance.
VARVARA [1.3K]

Work done is by the change in the potential energy of the system. The work done by gravity is 924.63 J.

<h3>What is the Kinetic Energy?</h3>
  • Potential energy in physics is the energy that an item retains as a result of its position in relation to other objects, internal tensions, electric charge, or other elements.
  • The gravitational potential energy of an object, which is based on its mass and distance from another object's center of mass, the elastic potential energy of an extended spring, and the electric potential energy of an electric charge in an electric field are examples of common types of potential energy. The joule, denoted by the letter J, is the energy unit in the International System of Units (SI).

Solution:

mass = 5.10 kg

height = 18.5 mm

We know that work done by the gravity on the watermelon is the change in the potential energy of the watermelon, therefore,

Work done due to gravity = change in the potential energy of the system

W = \Delta PE

W = mg (h₀ - h₁)

W = 5.10 × 9.8 × 18.5

W = 924.63 J

know more about potential energy brainly.com/question/24284560

#SPJ4

7 0
2 years ago
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