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The heat capacity of the silver spoon at the given temperature difference is 817.65 J.
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Heat capacity of the silver spoon</h3>
The heat capacity of theb silver spoon is the quantity of heat absorbed by the silver spoon. The heat capacity of the silver spoon at the given temperature difference is calculated as follows;
Q = mcΔθ
where;
- m is mass of the spoon
- c is specific heat capacity of silver = 0.237 J/g⁰C
- Δθ is change in temperature
Q = 50 x 0.237 x (89 - 20)
Q = 817.65 J
Thus, the heat capacity of the silver spoon at the given temperature difference is 817.65 J.
Learn more about heat capacity here: brainly.com/question/16559442
Answer:
The speed of proton when it emerges through the hole in the positive plate is 
.
Explanation:
Given that,
A parallel-plate capacitor is held at a potential difference of 250 V.
A A proton is fired toward a small hole in the negative plate with a speed of, 
We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :
So, the speed of proton when it emerges through the hole in the positive plate is .
There will not be enough momentum from the first hill to cross another hill if he same or larger size because of the way potential energy and kinetic energy works it will not be able go as high as it could go on he fist hill.
Answer:
Given:
Fundamental frequency: 470Hz
T1:310k,T2:315k
Calculating velocity
Recall v=(331m/s)✓[T1/273k)
V=331✓(310/273)
V1=331*(1.0656)=352.72m/s
V2=331✓(315/273)=355.5m/s
Fundamental frequency=4L
F2=F1(V2/V1)
F2=470(355.5/352.72)=474.4Hz
Beat=[F2-F1]=474.4-470=4.4Hz
Explanation:
