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3 years ago

Determine e when I = 0.50 A and R = 12 W.

Physics

1 answer:

sammy [17]3 years ago

7 0

Answer:

The correct answer is "24 V".

Explanation:

The given values are:

Current,

I = 0.50 A

Resistance,

R = 12 W

As we know,

⇒ $I = 0.5\times (\frac{E}{2R})$

On substituting the given values, we get

⇒ $0.5= (\frac{E}{4\times 12} )$

⇒ $0.5= (\frac{E}{48} )$

⇒ $E=24 \ V$

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Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

$[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\$

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$[\frac{m^2}{s^2} ]$

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

$v=\frac{dx}{dt} \\v = 0.75*2*t+5*t$

$(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]$

ii) replacing in the derivated equation.

$v=1.5*(4)+5\\v=11[m/s]$

iii) the average velocity is defined by the expresion v = x/t

$v = \frac{x-x_{0} }{t-t_{0} } \\$

$x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]$

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

$\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} } \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })$

b) One of the applications of the surface tension is the <u>capillarity</u> this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0

3 years ago

A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,

ICE Princess25 [194]

Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0

3 years ago