In electron capture, which particle would be the beta particle?

How do amplitude and frequency affect light energy? help asap !!!!!!

So, I’m assuming that we’re treating light as a propagating wave.

Amplitude measures the amount of energy transported by a wave, thus amplitude squared is directly proportional to the light’s energy. The higher the amplitude, the higher the energy.

Energy is also directly proportional to the frequency of a wave, the higher the frequency, the higher the energy.

I took my second answer from the formula below:

E=cf

7 0

3 years ago

What is the kinetic energy of a 1 kg pie if it is thrown at 10 m/s?

nignag [31]

Given

mass (m) = 1 kg

velocity (v) = 10 m/s

kinetic energy ( ke) = ?

,we know

K.E =1/2 m v²

= 1/2 * 1 * 10²

= 100/2

=50 joule

hope it helps :)

7 0

3 years ago

A 2200 kilogram car is accelerating at 3.4 m/s/s. what is the NET force?

sdas [7]

We Know, F = m*a
F = 2200 * 3.4
F = 7480 Kg m/s²

So, your final answer is 7480

7 0

3 years ago

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ $\frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}$

On multiplying both sides by "100", we get

⇒ $\frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}$

⇒ $\frac{\Delta T}{T}\times 100=0.005787$ (%)

∵ $T=2\pi\sqrt{\frac{l}{g} }$

On substituting the values, we get

⇒ $\frac{\Delta T}{T}$ % = $\frac{1}{2}\times \frac{\Delta l}{l}$ %

On applying cross multiplication, we get

⇒ $\frac{\Delta l}{l}$ % = $2\times \frac{\Delta T}{T}$ %

⇒ = $2\times 0.05787$

⇒ = $0.011574$

⇒ = $0.012$ %

6 0

3 years ago

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

Readme [11.4K]

Answer:

(a)0.0675 J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

$E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675 J$

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0

3 years ago