Answer:
student A or B
Explanation:
A common demonstration is to put a ringing alarm clock or bell in the bell jar, and when the vacuum is created, you can no longer hear the sound of the clock/bell.
The bell is connected to a lab pack or batteries and rung to show pupils it can be heard under normal circumstances. The bell jar is then connected to a vacuum pump using a vacuum plate (see Fig 2) and the air is removed from inside creating a near vacuum. The bell is then again rung. This time however, it cannot be heard.
Small low voltage buzzers can be used as a bell replacement for the bell and work in exactly the same way though teachers generally prefer bells as students may be able to see the hammer moving, proving that it is actually ringing even though they cannot hear it.
Some vacuum pumps are better than others at keeping a strong vacuum though if you cannot completely lose the sound, you will at least notice the volume decreasing.
Sound is simply a series of longitudinal waves travelling from the source, through the air to our ears. Without air present, these waves cannot form and therefore sound cannot be conveyed.
In a longitudinal wave the particles oscillate back and forth in the direction of the wave movement unlike transverse waves which like waves on the sea, single particles travel up and down and not in the direction of the wave.
Because you will not be able to create a perfect vacuum, you may still be able to hear the bell ring slightly. Vibrations from the ringing bell can also travel up to the bung in the bell jar which in turn may resonate the jar slightly. This means you may hear the bell ring, however strong the vacuum. To compensate for this, try to insulate the bell as much as possible from the bell jar. Hanging the bell using elastic cord means some of the vibrations will be absorbed by the cord and not be transferred to the bell jar.
Answer:
<em>T</em><em>h</em><em>e</em><em>r</em><em>e</em><em> </em><em>are</em><em> </em><em>t</em><em>wo hydrogen </em><em>atom</em><em> </em><em>in</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>reactants</em><em>.</em>
I think that number five is lithium
Answer:
a) time taken to catch up with speeding car is 12.25 secs
b) the police car will travel 273.8 m to catch up with the speeding car
Explanation:
Given that;
speed of car 
= 50 mi/hr = 22.352 m/s
acceleration of police car = 10 mi/hr = 4.47 m/s²
= 70 mi/hr = 31.29 m/s
Now time taken to reach maximum speed is t₁
so
= 
+ at₁
we substitute
31.29 = 0 + 4.47t₁
t₁ = 31.29 / 4.47
t₁ = 7 sec
now
d₁ = 0 + 1/2 × at₁²
d₁ = 0 + 1/2 × 0 + 4.47×(7)²
d₁ = 109.5 m
so distance travelled by the speeding car in time t₁ will be
= × t₁
we substitute
= 22.352 × 7
= 156.46 m
now distance between polive car and speeding car
Δd = - d₁
Δd = 156.46 - 109.5
Δd = 46.96 m
time taken to cover Δd will be
t₂ = Δd / ( - )
t₂ = 46.96 / ( 31.29 - 22.352 )
t₂ = 46.96 / 8.938
t₂ = 5.25 sec
distance travelled by the police in time t₂ will be
d₂ = × t₂
d₂ = 31.29 × 5.25
d₂ = 164.3 m
a) How long will it take before the officer catches up to the speeding car;
time taken to catch up with speeding car;
t = t₁ + t₂
t = 7 + 5.25
t = 12.25 secs
Therefore, time taken to catch up with speeding car is 12.25 secs
b) how far will it have travelled in order to do so;
distance = d₁ + d₂
distance = 109.5 + 164.3
distance = 273.8 m
Therefore, the police car will travel 273.8 m to catch up with the speeding car
Answer:
heat required in pan B is more than pan A
Explanation:
Heat required to raise the temperature of the substance is given by the formula
now we know that both pan contains same volume of water while the mass of pan is different
So here heat required to raise the temperature of water in Pan A is given as
Now similarly for other pan we have
So here by comparing the two equations we can say that heat required in pan B is more than pan A
