Coke has more fizz than Pepsi, because Coke has more carbonation in it. Pepsi contains more sugar (2 more tablespoons) than Coke, so it tastes slightly sweeter to many people.
<h3><u>Answer;</u></h3>
= 8.55 Joules
<h3><u>Explanation;</u></h3>
Work done is the product of force and the distance moved by an object.
Work done = Force × distance
Force = 95 Newtons
Distance = X2 -X1
= 4 - (-5)
= 9 cm
Thus;
work done = 95 × 9/100
<u>= 8.55 Joules </u>
The answer I'm going with is false
Answer:
the reflected wave is inverted and the transmitted wave is up
Explanation:
To answer this question we must analyze the physical phenomenon, with an wave reaching a discontinuity, we can analyze it as a shock.
Let's start when the discontinuity is with a fixed, very heavy and rigid obstacle, in this case the reflected wave is inverted, since the contact point cannot move
In the event that it collides with an object that can move, the reflected wave is not inverted, this is because the point can rise, they form a maximum at this point.
In the proposed case the shock is when the thickness changes, in this case we have the above phenomena, a part of the wave is reflected by being inverted and a part of the wave is transmitted without inverting.
The amplitude sum of the amplitudes of the two waves is proportional to the lanería that is distributed between them.
When checking the answers the correct one is the reflected wave is inverted and the transmitted wave is up
Answer:
Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:
Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium
Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:
The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
