Answer:
At 6 minutes, the soil temperature at a 45° angle of insolation is higher than at 0°.
At 15 minutes, the soil temperature at a 90° angle of insolation is higher than at 45°
Explanation:
First, foremost, and most critically, you must look at the graph, and critically
examine its behavior from just before until just after the 5-seconds point.
Without that ability ... since the graph is nowhere to be found ... I am hardly
in a position to assist you in the process.
Answer:
first object final velocity =2m/s
second = 12m/s
Explanation:
i hope this will help you,..,
The height at time t is given by
h(t) = -4.91t² + 34.3t + 1
When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s
Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.
The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
= 60.903 m
Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
The ball attains a maximum height of 60.9 m (nearest tenth)
Answer:
Explanation:
Given a parallel plate capacitor of
Area=A
Distance apart =d
Potential difference, =V
If the distance is reduce to d/2
What is p.d
We know that
Q=CV
Then,
V=Q/C
Then this shows that the voltage is inversely proportional to the capacitance
Therefore,
V∝1/C
So, VC=K
Now, the capacitance of a parallel plate capacitor is given as
C= εA/d
When the distance apart is d
Then,
C1=εA/d
When the distance is half d/2
C2= εA/(d/2)
C2= 2εA/d
Then, applying
VC=K
V1 is voltage of the full capacitor V1=V
V2 is the required voltage let say V'
Then,
V1C1=V2C2
V × εA/d=V' × 2εA/d
VεA/d = 2V'εA/d
Then the εA/d cancels on both sides and remains
V=2V'
Then, V'=V/2
The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2
