let the distance of pillar is "r" from one end of the slab
So here net torque must be balance with respect to pillar to be in balanced state
So here we will have
here we know that
mg = 19600 N
Mg = 400,000 N
L = 20 m
from above equation we have
so pillar is at distance 10.098 m from one end of the slab
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then
The extension of the spring due to the weight of the object on Moon is a value of , then
Recall that gravity on the moon is a sixth of Earth's gravity.
We have that the displacement at the earth was , then
Therefore the displacement of the mass on the spring on Moon is 0.05m
Answer:
a=0 v = v₀ + a t
a=0 line is horizontal
Explanation:
1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration
2, speed and relationship of a car is given by
v = v₀ + a t
where vo is the initial velocity, a is the acceleration and tel time
in this case I will calcograph velocity vs. time the constant acceleration is a straight line.
In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope
