The answer is "B" - If there are no windows then there will be no light coming in, and therefore you don't have to worry about what time of day you do the experiment at.
Answer:
Animals take in oxygen and give off carbon dioxide.
Explanation:
This is because plants take in our carbon dioxide and give off the oxygen that we as people and animals need to breathe.
Answer:
a)
= 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo = 
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s
Answer:
2.5 m/s²
Explanation:
Given,
Initial speed ( u ) = 10 m/s
Final speed ( v ) = 20 m/s
Time ( t ) = 4 seconds
To find : Acceleration ( a ) = ?
Formula : -
a = ( v - u ) / t
a = ( 20 - 10 ) / 4
= 10 / 4
= 5 / 2
a = 2.5 m/s²
Therefore,
The acceleration of the scooter is 2.5 m/s²
The current is defined as the amount of charge transferred through a certain point in a certain time interval:

where
I is the current
Q is the charge

is the time interval
For the lightning bolt in our problem, Q=6.0 C and

, so the average current during the event is