Answer:
6.39 J of energy is needed to generate 0.71 * 10⁻¹⁶ kg mass
Explanation:
According to the Equation: E = mc²
where the mass, m = 0.71 * 10⁻¹⁶ kg
the speed of light, c = 3 * 10⁸ m/s
The amount of energy needed to generate a mass of 0.71 * 10⁻¹⁶ kg is calculated as follows:
E = (0.71 * 10⁻¹⁶) (3 * 10⁸)²
E = 0.71 * 10⁻¹⁶ * 9 * 10¹⁶
E = 0.71 * 9
E = 6.39 J
Answer:
If it is not an object in motion, all forces are balanced.
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
Yes because you intoxicated
Hope this helped :)
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = 
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:
Plug in the given values and solve for 
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
