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Kamila [148]
3 years ago
5

What happens to an electron when energy is added?

Physics
1 answer:
Vladimir79 [104]3 years ago
7 0
When an electron<span> is hit by a photon of light, it absorbs the quanta of </span>energy<span> the photon was carrying and moves to a higher </span>energy<span> state. One way of thinking about this higher </span>energy<span> state is to imagine that the </span>electron<span> is now moving faster, (it has just been "hit" by a rapidly moving photon).</span>
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I'll mark brainless pictures down below​
julsineya [31]

The net force is (40 Newtons) (down the road).

But 40 Newtons is not going to move a piano very enthusiastically.

8 0
3 years ago
A man with a mass of 60 kg rides a bike with a mass of 13 kg. What is the force needed to accelerate the bike at 0.90 m/s2?
Scrat [10]
The total mass of the things on the road is 60 + 13, which is 73kg.

mass • acceleration = F

73 x 0.9m/s^2 = 65.7N

F = 65.7 N

7 0
2 years ago
A 50 mL graduated cylinder contains 25.0 mL of water. A 42.5040 g piece of gold is placed in the graduated cylinder and the wate
Elenna [48]

Answer:

19320 kg/m³

Explanation:

density: This can be defined as the ratio of the mass of a body to its volume. The S.I unit of Density is kg/m³.

The formula of density is given as,

D = m/v ......................... Equation 1.

Where D = Density of the gold, m = mass of the gold, v = volume of the gold.

Note: From Archimedes's Principle, the piece of gold displace an amount of water that is equal to it's volume.

Amount of water displace = 27.2 - 25 = 2.2 mL.

Given: m = 42.504 g = 0.042504 kg, v = 2.2 mL = (2.2/10⁶) m³ = 0.0000022 m³

Substitute into equation 1

D = 0.042504/0.0000022

D = 19320 kg/m³

Hence the density of the piece of gold = 19320 kg/m³

5 0
3 years ago
What are some common measurements that are given as a rate?
BabaBlast [244]

Answer:

taco bell

Explanation:

6 0
4 years ago
You’re an electrical engineer designing an alternator (the generator that charges a car’s battery). Mechanical engineers specify
makvit [3.9K]

Answer:

I will specify a value of 0.009T for the alternator’s magnetic field

Explanation:

E_peak = 14 V

d = 10cm = 0.1m, so r = 0.1/2 =0.05m

N = 250 turns

f = 1200rpm = (1200rp/m x 1m/60sec) = 20 revolutions per second

At peak performance, peak voltage is given by the equation;

E_peak = NABω

Let's make the magnetic field B the subject;

B = E_peak/(NAω)

Now we know that ω = 2πf

Thus, ω = 2π x 20 revs/s = 125.664 revs/s.

Let's convert it to the standard unit which is rad/s.

1 rev/s = 6.283 rad/s

Thus, 125.664 revs/s = 125.664 x 6.283 = 789.55 rad/s

Area (A) = πr² = π x 0.05² = 0.007854 m²

Thus, plugging in the relevant values to get;

B = 14/[(250 x 0.007854 x 789.55)] = 0.009T

8 0
3 years ago
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