HELPPPPP

What does a black asphalt road become hotter than a white cement sidewalk in the same amount of sunlight?

victus00 [196]

The black means that it is a great emitter/absorber of the electromagnetic spectrum. The electromagnetic radiation is reflected of the white and absorbed nurture black meaning that the temperature of the black tarmac increases to that greater the the white

7 0

3 years ago

Please someone help!!!

antoniya [11.8K]

I know it’s the Coulomb’s law and that I’m pretty sure the answer would be C.Inverse Square.

4 0

3 years ago

Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to

katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

$OB=0.40 km$

And the Max distance towards bookstore is,

$OA=3.65 km$

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

$AB=\sqrt{OB^{2}+OA^{2}}$

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

$AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km$

Therefore the distance between there destination is 3.67 km.

6 0

3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :