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guapka [62]
3 years ago
10

A - 28.4-μC charge is placed 16.4 cm from a charge q, the force between the two charges is 1240 N. What is the value of q?

Physics
1 answer:
anastassius [24]3 years ago
3 0

Answer:

1.31×10⁻⁴ C or 131 μC

Explanation:

From the question,

Applying Coulomb's Law,

F = kqq'/r²..................... Equation 1

Where F = Force between the charges, q = first charge, q' = second charge, r = distance between the charges, k = coulomb's constant.

make q' the subject of the equation

q' = F×r²/(kq)

q' = Fr²/kq................ Equation 2

Given: F = 1240 N, r = 16.4 cm = 0.164 m, q = 28.4 μC = 2.84×10⁻⁵ C,

Constant: k = 8.98×10⁹ Nm/C²

Substitute these values into equation 2

q' = (1240×0.164²)/[(2.84×10⁻⁵)×(8.98×10⁹)

q' = (33.35104)/(25.5032×10⁴)

q' = 1.31×10⁻⁴ C

q' = 131 μC

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What's difference between results in dot product and cross product of vectors.
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Answer:

A dot product of two vectors is also called the scalar product. ... The difference between the dot product and the cross product of two vectors is that the result of the dot product is a scalar quantity, whereas the result of the cross product is a vector quantity.

Explanation:

Hope it is helpful.....

6 0
3 years ago
If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.
Ksenya-84 [330]

Answer:

twice

Explanation:

From magnification = height of image / height of object

Distance of image/ distance of object = magnification

If the distance and height of the object represents the initial light distance and the exposed surface respectively.

And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.

Hence the new image exposure would be twice as large.

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H2= D2/D1× H1

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6 0
3 years ago
1. Dos puntos A y B están separados en línea recta 1475 m. Por el punto A pasa hacia el punto B un móvil con una velocidad de 15
ikadub [295]

Answer:

a) 35 sequndos   b) M1 = 600m   M2 = 875m

Explanation:

a)

Antes de que el otro móvil comience a moverse, el primer móvil ha viajado 5 segundos a una velocidad de 15 m / s.

5 segundos x 15 m / s = 75 metros.

Podemos establecer una ecuación con la información que conocemos. Juntos, en total, los dos recorren 1475 metros.

Voy a representar el tiempo como t. Un móvil ya ha recorrido 75 metros y hará otros 15 t. El otro aún no se ha movido y viajará 25t. Esos tres números se sumarán a 1475.

75 + 15t + 25t = 1475.

Resta 75 de ambos lados y suma términos similares de t: 40t = 1400

t = 35 segundos.

b)

Ahora tome la distancia recorrida por cada vehículo por separado y conecte 35 segundos para t.

Primer móvil: 75 + 15t = 75 + 15 (35) = 600m

Segundo móvil: 25t = 25 (35) = 875m

Usé un traductor para entender la pregunta y luego traduje lo que escribí. Lo siento si mi explicación no es clara.

6 0
4 years ago
Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes a
loris [4]

Answer:

Part a)

T = 342.5 \hat i + 675\hat j

Part b)

F_{net} = 342.5\hat i - 174\hat j

Part c)

F = 384.2 N

Part d)

\theta = 333 degree

Part e)

a = 4.4 m/s^2

Part f)

\theta = 333 degree

Explanation:

Part a)

Magnitude of tension force is given as

T = 757 N at 26.9 degree with vertical

T = 757 sin26.9 \hat i + 757 cos26.9 \hat j

T = 342.5 \hat i + 675\hat j

Part b)

Net force on Tarzen is given as

F_{net} = T + F_g

F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j

F_{net} = 342.5\hat i - 174\hat j

Part c)

magnitude of the force is given as

F = \sqrt{F_x^2 + F_y^2}

F = \sqrt{342.5^2 + 174^2}

F = 384.2 N

Part d)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-174}{342.5}

\theta = 333 degree

Part e)

Magnitude of the acceleration

a = \frac{F}{m}

m = \frac{849}{9.81} = 86.5 kg

tex]a = \frac{384.2}{86.5}[/tex]

a = 4.4 m/s^2

Part f)

Direction of acceleration is same as the direction of the force

\theta = 333 degree

6 0
3 years ago
A child lifts a box from the floor. The child then carries the box with constant speed to the other side of the room and puts th
Alenkasestr [34]

Answer:Zero

Explanation:

Given child moves the box with constant velocity over his head .

Force acting on box due to gravity is equal to his weight

Displacement of the box is Perpendicular to the direction of Applied force

Work done is given by

W=F\cdot s

W=Fs\cos \theta

here \theta is the angle between Force and displacement which is 90^{\circ}

so work done is zero

               

5 0
4 years ago
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