<span>If the pitch of a sound increased, then the wave frequency increased. (D)</span>
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:
we've the density of the water 
b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:
<u>Now the difference in the column is :</u>
Answer:
k = 9.6 x 10^5 N/m or 9.6 kN/m
Explanation:
First, we need to use the expression to calculate the spring constant which is:
w² = k/m
Solving for k:
k = w²*m
To get the angular velocity:
w = 2πf
The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:
f = V/x
f = 5.7 / 4.9 = 1.16 Hz
Now the angular velocity:
w = 2π*1.16
w = 7.29 rad/s
Finally, solving for k:
k = (7.29)² * 1800
k = 95,659.38 N/m
In two significant figures it'll ve 9.6 kN/m
If the corner is rounded and is perfectly circular, then the acceleration is centripetal and is always directed toward the center.
Answer: a) - 437.8° F, b) - 261°c.
Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.
5 (°F - 32) = 9 (k - 273)
Where °F = temperature in Fahrenheit and k = temperature in kelvin.
For question A, k = 12.0, by substituting to have the value for °F, we have
5(°F - 32) = 9 ( 12 - 273)
5(°F - 32) = 9(-261)
5(°F - 32) = - 2349
°F - 32 = - 2349/5
°F - 32 = - 469.8
°F = - 469.8 + 32
°F = - 437.8
Question B
The centigrade and kelvin scale are related by the formulae below
°c = k - 273
Where °c = temperature in centigrade and k = temperature in kelvin =12
°c = 12 - 273
°c = - 261
