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Mandarinka [93]
3 years ago
15

What is the unit of velocity ratio and mechanical advantage and why​

Physics
1 answer:
alexandr402 [8]3 years ago
8 0
Both don’t have units beacuse they are ratios
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Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

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Q = 0,06 × 10 ^ -3

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7 0
3 years ago
which of the folloing statements about ionization energy is true? A elements toward the bottom of a group periodic table general
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6 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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