Answer:
N1 / N2 = 1.0016
Explanation:
The apparent weight of the student is the value that a balance would have this corresponds in this case to the normal of the student. Let's write Newton's second law in the lower and upper part of the loop
For lower rotating wheel
N1 - W = m a
a = v² / r
N1 = W + v² / r
The relationship between linear and angular velocity is
v = w r
As the wheel rotates at a constant speed, we can use angular kinematics
w = θ / t
θ = 3 rev (2π rad / 1rev) = 6π rad
w = 6π / 150
w = 0.04π rad / s = 0.1257 rad / sec
calculate
N1 = mg + w² r
N1 = 55 9.8 + 0.1257² 27
N1 = 539 + 0.4267
N1 = 539.426 N
Now we perform the same calculation for the top
-N2 - W = -m a
N2 = -W + ma
N2 = 539 - 0.4267
N2 = 538.5733 N
The relationship between the weight at the bottom and top is
N1 / N2 = 539.4267 / 538.5733
N1 / N2 = 1.0016
we know the equation for the period of oscillation in SHM is as follows:
T = 2 * pi * sqrt(mass/k)
we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).
since d = v*T, we can say v = d/t = d * f
the final equation, after combining everything, is as follows:
v = d/(2 * pi) * sqrt(k/m)
by plugging everything in
v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))
We find our velocity to be:
v = 6.89 m/s
Answer:
W
Explanation:
= Temperature of the room = 22.0 °C = 22 + 273 = 295 K
= Temperature of the skin = 33.0 °C = 33 + 273 = 306 K
= Surface area = 1.50 m²
= emissivity = 0.97
= Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴
Rate of heat transfer is given as
W
Answer:
It remains constant
Explanation:
As we know that buoyant force on an object given as
Fb = ρ Vd g
ρ= Density of fluid
Vd=Volume displace by body
g=10 m/s²
Fb =buoyant force
So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.
So when rock gets deeper and deeper the buoyant force will remain constant.
It remains constant
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