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goblinko [34]
3 years ago
12

Which sound would be loudest?

Physics
1 answer:
Oliga [24]3 years ago
3 0
I think it might be C
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A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat
Natali [406]

centripetal acceleration is given by formula

a_c = \omega^2*R

given that

a_c = 34.1 m/s^2

R  =  5.91 m

now we have

\omega^2 R = 34.1

\omega^2 * 5.91 = 34.1

\omega^2 = 5.77

\omega = 2.4 rad/s

so the ratationa frequency is given by

\omega = 2 \pi f

2.4 = 2 \pi f

f = \frac{2.4}{2\pi}

f = 0.38 Hz

7 0
3 years ago
How will the force of friction affect a wood block being pushed over a table?
olasank [31]

Answer:

D

friction acts in the opposite direction of motion but does not affect the motion of the object

5 0
2 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
Why is argon gas used instead of air in light bulbs that contain a filament that is heated to glowing?
tamaranim1 [39]
<span>the heated filament will react with the oxygen in the air but now with the argon, which is a noble gas and hardly ever reacts.</span>
3 0
3 years ago
How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of ele
luda_lava [24]

Answer:

W = 1.5 x 10⁶ J = 1.5 MJ

Explanation:

First, we calculate the potential difference between the given 2 points. So, we have:

V₁ = Electric Potential at Initial Position = 6.7 V

V₂ = Electric Potential at Final Position = - 8.9 V

Therefore,

ΔV = Potential Difference = V₂ - V₁ = -8.9 V - 6.7 V = - 15.6 V

Since, we use magnitude in calculation only. Therefore,

ΔV = 15.6 V

Now, we calculate total charge:

Total Charge = q = (No. of Electrons)(Charge on 1 Electron)

where,

No. of Electrons = Avagadro's No. = 6.022 x 10²³

Charge on 1 electron = 1.6 x 10⁻¹⁹ C

Therefore,

q = (6.022 x 10²³)(1.6 x 10⁻¹⁹ C)

q = 96352 C

Now, from the definition of potential difference, we know that it is equal to the worked done on a unit charge moving it between the two points of different potentials:

ΔV = W/q

W = (ΔV )(q)

where,

W = work done = ?

W = (15.6 V)(96352 C)

<u>W = 1.5 x 10⁶ J = 1.5 MJ</u>

7 0
3 years ago
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