The work done by the shopping basket is 147 J.
<h3>When is work said to be done?</h3>
Work is said to be done whenever a force moves an object through a certain distance.
The amount of work done on the shopping basket can be calculated using the formula below.
Formula:
Where:
- W = Amount of work done by the basket
- m = mass of the shopping basket
- h = height of the shopping basket
- g = acceleration due to gravity.
Form the question,
Given:
- m = 10 kg
- h = 1.5 m
- g = 9.8 m/s²
Substitute these values into equation 2
- W = 10(1.5)(9.8)
- W = 147 J.
Hence, The work done by the shopping basket is 147 J.
Learn more about work done here: brainly.com/question/18762601
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
a E =
b E =
c E = 0 N/C
d 
e 
f V = 
g 
h 
i 
Explanation:
From the question we are given that
The first charge 
The second charge 
The first radius 
The second radius 

And ![Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]](https://tex.z-dn.net/?f=Potential%20%5C%20Difference%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%20%20%5B%5Cfrac%7Bq_1%20%7D%7Br%7D%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%5D)
The objective is to obtain the the magnitude of electric for different cases
And the potential difference for other cases
Considering a
r = 4.00 m


Considering b

This implies that the electric field would be

This because it the electric filed of the charge which is below it in distance that it would feel

= 
Considering c
r = 0.200 m
=> 
The electric field = 0
This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field
Considering d
r = 4.00 m
=> 
Now the potential difference is

This so because the distance between the charge we are considering is further than the two charges given
Considering e
r = 1.00 m 
![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5Cfrac%7B1.00%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2026.79%20%2A10%5E3%20V)
Considering f

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.700%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2034.67%20%2A10%5E3%20V)
Considering g

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
Considering h

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7BR_1%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
Considering i

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7BR_1%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.
so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain
but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.
there hoped this helped I guess
Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol
<h3>
<em>Al(OH)3 = 78.01 g/mol</em></h3>
Answer:
The helicopter uses 35 gallons to fly for 5 hours.
Explanation:
The amount of gas that a helicopter uses for flying varies directly proportional to the number of hours spent flying.
g ∝ T
where g represents amount of gas and T time of flight.
Then,

The helicopter files 4 hours and uses 28 gallons of fuel.
Here, g₁= 28 gallons, T₁=4 hours
g₂=?, T₂=5 hours.


⇒28×5= g₂×4
⇒ g₂×4=28×5

gallons
The helicopter uses 35 gallons to fly for 5 hours.