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3 years ago

The law of conservation of energy states...

Physics

2 answers:

Alecsey [184]3 years ago

6 0

The law of conservation of energy states that energy can not be created nor destroyed.

iris [78.8K]3 years ago

3 0

Can neither be created or destroyed.

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A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude

Pani-rosa [81]

The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
$y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])$
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
$y(x,t)=0.750 cm$
and therefore this value corresponds to the amplitude of the wave.

4 0

3 years ago

Read 2 more answers

What is the minimum mass needed for a star to be on the main sequence? What happens to stars that do not meet the minimum mass?

Mariana [72]

Main sequence stars are characterised by the source of their energy.They are all undergoing fusion of hydrogen into helium within their cores. The mass of the star is the main element for such process or phenomenon to take place for it is a determinant of both the rate at which they perform the said activity and the amount of fuel available.

To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.

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2 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

oee [108]

Answer:

3.25 × 10^7 m/s

Explanation:

Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)

Ek = 1/2 mv2 = qΔV .................. 1

Given that V is the electron speed in m/s

Charge of electron = 1.60217662 × 10-19 coulombs

Mass of electron = 9.109×10−31 kilograms

ΔV = 3.0kV = 3000V

Make V the subject of the formula in eqaution 1

V = sqr root 2qΔV/m

V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31

V = 3.25 × 10^7 m/s

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3 years ago

Como foi a despedida do bilhete?

Dmitrij [34]

Can you translate to english ?

8 0

2 years ago

A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int

Inessa05 [86]

Let F be the magnitude of the force applied to the cart, m the mass of the cart, and a the acceleration it undergoes. After time t, the cart accelerates from rest v₀ = 0 to a final velocity v. By Newton's second law, the first push applies an acceleration of

F = m a → a = F / m

so that the cart's final speed is

v = v₀ + a t

v = (F / m) t

If we force is halved, so is the accleration:

a = F / m → a/2 = F / (2m)

So, in order to get the cart up to the same speed v as before, you need to double the time interval t to 2t, since that would give

(F / (2m)) (2t) = (F / m) t = v

3 0

2 years ago