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borishaifa [10]
3 years ago
11

How do you solve this?

Physics
1 answer:
Andrews [41]3 years ago
5 0

Answer:

The circuit is in series connection,

the same current will flow through i.e

I1 = I2 = I3 = 2A.

Explanation:

We'll begin by calculating the total resistance in the circuit. This is shown below:

R1 = 4Ω

R2 = 3Ω

RT =..?

Total resistance in series can be obtained as follow:

RT = R1 + R2

RT = 4 + 3

RT = 7Ω

Next we shall determine the total current flowing in the circuit:

Voltage (V) = 14V

Resistance (R) = 7Ω

Current (I) =..?

V = IR

14 = I x 7

Divide both side by 7

I = 14/7

I = 2A.

Since the circuit is in series connection,

the same current will flow through i.e

I1 = I2 = I3 = 2A.

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Answer:

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3 years ago
A 1.94-m-diameter lead sphere has a mass of 5681 kg. A dust particle rests on the surface. What is the ratio of the gravitationa
scoundrel [369]

To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,

\text{Diameter of the lead sphere} =  D=1.940m

\text{Mass of the lead sphere} =  m_1 = 5681kg

\text{Mass of the dust particle} = m_2

Distance between the center of lead sphere to dust particle

r = \frac{D}{2}

r = \frac{1.940m}{2}

r = 0.97m

Gravitational force of the sphere on the dust particle:

F = \frac{Gm_1m_2}{r^2}

F = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5681kg)(m_2)}{(0.97m)^2}

F = (4.027*10^{-7} N/kg)m_2

Weight of the dust particle

W = m_2 g

W = m_2 (9.8m/s^2)

Ratio of F and W:

\frac{F}{W} = \frac{(4.07*10^{-7}N/kg) m_2)}{m_2(9.8m/s^2)}

\frac{F}{W} = 4.153*10^{-8}

Therefore the ratio is 4.153*10^{-8}

3 0
3 years ago
True or false the number of times a machine multipies effort force is a mechanical atlantis
lesantik [10]

Answer: i think its true

Explanation:

7 0
3 years ago
Name at least three fundamental differences between the harmonic oscillator dynamics and the simple pendulum dynamics
xenn [34]

Answer: please find the answer in the explanation.

Explanation:

Harmonic can be experienced by any body that repeats itself. The pattern can be sinusoidal, square, tooth etc.

The fundamental differences between the harmonic oscillator dynamics and the simple pendulum dynamics are:

1.) The harmonic oscillator dynamics can be sinusoidal or square wave so far the motion is periodic while the simple pendulum dynamics is always sinusoidal.

2.) In simple pendulum dynamics, the period of oscillation is independent of the amplitude. While the period in harmonic oscillator dynamics depends on the amplitude.

3.) Differential equation is only one method to analyze the simple pendulum dynamics where there are several methods to analyze the harmonic oscillator dynamics.

8 0
4 years ago
The resistance, R, of a wire varies directly as its length and inversely as the square of its diameter. If the resistance of a w
lbvjy [14]

R is proportional to the length of the wire:

R ∝ length

R is also proportional to the inverse square of the diameter:

R ∝ 1/diameter²

The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.

Calculate the scale factor due to the changed length:

k₁ = 2800/2700 = 1.037

Scale factor due to changed diameter:

k₂ = 1/(0.1/0.26)² = 6.76

Multiply the original resistance by these factors to get the new resistance:

R = R₀k₁k₂

R₀ = 9850Ω, k₁ = 1.037, k₂ = 6.76

R = 9850(1.037)(6.76)

R = 69049.682Ω

Round to the nearest hundredth:

R = 69049.68Ω

8 0
3 years ago
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