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Stolb23 [73]
3 years ago
5

What are 3 common sources of voltage difference?

Physics
1 answer:
rodikova [14]3 years ago
3 0
We are asked to enumerate 3 different sources of potential difference and the answers are listed below:
1. There was two parallel sources and voltage of each source is different from each other.
2. Different batteries are connected.
3, Different values of resistors were used in the circuit.

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Balance following equation then classify the reaction as synthesis (s)decomposition (d) single replacement (SD)double replacemen
stiv31 [10]
2H₂ + O₂ ⇒ 2H₂O
Classify: S (=synthesis)
6 0
3 years ago
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

F_{n} = 30.11N

Direction of the net force (β)

\beta =tan^{-1} (\frac{29.13}{7.62} )

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0
3 years ago
The speed of water flowing through the 4in diameter section of the piping system is 3ft/s. What is the volume flow rate of water
olga_2 [115]
1 ft =12 in 
4 in = 0.333 ft 
volume = (п/4)*(0.333)² = 0.087 ft²
vol. flow = spead *volume
              =3 ft/s * 0.087 ft² 

vol flow = 0.261 ft³/s
3 0
3 years ago
(a) Calculate the acceleration due to gravity on the surface of the Sun.
Ira Lisetskai [31]
<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

                      g=\frac{GM}{r^2}

 Here we need to find acceleration due to gravity of Sun,

                G = 6.67259 x 10⁻¹¹ N m²/kg²

    Mass of sun, M = 1.989 × 10³⁰ kg

    Radius of sun, r = 6.957 x 10⁸ m

Substituting,

                g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

   Ratio of gravity = 274.21/9.81 = 27.95

   Weight = mg

  Factor of increase in weight = 27.95

8 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
2 years ago
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