Multiplying the power of any signal by 5 can be described as
an increase of 6.99 dB .
If the whistle blew at 70 dB initially, and its sound power became
multiplied by 5, and the whistle and the listener both stayed in
the same places, then the listener would tell you that the whistle
was now blowing at 76.99 dB .
(More likely, he would report "77 dB" as he held his ears and winced.)
Answer:
the thermistor temperature = 
Explanation:
Given that:
A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.
i.e Temperature
Resistance of the thermistor
20,000 ohms
Material constant
= 3650
Resistance of the thermistor
= 500 ohms
Using the equation :


Taking log of both sides





Replacing our values into the above equation :






Thus, the thermistor temperature = 
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
Answer:
convex lens
Explanation:
An image is form in retina with light rays converging most at cornea and upon entering and existing the lens.Rays from top and bottom of the object are traced and produce an inverted image on the retina