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miv72 [106K]
2 years ago
15

The signal to a speaker causes .5 amps of current. The signal level is increased to 2.5 amps. What is the gain, rounded to the n

earest decibel?
Physics
1 answer:
devlian [24]2 years ago
7 0

Explanation:

It is given that,

The signal to a speaker causes 0.5 amps of current, I_1=0.5\ A

The signal level is increased to 2.5 amps, I_2=2.5\ A

We need to find the gain to the nearest decibel. The decibel gain formula of the speaker is given by :

db=20\ log(\dfrac{I_2}{I_1})

db=20\ log(\dfrac{2.5}{0.5})

db = 13.97 db

or

db = 14 db

So, the gain pf the speaker is 14 db. Hence, this is the required solution.

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A wire of 1mm diameter and 1m long fixed at one end is stretched by 0.01mm when a lend of 10 kg is attached to its free end.calc
Otrada [13]

Answer:

E = 1.25×10¹³ N/m²

Explanation:

Young's modulus is defined as:

E = stress / strain

E = (F / A) / (dL / L)

E = (F L) / (A dL)

Given:

F = 10 kg × 9.8 m/s² = 98 N

L = 1 m

dL = 10⁻⁵ m

A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²

Solve:

E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)

E = 1.25×10¹³ N/m²

Round as needed.

5 0
2 years ago
Determine the torque applied to the shaft of a car that transmits 225 hp
Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu

8 0
2 years ago
Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (
RideAnS [48]

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0
2 years ago
A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold
tino4ka555 [31]

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



6 0
2 years ago
Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi
Serga [27]

Answer:

Flow Rate = 80 m^3 /hours  (Rounded to the nearest whole number)

Explanation:

Given

  • Hf = head loss
  • f = friction factor
  • L = Length of the pipe = 360 m
  • V = Flow velocity, m/s
  • D = Pipe diameter = 0.12 m
  • g = Gravitational acceleration, m/s^2
  • Re = Reynolds's Number
  • rho = Density =998 kg/m^3
  • μ = Viscosity = 0.001 kg/m-s
  • Z = Elevation Difference = 60 m

Calculations

Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

7 0
2 years ago
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