Ask question

Ask question

All categories

3 years ago

12

https://brainly.com/question/add?entry=1642&task_content=Tania+was+asked+to+describe+the+relashio+of+electrity+and+magnetism

+as+it+relates+to+earth+atmosphere+

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

wat

Explanation:

You might be interested in

Small fragments of orbiting bodies that have fallen on Earth's surface are known as

kolezko [41]

There are known as B. meteorites: asteroids are much larger.

6 0

3 years ago

Fused quartz possesses an exceptionally low coefficient of linear expansion, 5.50 × 10 − 7 ( ∘ C ) − 1 . Suppose a bar of fused

kozerog [31]

To solve this problem we will use the concepts related to thermal expansion in a body for which the initial length, the coefficient of thermal expansion and the temperature change are related:

$\Delta L = L0\alpha\Delta T$

Where,

$\Delta L$ = Change in Length

$\alpha$ = Coefficient of linear expansion

$\Delta T$ = Change in temperature

$L_0$ = Initial Length

Our values are:

$L_0 = 3.45m$

$\alpha = 5.5*10^{-7} \°C^{-1}$

$\Delta T = 235-20 = 215\°C$

Replacing we have,

$\Delta L = (3.49) (5.5*10^{-7}) [(215)$

$\Delta L = 0.0004126m$

$\Delta L = 0.4126mm$

Therefore the change in milimiters was 0.4126mm

5 0

3 years ago

Which of the following prefixes represents the biggest number?

Komok [63]

<span>kilo
hecto
deco
deci
centi
milli

These are the orders from largest to smallest.

Answer: Kilo represents the biggest number. </span>

3 0

4 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

Ne4ueva [31]

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

t = v / a

t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

t = v / g

t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0

3 years ago

A river flows due south with a speed of 2.0 m/s .You steer a motorboat across the river; your velocity relative to the water is

mihalych1998 [28]

Answer:

a) $v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

b) $\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

c) $t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

d) $Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

Explanation:

Part a

For this case the figure attached shows the illustration for the problem.

We know that $v_y = 2 m/s$ represent the velocity of the river to the south.

We have the velocity of the motorboard relative to the water and on this case is $V_x= 4.8 m/s$

And we want to find the velocity of the motord board relative to the Earth $v_m$

And we can find this velocity from the Pythagorean Theorem.

$v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

Part b

We can find the direction with the following formula:

$\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

Part c

For this case we can use the following definition

$D = Vt$

The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:

$t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

Part d

For this case we can use the same definition but now using the y compnent we have:

$Y = v_y t$

And replacing we got:

$Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

7 0

3 years ago