The wavelengths for visible light rays correspond to which of these options?

The lower atmosphere is mostly warmed by radiated heat from Earth's surface. However, water heats up and cools down more slowly

JulsSmile [24]

The answer is B. On a sunny day, the air over a lake will be cooler than the air over the bordering land.

6 0

3 years ago

Read 2 more answers

11. You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the gla

Flauer [41]

Explanation:

it happens in the morning time

3 0

3 years ago

A smith needs to melt 0.0500 kg of gold at 21.0°C. How much heat must be added? (Remember, she has to heat it to the melting poi

baherus [9]

Answer:

9704.6 J

Explanation:

Total Thermal Energy

= Energy required to bring gold to melting point + Energy required to change the state of gold from solid to liquid

= mcT + mlf [bolded is energy used to bring gold to melting point, italicised is state change of gold from sold to liquid]

= (0.0500)(126)(1063 - 21) + (0.0500)(6.28 × 10^4)

= 9704.6 J

4 0

3 years ago

Please help me .finish this paper

Evgesh-ka [11]

Solution-1:-

$\boxed{\sf \dfrac{10\times 1000}{60\times 60}}$

Solution:-2

$\boxed{\sf Sodium\:and\:Potassium}$

Solution:-3

$\boxed{\sf 320m}$

Solution:-4

$\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}$

Solution:-5

$\boxed{\sf Mg_3N_2}$

Solution:-6

$\boxed{\sf Grapes\:and\:Rambutan}$

Solution:-7

$\boxed{\sf {}^{}_{}N}$

Solution:-8

$\boxed{\sf Galactuse}$

Solution:-9

$\boxed{\sf Y-X}$

Solution:-10

$\boxed{\sf Cell\:wall}$

7 0

3 years ago

At what height h above the ground does the projectile have a speed of 0.5v?

maw [93]

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

6 0

2 years ago