Answer: the sun
Explanation:
The sun's radiant energy reaches the earth's surface either directly through radiation, indirectly through convection, or it can move "across" or "through" objects or materials on the surface via conduction. Let's look more closely at each case. We've probably experienced the feeling of "warmth" of the sun on our skin on a sunny day. Light energy from the sun is reaching us across space and down through the atmosphere through radiation. A dark colored vehicle in the sun quickly becomes warm (or hot!) to the touch because of radiation. The light energy from the sun heats the air in the earth's atmosphere, and this drives convection and transfers thermal energy around. It is possible that we've felt a "hot breeze" on our skin on sunny days. The thermal energy in the air will be carried to objects in its path, and it will warm them.
Answer:
Ben's average speed was twice Debby's average speed.
Explanation:
Ben covered a total distance of 16 miles (10+4+2) and Debby covered 8 miles (3+2+2+1) which is half of what Ben covered. As they both reached the place in the same amount of time it tells us Ben was faster.
A red apple absorbs all colors of visible light except red, so red light
is the only light left to bounce off of the apple toward our eyes.
(This is a big part of the reason that we call it a "red" apple.)
Here's how the various items on the list make out when they hit the apple:
<span>Red . . . . . reflected
Orange . . absorbed
Yellow . . . </span><span><span>absorbed
</span>Green . </span><span><span>. . absorbed
</span>Blue . . </span><span><span>. . absorbed
</span>Violet .</span><span> . . absorbed</span>
<span>Black . . . no light; not a color
White . . . has all colors in it</span>
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
