How does earth move'

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

A sample of helium (He) occupies 8.0 liters at 1 atm and 20.0◦C. What pressure is necessary to change the volume to 1.0 liters a

nevsk [136]

Apply the combined gas law

PV/T = const.

P = pressure, V = volume, T = temperature, PV/T must stay constant.

Initial PVT values:

P = 1atm, V = 8.0L, T = 20.0°C = 293.15K

Final PVT values:

P = ?, V = 1.0L, T = 10.0°C = 283.15K

Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:

1(8.0)/293.15 = P(1.0)/283.15

P = 7.7atm

7 0

3 years ago

A 55.2 kg softball player moving 3.11 m/s slides across dirt with uk=0.310. How far does she slide before coming to a stop

hoa [83]

Refer to the attachment

5 0

2 years ago

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp

Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

so

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_{i}$ = 60 mph = 26.8224 m/s

Final velocity $V_{f}$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $\frac{1}{2}$ m( $V_{i}$ ² - $V_{f}$ ² )

we substitute

Δk = $\frac{1}{2}$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $\frac{1}{2}$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

so

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0

2 years ago

At what height is an object that weighs 490 newtons if its gravitational potential energy is 4900 J?

Sergio [31]

Answer:

The height at which the object is moved is 10 meters.

Explanation:

Given that,

Force acting on the object, W = F = 490 N

The gravitational potential energy, P = 4900 J

We need to find the height at which the object is moved. We know that the gravitational potential energy is possessed due to its position. It is given by :

$P=mgh\\\\h=\dfrac{P}{mg}\\\\h=\dfrac{4900\ J}{490\ N}\\\\h=10\ m$

So, the height at which the object is moved is 10 meters. Hence, this is the required solution.

5 0

3 years ago