Answer:
a. 5 batteries b. 1050 mAh
Explanation:
Here is the complete question
A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed? B) What mAh capacity should the batteries have?
Solution
A) How many batteries are needed?
Since the nominal voltage for a single NiMH battery is 1.2 V per battery and we require 6V DC power, we combine the batteries in series to obtain a total voltage of 6 V. The number of batteries required, n = total voltage/voltage per cell = 6V/1.2V per battery = 5 batteries
So, the number of batteries needed is 5.
B) What mAh capacity should the batteries have?
Since the batteries are in series, they would each deliver a current of 150 mA. Since we require a current of 150 mA for 7 hours, the number of milliampere-hours capacity mAh of batteries required is Q = It where I = current = 150 mA and t = time = 7 hours.
So, Q = It = 150 mA × 7 h = 1050 mAh.
So, the batteries should have a mAh of 1050 mAh
Answer:
w = 706.32 [N]
Explanation:
The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.
w = m*g
where:
w = weight [N] (units of Newtons)
m = mass = 72 [kg]
g = gravity acceleration = 9.81 [m/s²]
Then we have:
![w = 72*9.81\\w = 706.32 [N]](https://tex.z-dn.net/?f=w%20%3D%2072%2A9.81%5C%5Cw%20%3D%20706.32%20%5BN%5D)
Answer:
How Urgent are we talking?
Explanation:
Answer:
There are no gaps in space between the photons as they travel. If you were to look at a wave then you'd come to a conclusion that indeed that there aren't any gaps unless they are specifically placed.The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.
Answer:
1) 57 revolutions.
2) 4.52 m
3) 1531.2 m
Explanation:
Question 1:
Given:
Initial velocity, 
Final velocity, 
Time, 
Acceleration, 
Now, displacement of the tire is given as:

Displacement of tire in 1 revolution is equal to its circumference.
Therefore, displacement in 1 revolution = 
Now, number of revolutions is given as:

Therefore, the number of revolutions are 57.
Question 2:
Given:
Radius of the wheel is, 
Angle of rotation is,
°
Converting degree to radians, we get:

Now, path length is given as:

Therefore, the path length of a point on the wheel is 4.52 m
Question 3:
Radius of the wheel is, 
Angle of rotation is,
radians
Now, path length is given as:

Therefore, the path length of a point on the wheel is 1531.2 m.