Answer:
P=9.58 W
Explanation:
According to Newton's second law, and assuming friction force as zero:
![F_m=m.a\\F_m=0.875kg*a](https://tex.z-dn.net/?f=F_m%3Dm.a%5C%5CF_m%3D0.875kg%2Aa)
The acceleration is given by:
![a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7Bt%7D%5C%5Ca%3D%5Cfrac%7B0.685m%2Fs%7D%7B21.5%2A10%5E%7B-3%7Ds%7D%5C%5C%5C%5Ca%3D31.9m%2Fs%5E2)
So the force exerted by the motor is:
![F_m=0.875kg*31.9m/s^2\\F_m=27.9N](https://tex.z-dn.net/?f=F_m%3D0.875kg%2A31.9m%2Fs%5E2%5C%5CF_m%3D27.9N)
The work done by the motor is given by:
![W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m](https://tex.z-dn.net/?f=W_m%3DF_m%2Ad%5C%5C%5C%5Cd%3D%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E2%5C%5Cd%3D%5Cfrac%7B1%7D%7B2%7D%2A31.9m%2Fs%5E2%2A%2821.5%2A10%5E%7B-3%7Ds%29%5E2%5C%5C%5C%5Cd%3D7.37%2A10%5E%7B-3%7Dm)
![W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J](https://tex.z-dn.net/?f=W_m%3D27.9N%2A7.37%2A10%5E%7B-3%7Dm%5C%5CW_m%3D0.206J)
And finally, the power is given by:
![P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BW_m%7D%7Bt%7D%5C%5CP%3D%5Cfrac%7B0.206J%7D%7B21.5%2A10%5E%7B-3%7Ds%7D%5C%5C%5C%5CP%3D9.58W)
I don't know this 1 I'm sorry I can't help you
Answer:
hello the diagram related to this question is missing attached below is the missing diagram
Answer :
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Explanation:
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Answer:
1. 38,500
2. 308,000
Explanation:
This would require a calculator. To find momentum, you multiply mass and velocity. You always want your mass to be measure in kilograms, but that is irrelevant in this question because they already are, it is just something to remember.
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.