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3 years ago

Match the events related to the formation of the universe with the stages during which they occurred.

Physics

1 answer:

igomit [66]3 years ago

8 0

Answer:

does this even make sense

Explanation:

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If a car drives 10 miles NE and then due East for 10 minutes traveling 8

Nimfa-mama [501]

I think the answer is 28 miles

7 0

2 years ago

Read 2 more answers

A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w

son4ous [18]

Answer:

5773.50269 Hz

23 A

Explanation:

$L$ = Inductance = 6 mH

$C$ = Capacitance = 5 μF

$R$ = Resistance = 3 Ω

$\epsilon$ = Maximum emf = 69 V

Resonant angular frequency is given by

$\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz$

The resonant angular frequency is 5773.50269 Hz

Current is given by

$I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A$

The current amplitude at the resonant angular frequency is 23 A

7 0

3 years ago

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0

3 years ago

Which sphere is NOT a part of the cycling of oxygen through Earth's systems? A. biosphere B. exosphere C. atmosphere D. hydrosph

postnew [5]

Exosphere is the answer . I have done this lesson in my physics class it's exosphere.

8 0

3 years ago

Read 2 more answers

As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet

Pani-rosa [81]

Answer:

new atmospheric pressure is 0.9838 × $10^{5}$ Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × $10^{5}$ = 13.6 × 10³ × 9.81 × h

h = 0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × $10^{5}$ Pa

6 0

3 years ago