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3 years ago

A 1.5m wire carries a 5 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

Physics

1 answer:

Natali [406]3 years ago

8 0

Resistance = (voltage) / (current)

Resistance = (68 V) / (5 A)

<em>Resistance = 13.6 ohms</em>

Since we already know the voltage and the current, we don't need to know the length of the wire. It makes no difference.

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This gradual change in beak size over time is known as _______.

yan [13]

The correct answer is evolution

4 0

4 years ago

Read 2 more answers

An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle

shtirl [24]

Answer:

$-9.45\times10^{-16} \text{ J}$

Explanation:

According to Coulomb's law, the force of attraction between two point charges, $q_1$ and $q_2$ , separated by a distance $d$ is given by

$F = k\dfrac{q_1q_2}{d^2}$

$k$ is a constant with a value of $9\times10^9\text{ F/m}$ .

When we substitute the values from the question,

$F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}$

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

$W = F\times r$

$W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}$

This is negative because work is done on the electron, not by it.

3 0

3 years ago

Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay abo

Luba_88 [7]

Answer:

#_time = 7.5 10⁴ s

Explanation:

In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.

t = $\frac{t_p}{ \sqrt{1- (v/c)^2} }$

where t_p is the person's own time in an immobile reference frame,

$t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }$

let's calculate

we assume that the speed of the space station is constant

$t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }$

$t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}$

t_ = 0.99998666657 s

therefore the time change is

Δt = t - t_p

Δt = 1 - 0.9998666657

Δt = 1.3333 10⁻⁵ s

this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s

#_time = 1 / Δt

#_time = $\frac{1}{1.3333 \ 10^{-5}}$