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3 years ago

How much charge flows from a 6.0 v battery when it is connected across a completely discharged 15.7 μf capacitor?

1 answer:

4 0

The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 6.0V (assuming circuit resistance is negligable) and it has a capacitance of 15.7<span>μf or </span> 15.7x10^-6f, therefore charge equals (15.7x10^-6)x6=9.42x10^-5C (Coulombs).

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I a liter jar contains 0.5 moles of gas at a pressure of 2 ATM what is the temperature of the gas

sweet [91]

Answer:

<h2>The temperature of the gas is 48.75 Kelvin.</h2>

Explanation:

Using the ideal gas equation as shown

PV = nRT where;

P is the pressure of the gas in ATM

V is the volume of the gas

n is the number of moles

R is the ideal gas constant

T is the temperature in Kelvin

From the formula, $T = \frac{PV}{nR}$

Given the following parameters V = 1litre, n = 0.5moles. pressure = 2ATM

R = 0.08206 atm L/molK

On substituting to get the temperature we have:

$T = \frac{2*1}{0.08206*0.5} \\T = \frac{2}{0.04103}\\T = 48.75Kelvin$

5 0

3 years ago

What is the heart rate recorded a few minutes after completing a workout?

DochEvi [55]

Correct answer should be C

8 0

3 years ago

Read 2 more answers

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

erica [24]

Answer:

0.235 nC

Explanation:

Given:

$E$ = the magnitude of electric field = $8.50\ kN/C =8.50\times 10^{3}\ N/C$

$F$ = the magnitude of electric force on each antenna = $2.00\ \mu N =2.00\times 10^{-6}\ N$

$q$ = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

$\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC$

Hence, the value of q is 0.235 nC.

4 0

3 years ago

A heat pump has a coefficient of performance of 4. if the heat pump absorbs 20 cal of heat from the cold outdoors per cycle, how

Lubov Fominskaja [6]

A heat pump has a coefficient of performance of 7.05. If the heat pump absorbs 20 cal of heat from the cold outdoors in each cycle, find the heat expelled to the warm indoors. Answer in units of cal. I believe COP (heating mode) : 7.05 and COP = Qh/W Qc = 20 cal so I have to find Qh= ? cal I dont know an equation to put all this together? Please help, thank you.

5 0

3 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

4 years ago