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Dmitry_Shevchenko [17]
3 years ago
14

F. A 50. A resistor (R2), and unknown resistor R2, a 120 Volt source, and an ammeter are connected in a

Physics
1 answer:
Archy [21]3 years ago
6 0

Complete question is;

A 50.-ohm resistor, an unknown resistor R, a 120-volt source, and an ammeter are connected in a complete circuit. The ammeter reads 0.50 ampere.

A) Calculate the equivalent resistance of the circuit shown.

B) Determine the resistance of resistor R shown in the diagram.

Answer:

A) R_eq = 240 Ω

B) R = 190 Ω

Explanation:

A) To get the equivalent resistance, we will use the formula;

R = V/I

Where;

V is Voltage

I is current

R is equivalent resistance

From the question, V = 120 V and I = 0.5A

Thus;

R_eq = 120/0.5

R_eq = 240 Ω

B) From the image, we see that the resistors are connected in series.

Formula for resistors in series is;

R = R1 + R2 +..... Rn

Thus;

240 = 50 + R

R = 240 - 50

R = 190 Ω

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Pani-rosa [81]

Answer:

\frac{1}{2}  \times m \times  {v}^{2}  \\  \frac{1}{2}  \times m \times  {2}^{2} \\ 2 \times m

I dont the child mass...you should substitute that value to (m)

then you can get your answer

5 0
3 years ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
Gennadij [26K]

Answer:

18.1 × 10⁻⁶ A = 18.1 μA

Explanation:

The current I in the wire is I = ∫∫J(r)rdrdθ

Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.

So, dI = J(r)rdrdθ

dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²

Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m

I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m  =  0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA

3 0
4 years ago
If the slowest instruction in the SCA executes in 12.5 ns, then what is maximum system clock frequency in MHz
Margaret [11]

Since the slowest instruction in the SCA executes in 12.5 ns, the maximum system clock frequency is 80 MHz

To answer the question, we need to know what frequency is.

<h3>What is frequency?</h3>

Frequency is the number of oscillations per second of a wave.

It is given by f = 1/T where T = period of wave

Now, given that the slowest instruction in the SCA executes in t = 12.5 ns, we need to calculate maximum system clock frequency, f.

<h3>What is the maximum system clock frequency?</h3>

So, f = 1/t

= 1/12.5 ns

= 1/(12.5 × 10⁻⁹ s)

= 1/12.5 × 10⁹ Hz

= 0.08 × 10⁹ Hz

= 80 × 10⁻³ × 10⁹ Hz

= 80 × 10⁶ Hz

= 80 MHz

So, the maximum system clock frequency is 80 MHz

Learn more about maximum system clock frequency here:

brainly.com/question/14636488

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5 0
2 years ago
The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hy
Aloiza [94]

Answer:

3x10^5m/s

Explanation:

See attached file

Explanation:

4 0
3 years ago
a large cargo trucks needs to cross a bridge the truck is 30m long, and 3.2 wide, the cargo exerts a force of 54,000 N the bridg
otez555 [7]

Answer:

It isn't safe for the truck to cross the bridge because the pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 Pa > 450 Pa.

Explanation:

Pressure is expressed in Force/Area.

So, for the truck, force exerted = 54000 N

Area covered = 30 × 3.2 = 96 m²

Pressure exerted by the truck = 54000/96 = 562.5 Pa

The pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 > 450, hence, it isn't safe for the truck to cross the bridge.

5 0
4 years ago
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