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liberstina [14]
2 years ago
14

Find the impedance of the load for maximum power transfer in the attached circuit.

Physics
1 answer:
Tanya [424]2 years ago
7 0
I can't find the attached circuit.

Maximum power transfer occurs when the load impedance
is equal to the internal impedance of the power source.
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I meed help with these 2 questions plz
ludmilkaskok [199]

-- 30N

-- the sum of all the forces

4 0
3 years ago
4. What does doubling the voltage do to the strength of the electromagnet?​
Naya [18.7K]

Answer:

it can make it stronger!

5 0
2 years ago
Read 2 more answers
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

5 0
2 years ago
A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
2 years ago
A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr
NARA [144]
To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds

Horizontal distance
=0.335s*1.2m/s
=0.402 meters
8 0
3 years ago
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