To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.
Where
A = Cross-sectional Area
Y = Young's modulus
= Coefficient of linear expansion for steel
= Temperature Raise
Our values are given as,
Replacing we have,
Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N
<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>
Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.
(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]
= (64.2 m/s) - [ 14.7 m/s ]
= 49.5 m/s . (upward)
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Answer: A. Greater than 384 Hz
Explanation:
The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.
Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.
Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.
So option a is correct. i.e. now frequency will be greater than 384 Hz.
