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Elenna [48]
3 years ago
8

What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.

Physics
1 answer:
Thepotemich [5.8K]3 years ago
4 0

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

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Si un cuerpo adquiere una carga de -0,02 C, ha ganado o ha perdido electrones? Cuantos?
sammy [17]

Question :

If a body acquires a charge of -0.02 C, has it gained or lost electrons? Many?

Solution :

We know, charge gained is shown by negative sign.

Since, charged acquired is given as -0.02 C .

Therefore, it is body has gained electrons.

Now, number of electrons is given by :

n = \dfrac{net\ charge}{charge \  on \ one \  electron}\\\\n = \dfrac{-0.02}{-1.60 \times 10^{-19}}\\\\n = 1.25\times 10^{17}

Hence, this is the required solution.

7 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
3 years ago
Kon'nichiwa~<br>please help me with this question!!​
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If the refractive index of water is 1.33, then its critical angle is...
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1. A sample of gas has a constant temperature and number of particles. As the volume of the gas sample is increased, the pressur
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Answer:

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Explanation:

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