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dezoksy [38]
3 years ago
8

10. How does surface area affect the amount of air resistance an object experiences?

Physics
1 answer:
ikadub [295]3 years ago
6 0

Answer:cross-sectional area, and thus surface area, increases the amount of air resistance an object experiences

Explanation:

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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

3 0
3 years ago
What is physical change ?​
ss7ja [257]

Answer:

A physical change is a change to the physical—as opposed to chemical—properties of a substance. They are usually reversible. The physical properties of a substance include such characteristics as shape (volume and size), color, texture, flexibility, density, and mass.

3 0
2 years ago
What can happen to solar radiation when it enters atmosphere​
grandymaker [24]

Answer:

The solar radiation is first intercepted by Earth's atmosphere, just a small part of the radiation is absorbed by gases such as water vapor. Some of the radiation is reflected back to space by the clouds and Earth's surface.

3 0
3 years ago
How fast was the storm that hit Galveston and how men died?​
vodomira [7]

Answer:

The storm was a category 4 hurricane that struck Galveston, Texas, on September 8, 1900, bringing winds of 130 miles (210 km) per hour and high tides that overwhelmed the low-lying coastal city, demolishing buildings and claiming more than 8,000 lives.

00p- now I can actually answer :)

Hope that I helped you a little :0

3 0
3 years ago
An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the sign
Temka [501]

Answer:

Explanation:

Given that,

Frequency of radio signal is

f = 800kHz = 800,000 Hz.

Distance from transmitter

d = 8.5km = 8500m

Electric field amplitude

E = 0.9 V/m

The average energy density can be calculated using

U_E = ½•ϵo•E²

Where ϵo = 8.85 × 10^-12 F/m

Then,

U_E = ½ × 8.85 × 10^-12 × 0.9²

U_E = 3.58 × 10^-12 J/m²

The average electromagnetic energy density is 3.58 × 10^-12 J/m²

6 0
3 years ago
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