The correct answer is D. I alread took this test.
A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is
Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:
and substituting t = 75 seconds, we find
In degrees, it is
So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by
where we have
r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find
Answer:
4960 N
Explanation:
First, find the acceleration.
Given:
v₀ = 6.33 m/s
v = 2.38 m/s
Δx = 4.20 m
Find: a
v² = v₀² + 2aΔx
(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)
a = -4.10 m/s²
Next, find the force.
F = ma
F = (1210 kg) (-4.10 m/s²)
F = -4960 N
The magnitude of the force is 4960 N.
Answer:
The correct answer is B-25 V
Explanation:
We apply Ohm's Law, according to which:
V = i x R
V = 5A x 5Ω
V= 25 V
Being V the potential difference whose unit is the VOLT, i the current intensity (Ampere) and R the electrical resistance (ohm)
Answer:
Explanation:
a )
Reaction force of the ground
R = mg
= 160 N
Maximum friction force possible
= μ x R
= μ x 160
= .4 x 160
= 64 N .
b )
160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,
Taking moment about top point of ladder
160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3
240 + 444 + 4f = 2700
f = 504 N
c )
Let x be the required distance.
Taking moment about top point of ladder
160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )
240 + 444 x + 1440 = 2700
x = 2.3 m
so man can go upto 2.3 at which maximum friction acts .
