Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
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L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
Specific gravity of other fluid = .854 (Approx)
Explanation:
Given:
Mass of water = 35 g
Mass of filled bottle with water = 98.44 g
Mass of filled bottle with fluid = 89.22 g
Computation:
Mass of water = 98.44g - 35g = 63.44g
Density of water = 1000 g/L
Volume of bottle = 63.44/1000 = 0.06344L
Mass of other liquid = 89.22g - 35g = 54.22g
Density of other liquid = 54.22g/0.06344L = 854.665826 g/L
Water has a specific gravity = 1
So , specific gravity of other fluid
1000 / 854.665826 = 1 / specific gravity of other fluid
Specific gravity of other fluid = .854 (Approx)
Answer:
Explanation:
i = Imax sin2πft
given i = 180 , Imax = 200 , f = 50 , t = ?
Put the give values in the equation above
180 = 200 sin 2πft
sin 2πft = .9
sin2π x 50t = .9
sin 360 x 50 t = sin ( 360n + 64 )
360 x 50 t = 360n + 64
360 x 50 t = 64 , ( putting n = 0 for least value of t )
18000 t = 64
t = 3.55 ms .
Answer:
B. W is positive and a is negative
Explanation:
As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have
initial angular velocity is termed as positive angular velocity

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed
so we will have

so here correct answer is
B. W is positive and a is negative
Answer:

Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is

Now the distance from the charge is

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.



