A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)
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How is the change in electric potential energy of the proton-field system calculated?</h3>
- Work done on the proton =Negative of the change in the electric potential energy of the proton field
- In the given case, W = -qΔV
- -W = qΔV
- = qEcosθ
- Therefore, work done on the proton = -e(8.50×
N/C)(2.5m)(1) - = -3.40×
J - Any change in the potential energy indicates the work done by the proton.
- Therefore the positive sign shows that the potential energy increases when the proton does the work.
- The negative sign shows that the potential energy decreases when the proton does the work.
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Answer:
252J
Explanation:
Given parameters:
Distance = 72m
Force = 3.5N
Unknown:
Work done on the house = ?
Solution:
Work done is the force applied to move a body through a particular distance.
Work done = Force x distance
Now insert the parameters and solve;
Work done = 3.5 x 72 = 252J
Since 1m/s goes to 4m/s in 3 seconds,
then 4m/s goes to 8m/s in 6 seconds.
So your answer would be:
4m/s to 8m/s in 6 seconds
Answer:
At 3.86K
Explanation:
The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:
gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2
Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .
We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =
√0.7/ 0.0469 = 3.86K.
Acceleration is change is velocity over time. So the change in velocity is 40 and time is 4.5 sec. So the acceleration is 40/4.5=8.89 m/s^2
