Question 6 of 10

Scilla [17]

I think the number 2, not sure

3 0

2 years ago

The surface of an incline is coated with an experimental substance that is intended to reduce the frictional force between a blo

nikklg [1K]

Answer:

粗糙錯愕額外此時自由KK預約為我

Explanation:

因為嗚嗚嗚粗糙此次四月蘇澳碩果僅存有一位我也在鶯鶯燕燕

4 0

2 years ago

Read 2 more answers

When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where i

Gnoma [55]

Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.

The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

4 0

3 years ago

Is there gravity in water?

Alexus [3.1K]

Answer: Yes.

Explanation:

There is gravity in water because How does the water go up in the cycle? the garavity with the wind pull it up.

3 0

2 years ago

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Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o

o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

$\displaystyle V_1=\frac{Q_1}{C_1}$

$\displaystyle V_2=\frac{Q_2}{C_2}$

They are both the same after connecting them, thus

$\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}$

Or, equivalently

$\displaystyle Q_2=\frac{C_2Q_1}{C_1}$

The total charge of both capacitors is

$\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)$

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

$Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C$

Now we compute Q1 from the equation above

$\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C$

The final voltage of any of the capacitors is

$\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V$

7 0

2 years ago