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Ksivusya [100]
2 years ago
9

A 16-Ω loudspeaker, an 8.0-Ω loudspeaker, and a 4.0-Ω loudspeaker are connected in parallel across the terminals of an amplifier

. Determine the equivalent resistance of the three speakers, assuming that they all behave as resistors.
Physics
1 answer:
kolbaska11 [484]2 years ago
6 0

Answer:

2.286 ohm

Explanation:

R1 = 16 ohm

R2 = 8 ohm

R3 = 4 ohm

They all are connected in parallel combination

Let the equivalent resistance is R.

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/16 + 1/8 + 1/4

1/R = (1 + 2 + 4) / 16

1/R = 7 / 16

R = 16/7 = 2.286 ohm

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Please help! will give brainlest!!!!!!!!!!!!
eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

a=0

Therefore the net force is zero as well:

\sum F = 0

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

\sum F = F cos \theta - F_f = 0

where

F cos \theta is the horizontal component of the applied force, with

F = 22.5 N

\theta=32.0^{\circ}

F_f is the force of friction

And solving for F_f, we find

F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 0
2 years ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
2 years ago
A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initia
prisoha [69]

Answer:

The water will flow at a speed of 3,884 m/s

Explanation:

Torricelli's equation

v = \sqrt{2gh}

*v = liquid velocity at the exit of the hole

g = gravity acceleration

h = distance from the surface of the liquid to the center of the hole.

v = \sqrt{2*9,8m/s^2*0,77m} = 3,884 m/s

6 0
2 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
There is an electric field in the region between the two plates. The magnitude of this electric field is ed. This imposes anothe
krok68 [10]

Answer:

it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

Explanation:

The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.

 To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates

5 0
2 years ago
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