Answer:
less than 20580 N
Explanation:
According to the newton's second law of motion
Force = mass * acceleration
(assuming gravitational acceleration =9.8 m/s2 )
acceleration = 30*9.8 = 294 m/s2
acting Force = 70 * 294
= 20580 N
Since the acceleration was less than 30g , acting force should also be less than 20580 N
Answer:
0.426 volts
Explanation:
It is given that,
The radius of a circular loop, r = 11.2 cm = 0.112 m
An elastic conducting material is stretched into a circular loop.
It is placed with its plane perpendicular to a uniform 0.880 T magnetic field.
The radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s, dr/dt = 0.688 m/s
We need to find the emf induced in the loop at that instant.
So, the magnitude of induced emf is 0.426 volts.
The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.
The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N
The other free body diagram is around the joint of the three strings.
In this case, you can do the horizontal forces equilibrium equation as:
T1* cos(60) - T2*cos(40) = 0
And the vertical forces equilibrium equation:
Ti sin(60) + T2 sin(40) = T3 = 325 N
Then you have two equations with two unknown variables, T1 and T2
0.5 T1 - 0.766 T2 = 0
0.866 T1 + 0.643T2 = 325
When you solve it you get, T1 = 252.8 N and T2 = 165 N
Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N
