We are given that,
We need to find when
The equation that relates x and can be written as,
Differentiating each side with respect to t, we get,
Replacing the value of the velocity
The value of could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to
Search light is rotating at a rate of 0.96rad/s
Answer:
So the force of attraction between the two objects is 3.3365*10^-6
Explanation:
m1=10kg
m2=50kg
d=10cm=0.1m
G=6.673*10^-11Nm^2kg^2
We have to find the force of attraction between them
F=Gm1m2/d^2
F=6.673*10^-11*10*50/0.1^2
F=3.3365*10^-8/0.01
F=3.3365*10^-6
Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B
Where =Mass of block B
Substitute the values
For block A
Where Mass of block A
Substitute the values
Hence, the friction force acting on block A=40 N
B. Mentally represent their environment