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slega [8]
3 years ago
14

The sun is directly overhead at the equator on what day of the year

Physics
2 answers:
Aleks [24]3 years ago
8 0

Equinox...

The Sun is directly overhead at "high-noon" on the equator twice per year, at the two equinoxes. Spring (or Vernal) Equinox is usually March 20, and Fall (or Autumnal) equinox is usually September 22.

Phoenix [80]3 years ago
7 0
Both equinoxes ... September 21 and March 21.
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How many Joules of potential
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Answer:

40 j, 80j.

Explanation:

P.E= mgh. G=10 m/s².

For 4m, P.E=1*10*4=40 joules.

For 8m, P.E=1*10*8=80 joules.

4 0
3 years ago
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.
saveliy_v [14]

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

6 0
3 years ago
Two samples of dirt are collected from a suspect's tread in his shoe and a crime scene. She notes very similar characteristics.
elena-14-01-66 [18.8K]

brainly.com/question/11542618?answering=true&answeringSource=greatJob%2FquestionPage

8 0
3 years ago
The generation of a magnetic field by an electric current is___ .
Alex787 [66]

Answer:

Electromagnetic induction

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The process of generating electric current with a magnetic field. It occurs whenever a magnetic field and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic field.

5 0
2 years ago
The driver of a car slams on the brakes, causing the car to slow down at a rate of
sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 years ago
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