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3 years ago

These clouds appear long, flat, and spread across the sky.

2 answers:

8 0

Stratus –layered or sheet like, low (50 - 6500 feet) Cumulus – puffy cotton balls, vertical development Nimbus – dark and rainy Cirrus – feather-like, high (20 000 feet) Alto – medium altitude (6500 - 20 000 feet)

photoshop1234 [79]3 years ago

4 0

I think the answer is Cirrus. They are flat feathery clouds that usually stretch long distances over the sky.

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A rocket travels 1.3 km in 62 ms. What is its average speed in m⋅s−1? Do not give your answer in scientific notation. The answer

hjlf

Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

$Speed = \frac {distance}{time}$

$Speed = \frac {1300}{3720}$

Average speed = 0.35 m/s

7 0

4 years ago

A small, spring-loaded cannon launches a tennis ball from level ground with an initial speed vi at an angle θi with the horizont

lora16 [44]

Answer:

a. $T\ =\ \dfrac{2v_isin\theta_i}{g}$

b. $v_icos\theta_i$

c. $v_1sin\theta_i$

Explanation:

Given,

*initial velocity of the ball = $v_i$

*angle of projection = \theta_i

Horizontal component of the initial velocity of the ball = $v_x\ =\ v_icos\theta_i$

vertical initial component of the initial velocity of the ball = $v_y\ =\ v_isin\theta_i$

part a

From the kinematics,

In the y-direction motion,

total vertical displacement of the ball during the whole motion is zero.

Ball is moving under the gravitational acceleration, therefore the acceleration of the ball = -g, because gravitational acceleration always acts in the downward direction,

Let t be the time of flight of the whole motion,

$y\ =\ v_yt\ -\ \dfrac{1}{2}gt^2\\\Rightarrow 0\ =\ v_isin\theta_i t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow t\ =\ \dfrac{2v_isin\theta_i}{g}\\$

part b.

At the peak of the path of the ball, the vertical component of the velocity of the ball becomes zero, only horizontal component of the velocity acts on the ball is equal to = $v_x\ =\ v_icos\theta_i$

part c.

Initial vertical component of the velocity of the ball = $v_y\ =\ v_isin\theta_i$

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3 years ago

An eye is corrected by a -5.50 D lens, 2.0 cm from the eye. What is this eye's far point without glasses?

Paraphin [41]

Answer:

The relationship between diopter D and focal length f is D=1fD=1f So, f=−15.5=−0.182mf=−15.5=−0.182m

Explanation:

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densk [106]

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hram777 [196]

If the machine works at a 70% efficiency it must complete 14,000 work out of the 20,000

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