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
<span>238,900 mi hope it helps :)</span>
Answer:
387 volts
Explanation:
Ohm's law is used to relate voltage, current and resistance.
The formula is as follows:V = I * R
where:
V is the applied voltage (measured in volts)
I is the current flowing (measured in amperes)
R is the resistance (measured in ohm)
In the given, we have:
current (I) = 9 amperes
resistance (R) = 43 ohm
Substitute with the givens in the above formula to get the voltage as follows:
V = 9 * 43
V = 387 volts
Hope this helps :)
Answer:
![\frac{M_e}{M_s} = 3.07 \times 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7BM_e%7D%7BM_s%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-6%7D)
Explanation:
As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula
![T = 2\pi \sqrt{\frac{r^3}{GM}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Br%5E3%7D%7BGM%7D%7D)
now for the time period of moon around the earth we can say
![T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}](https://tex.z-dn.net/?f=T_1%20%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7Br_1%5E3%7D%7BGM_e%7D%7D)
here we know that
![T_1 = 0.08 year](https://tex.z-dn.net/?f=T_1%20%3D%200.08%20year)
![r_1 = 0.0027 AU](https://tex.z-dn.net/?f=r_1%20%3D%200.0027%20AU)
= mass of earth
Now if the same formula is used for revolution of Earth around the sun
![T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}](https://tex.z-dn.net/?f=T_2%20%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7Br_2%5E3%7D%7BGM_s%7D%7D)
here we know that
![r_2 = 1 AU](https://tex.z-dn.net/?f=r_2%20%3D%201%20AU)
![T_2 = 1 year](https://tex.z-dn.net/?f=T_2%20%3D%201%20year)
= mass of Sun
now we have
![\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7BT_1%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Br_2%5E3%20M_e%7D%7Br_1%5E3%20M_s%7D%7D)
![\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B0.08%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B1%20M_e%7D%7B%280.0027%29%5E3M_s%7D%7D)
![12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}](https://tex.z-dn.net/?f=12.5%20%3D%20%5Csqrt%7B%285.08%20%5Ctimes%2010%5E7%29%5Cfrac%7BM_e%7D%7BM_s%7D%7D)
![\frac{M_e}{M_s} = 3.07 \times 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7BM_e%7D%7BM_s%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-6%7D)
Answer:
Explanation:
The charge alters that space, causing any other charged object that enters the space to be affected by this field. The strength of the electric field is dependent upon how charged the object creating the field is and upon the distance of separation from the charged object.