700 J is the work done by the system.
<u>Explanation:</u>
The first law of thermodynamics is that the change in internal energy of the system is equal to the net heat transfer to the system minus the complete work performed by the system.
Where,
∆U – Change in internal energy
Q – Heat transfer to the system
Q – Work done
Here,
<u>Given data:</u>
∆U - 400 J
Q - 1100 J
We need to the work done by the system (W)
By applying the given values in the above equation, we get
400 = 1100 - W
W = 1100 - 400 = 700 J
Answer: The foot kicked the ball with alot of force the ball had force too so their toes are left stinging
Explanation:
Explanation:
The fluctuating sound heard when two objects vibrate with different frequencies is called beats. It is given that guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beats/s when sounded with a 357 Hz tuning fork.
It is assumed to find the vibrational frequency of the string.
For 3 beats/s, beat frequency can be :
352 - 3 or 352 + 3 = 349 Hz or 355 Hz
For 8 beats/s, beat frequency can be :
357 - 8 or 357 + 8 = 349 Hz or 365 Hz
It means that the vibrational frequency is 349 Hz.
Answer:
The object's maximum speed remains unchanged.
Explanation:
The speed of a particle in SHM is given by :
Maximum speed is, 
If A' = 2A and T' = 2T
So, the maximum speed of the object remains the same i.e. it remains unchanged. Hence, this is the required solution.
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V
