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Rudiy27
3 years ago
5

If a car has a centripetal acceleration of 7m/s2 over a radius of 7m. How fast is it going

Physics
1 answer:
dangina [55]3 years ago
3 0

Answer:

7 m/s (agrees with answer a in your list)

Explanation:

Recall that the centripetal acceleration is defined by the square of the tangential velocity divided by the radius of the rotational motion:

a_c=\frac{v_t^2}{R}

then the tangential velocity is extracted from here as:

a_c=\frac{v_t^2}{R} \\v_t^2=a_c * R\\v_t=\sqrt{a_c * R}

in our case, this becomes:

v_t=\sqrt{7*7} = 7 \,\,m/s

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Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon
posledela

Answer:

Q1_new = 515.68 µC

Q2_new = 246.82 µC

Explanation:

Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.

Then it is only necessary to calculate the charge on each capacitor:

Q1 = 5.85 µF * 250 V = 1462.5 µC

Q2 = 2.8 µF * 250 V = 700 µC

Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.

When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:

1462.5 µC - 700 µC = 762.5 µC

Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.

This 762.5 µC will be divided proportionately between the two capacitors.

Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC

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