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netineya [11]
4 years ago
15

High-level clouds usually occur above 20,000 ft. Which is the best description of the clouds’ appearance?

Physics
1 answer:
PSYCHO15rus [73]4 years ago
8 0
D. Thin and composed of Ice crystals
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B. Study the given figure and calculate the area of the leaf
yanalaym [24]

Answer:

19 cm sq

Explanation:

you have to count one full and half area occupied by leaf in each box, then you have to multiply with 1 cm sq.

4 0
3 years ago
Which describes how chemical changes are different from physical changes?
lions [1.4K]

Answer:

B. Chemical changes involve the formation of a new substance.

Explanation:

They can both release energy, both can be measured, and chemical changes can be caused by oxygen, so I believe it would be B.

6 0
4 years ago
Identify the motion when a car is moving with constant speed along a curve
zavuch27 [327]

Answer:

Uniform circular motion

Explanation:

When a car is moving with constant speed along a curve, it will said to have uniform circular motion. The speed of the car remains constant in this case. But its velocity keeps on changing. The velocity can be calculated by drawing the tangent on the circle. Hence, the motion of the car is uniform circular motion.

5 0
3 years ago
Read 2 more answers
What do the arrows in this photograph represent
babunello [35]

Answer: D. An action-reaction force pair

Explanation: When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body. There are two forces resulting from this interaction - a force on the chair and a force on your body. Another example would be a person pushing against a wall (action force), and the wall exerts an equal and opposite force against the person.

7 0
3 years ago
A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.
umka21 [38]

Explanation:

Given that,

The disintegration constant of the nuclide, \lambda=0.0178\ h^{-1}

(a) The half life of this nuclide is given by :

t_{1/2}=\dfrac{ln(2)}{\lambda}

t_{1/2}=\dfrac{ln(2)}{0.0178}

t_{1/2}=38.94\ h

(b) The decay equation of any radioactive nuclide is given by :

N=N_oe^{-\lambda t}

\dfrac{N}{N_o}=e^{-\lambda t}

Number of remaining sample in 4.44 half lives is :

t_{1/2}=4.44\times 38.94

t_{1/2}=172.89\ h^{-1}

So, \dfrac{N}{N_o}=e^{-0.0178\times 172.89}

\dfrac{N}{N_o}=0.046

(c) Number of remaining sample in 14.6 days is :

t_{1/2}=14.6\times 24

t_{1/2}=350.4\ h^{-1}

So, \dfrac{N}{N_o}=e^{-0.0178\times 350.4}

\dfrac{N}{N_o}=0.0019

Hence, this is the required solution.

4 0
4 years ago
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