Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
Answer:
Total impulse =
= Initial momentum of the car
Explanation:
Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.
The final velocity of the car is 0 m/s as it is brought to rest.
Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.
The magnitude of impulse is the absolute value of the change in momentum.
![|J|=|p_f-p_i|](https://tex.z-dn.net/?f=%7CJ%7C%3D%7Cp_f-p_i%7C)
Momentum of an object is equal to the product of its mass and velocity.
So, the initial momentum of the car is given as:
![p_i=mv](https://tex.z-dn.net/?f=p_i%3Dmv)
The final momentum of the car is given as:
![p_f=m(0)=0](https://tex.z-dn.net/?f=p_f%3Dm%280%29%3D0)
Therefore, the impulse is given as:
![|J|=|p_f-p_i|=|0-mv|=|-mv|=mv](https://tex.z-dn.net/?f=%7CJ%7C%3D%7Cp_f-p_i%7C%3D%7C0-mv%7C%3D%7C-mv%7C%3Dmv)
Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.
Answer:
2.5m
Explanation:
Torque is defined as the rotational effect of a force on a body.
The torque T for the maximum shear stress is given as 0.1 Nm
Frictional torque is the torque caused by a frictional force
The frictional torque F is given as 0.04 Nm/m
The maximum length of the shaft is thus given as
L = T / F
= 0.1/0.04
L= 2.5 m
Answer:
A. 0.199 J
B. 0.0663 C
C = 0.0221 F
D. 12.68 ohms
Explanation:
From the question:
time duration, t = 0.28 seconds
Average power, P = 0.71 W
Average voltage, V = 3 V
A) Energy is given as:
E = P * t
=> E = 0.71 * 0.28 = 0.199 J
B) Electrical energy is also given as:
E = qV
where q = charge
=> q = E / V
∴ q = 0.199 / 3 = 0.0663 C
C) Capacitance is given as charge over voltage:
C = q / V
=> C = 0.0663 / 3 = 0.0221 F
D) Electrical power, P, can also be given as:
P = ![V^2 / R](https://tex.z-dn.net/?f=V%5E2%20%2F%20R)
where R = resistance
=> R = ![V^2 / P](https://tex.z-dn.net/?f=V%5E2%20%2F%20P)
R = ![3^2 / 0.71 = 9 / 0.71 = 12.68 ohms](https://tex.z-dn.net/?f=3%5E2%20%2F%200.71%20%3D%209%20%2F%200.71%20%3D%2012.68%20ohms)