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2 years ago

What is the total charge of 1.0g of electrons?

1 answer:

8 0

<span>1.0 gram (g) of electrons would contain 10^27 electrons Electrons have an electric charge of −1.602×10−19 coulomb so total charge of 1 g electrons = -1.602 x 10^-19 x 10^27 = -1.602 x 10^8</span>

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What is one joule work

natta225 [31]

It's a Newton Meter Those two are multiplied to get a joule.

8 0

3 years ago

The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th

Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

$v=\sqrt{\frac{2GM}{R}}$

Here we have

Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

R = 1 AU = 1.496 × 10¹¹ m

M = 2.3 × 10³⁰ kg

Substituting

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s$

Approximate escape speed = 45.3 km/s

6 0

3 years ago

Science questions!! Please help!!

kondor19780726 [428]

Answer:

1. b

2. c

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6. c

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7 0

2 years ago

A ball is thrown straight upward and rises to a maximum

Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

$v_f^2 - v_i^2 = 2 a d$

here we know that

$v_f = 0$

also we have

$a = -9.81 m/s^2$

so we have

$0 - v_i^2 = 2(-9.81)(16)$

$v_i = 17.72 m/s$

Now we need to find the height where its speed becomes half of initial value

so we have

$v_f = 0.5 v_i$

now we have

$v_f^2 - v_i^2 = 2 a d$

$(0.5v_i)^2 - v_i^2 = 2(-9.81)h$

$-0.75v_i^2 = -19.62 h$

$0.75(17.72)^2 = 19.62 h$

$h = 12 m$

3 0

3 years ago

A circuit component with a high resistance will have a low electric current.

Zepler [3.9K]

You haven't said what 'high' resistance or 'low' current means, so there's way not enough info to nail the statement as true or false. The most precise answer is "certainly could be but not necessarily". Anyway, the current in the circuit depends on BOTH the resistance AND the voltage. So without knowing the voltage too, you can't say anything about the current.

6 0

2 years ago