<span>1.5 minutes per rotation.
The formula for centripetal force is
A = v^2/r
where
A = acceleration
v = velocity
r = radius
So let's substitute the known values and solve for v. So
F = v^2/r
0.98 m/s^2 = v^2/200 m
196 m^2/s^2 = v^2
14 m/s = v
So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin.
Its circumference is 2*pi*r, so
C = 2 * 3.14159 * 200 m
C = 1256.636 m
And we need a velocity of 14 m/s, so
1256.636 m / 14 m/s = 89.75971429 s
Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.</span>
Answer:
ΔLa/ΔLb = 1
Explanation:
The change in length of a solid is given by the following formula:
ΔL = α L ΔT
where,
ΔL = Change in length
α = coefficient of linear expansion
L = Original Length
ΔT = Change in Temperature
Since, the length and change in temperature for both rods are same. Also, the material of each rod is same, which implies that coefficient of linear expansion for both rods is same. Hence, the ratio of change in length of both rods will be:
<u>ΔLa/ΔLb = 1</u>
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
a. mass density
Explanation:
<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>
- When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.
<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>
Answer:
0.243
Explanation:
<u>Step 1: </u> Identify the given parameters
Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,
collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09
Frictional diameter =45mm
<u>Step 2:</u> calculate the torque required to raise the load
![T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}](https://tex.z-dn.net/?f=T_%7BR%7D%20%3D%20%5Cfrac%7B5%2825%29%7D%7B2%7D%20%5B%5Cfrac%7B5%2B%5Cpi%280.09%29%2825%29%7D%7B%5Cpi%2825%29-0.09%285%29%7D%5D%2B%5Cfrac%7B5%280.06%29%2845%29%7D%7B2%7D)
= (9.66 + 6.75)N.m
= 16.41 N.m
<u>Step 3:</u> calculate the torque required to lower the load
![T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}](https://tex.z-dn.net/?f=T_%7BL%7D%20%3D%20%5Cfrac%7B5%2825%29%7D%7B2%7D%20%5B%5Cfrac%7B%5Cpi%280.09%29%2825%29%20-5%7D%7B%5Cpi%2825%29%2B0.09%285%29%7D%5D%2B%5Cfrac%7B5%280.06%29%2845%29%7D%7B2%7D)
= (1.64 + 6.75)N.m
= 8.39 N.m
Since the torque required to lower the thread is positive, the thread is self-locking.
The overall efficiency = 
= 
= 0.243
