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jonny [76]
4 years ago
14

Which statements about cirrus clouds are true? Choose all answers that are correct. A. Cirrus clouds form high in the sky. B. Ci

rrus clouds usually mean fair weather. C. Cirrus clouds near the ground are called fog. D. Cirrus clouds are ice crystals that look like wispy streamers. please dont report
Physics
2 answers:
devlian [24]4 years ago
6 0
If i am correct the answer should be C
Vesna [10]4 years ago
5 0
A is the answr hope i helped
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Your science teacher gives you three liquids to pour into a jar. After pouring, the liquids layer as seen here. What property of
Blizzard [7]
A. mass
b. color
c. density
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i believe A
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3 years ago
Read 2 more answers
Please do it to me thank alot
faltersainse [42]

Answer:

See explanations below

Explanation:

For resistors in series, the will be added for those in parallel, we will take the sum of their reciprocal

A) 3 2ohms resistors in series

Equivalent resistance = 2 + 2 + 2

Equivalent resistance = 6ohms

B) 2 4 ohms resisstors in parallal

1/R = 1/4 + 1/4

1/R = 2/4

R = 4/2

R = 4ohms

Equivalent resistance is 2ohms

C) 4 ohms with in series with 8ohms = 4 + 8

Equivalent resistance = 4 + 8

Equivalent resistance = 12ohms

D) 4 ohms and 2 ohms in parallel is expressed as;

1/R = 1/4 + 1/2

1/R = 1+2/4

1/R = 3/4

R = 4/3

4/3 ohms in series with 6ohms

Equivalent resistance = 4/3 + 6

Equivalent resistance =4+18/3

Equivalent resistance = 22/3 ohms

8 0
3 years ago
What is a transverse wave? How do the particles in the medium move in relation to the energy of the wave? Does this wave require
Pani-rosa [81]
A transverse wave is a wave where the particles in the medium move perpendicular (at right angles) to the direction of the source or its propagation (think of a snake slithering through grass) an example of a transverse wave could be a light wave. Light waves for instance don’t need a medium in order to propagate but transverse waves in general do need a medium.
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3 years ago
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An electric vehicle starts from rest and accelerates at a rate a1 in a straight line until it reaches a speed of v. The vehicle
levacccp [35]

(a) t=\frac{v}{a_1}+\frac{v}{a_2}

In the first part of the motion, the car accelerates at rate a_1, so the final velocity after a time t is:

v = u +a_1t

Since it starts from rest,

u = 0

So the previous equation is

v= a_1 t

So the time taken for this part of the motion is

t_1=\frac{v}{a_1} (1)

In the second part of the motion, the car decelerates at rate a_2, until it reaches a final velocity of v2 = 0. The equation for the velocity is now

v_2 = v - a_2 t

where v is the final velocity of the first part of the motion.

Re-arranging the equation,

t_2=\frac{v}{a_2} (2)

So the total time taken for the trip is

t=\frac{v}{a_1}+\frac{v}{a_2}

(b) d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}

In the first part of the motion, the distance travelled by the car is

d_1 = u t_1 + \frac{1}{2}a_1 t_1^2

Substituting u = 0 and t_1=\frac{v}{a_1} (1), we find

d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}

In the second part of the motion, the distance travelled is

d_2 = v t_2 - \frac{1}{2}a_2 t_2^2

Substituting t_2=\frac{v}{a_2} (2), we find

d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}

So the total distance travelled is

d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}

7 0
3 years ago
PLEASE HELP NEED ANSWERS NOW! Time-Distance Graph
Anastaziya [24]

Answer: I added a picture of the answer

Explanation: its right

4 0
4 years ago
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