Explanation:
(a) It is known that relation between charge and volume is as follows.
= 
= 
= 
Now, according to Gauss's law
= 
= 3.68 
Hence, the electric flux through this cubical surface if its edge length is 4.00 cm is 3.68 .
(b) Similarly, we will calculate the electric flux when edge length is 16.8 cm as follows.
= 
= 
= 
Now, according to Gauss's law
= 
= 2.72
Therefore, the electric flux through this cubical surface if its edge length is 4.00 cm is 2.72 .
<u>Answer:</u>
<em>The required radius of its motion is 
.</em>
<u>Explanation:</u>
The formula for calculating the required radius of its motion is given by
Where <em>m= mass </em>
<em>V= moving velocity
</em>
<em>F=frictional force
</em>
<em>r = radius of its motion
</em>
Then the required radius of its motion is given by
<u>Given that
</u>
<em>mas =0.0818 kg
</em>
<em>Frictional force= 0.108 N
</em>
<em>Moving with Velocity of = 0.333 m/s
</em>
<em>radius of its motion = ![\frac{[0.0818 kg \times (0.333 m/s)^2]}{0.108 N}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0.0818%20kg%20%5Ctimes%20%280.333%20m%2Fs%29%5E2%5D%7D%7B0.108%20N%7D)
</em>
<em>Hence the required radius of its motion is r = 
</em>
Just write a summary don't do the little details and don't explain everythinh
Answer:
The tip of her shadow is moving at the speed of 9.66 ft/sec
Explanation:
The height of the street light = 19 feet
The height of the woman = 5.25 feet
Distance between the woman and the base of the pole, x = 35 ft
The speed of the woman towards the pole, dx/dt = 7ft/sec
The distance from the base of the streetlight to the tip of the woman's shadow = y
The distance from the woman to the tip of her shadow = y - x
The diagram illustrating this description is shown below
Using similar triangle:
Find the derivative of both sides with respect to time, t
The tip of her shadow is moving at the speed of 9.66 ft/sec
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
