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rjkz [21]
3 years ago
6

Dos personas aplican sendas fuerzas de 50N sobre una mesa, en direcciones horizontales perpendiculares entre sí. ¿Cuanto vale el

módulo e la resultante de estas dos fuerzas? Ilustra tu respuesta con un dibujo.
Physics
1 answer:
kolezko [41]3 years ago
5 0

Answer:

|Fr| = 50*√2  [N]

Explanation: (See Annex )

En el diagrama de cuerpo libre ( mostrado en el anexo) se ve que, al ser las dos fuerzas iguales (50 N) el paralelogramo formado para encontrar la resultante de las fuerzas (Fr)  es un cuadrado, y la diagonal de ese cuadrado ( que es al mismo tiempo la hipotenusa del triangulo recto OPA ) es igual a:

Hipotenusa = |Fr| = √ (50)² + (50)²

|Fr| = √2* (50)²

|Fr| = 50*√2  [N]

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what is the force of gravity attraction between an object with a mass of 0.5 kg and another object has a mass of 0.33 kg and a d
elena55 [62]

Answer:

from \: newton \: law \: of \: gravitation \\ F =  \frac{GMm}{ {r}^{2} }  \\ G = 6.67 \times  {10}^{ - 11}  \\ M = 0.5 \: kg \\ m = 0.33 \: kg \\ r = 0.002 \: m \\ substitute \\ F =  \frac{(6.67 \times  {10}^{ - 11}) \times (0.5) \times (0.33)) }{ {(0.002)}^{2} }  \\  = 2.75 \times  {10}^{ - 6} N

4 0
3 years ago
The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the
Xelga [282]

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

8 0
3 years ago
PLEASE HELP <br> IS THIS CORRECT?
Umnica [9.8K]

Answer:

yes!

Explanation:

7 0
3 years ago
Read 2 more answers
A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im
Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°,  I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0
4 years ago
Radioactive isotopes of an atom are:
Len [333]

Radioactive isotopes of an atom aren't as stable as you would think, so the correct answer is: C. Less Stable

(I can verify this because I took the test.)

Hope this helps! Have a wonderfully wonderful day!

Cheers mate!

8 0
3 years ago
Read 2 more answers
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