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3 years ago

Which variable stars have pulsation periods between 1.5 hours and 1.2 days?

2 answers:

7 0

D. Nebulas is a cloud of dust,
B. Protostars & A. Cepheid variables have longer pulsation periods.
Whereas, C. RR Lyrae variables have pulsation periods between 1.5 hours and 1.2 days. So the answer is C.

maria [59]3 years ago

3 0

Correct answer choice is :

C) RR Lyrae variables

Explanation:

RR Lyrae variables are periodical unsteady stars, usually discovered in globular groups. They are accepted as regular lights to measure galactic distances, as a part of the infinite distance ladder. This class of unsteady star is identified after the prototype and glorious example, RR Lyrae. R Lyraes are vibrating parallel section aging stars of phantom class A or F, with a mass of around half the Sun's.

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derive formula of kinetic energy of particular having mass volume and velocity of dimensional analysis

natita [175]

Answer:

E=1/2kmv^2

Explanation:

To get the formula...

The dimension of energy is [ M^1 L^2 T^-2]

The dimension of mass [M]=[M^1 L ^0T ^0]

Dimension of velocity[V]=[M ^0 L ^1 T ^−1]

Let [E]=k[M] ^x[V] ^y

K is constant and it remain a dimensionless quantity.

Therefore, M ^1L ^2T ^−2]=k[M ^1L ^0T ^0] x[M ^0L ^1 T ^−1] y

So, x=1,y=2

E= Kmv^2

K=1/2

Therefore E is 1/2kmv^2.

4 0

3 years ago

A 20 kg crate of books is at rest on a table. what is the normal force on the books?

Umnica [9.8K]

The weight of 20 kg is

(mass) x (gravity) =

(20 x 9.8) = 196 newtons, downward.

There must be a force of 196 newtons upward on the mass.
Otherwise the mass is accelerating either up or down.

4 0

3 years ago

A person struggles very hard to lift a large boulder. He puts in so much effort, he starts to sweat, his heart rate increases, a

Lesechka [4]

No, he did not perform any work. Work is when you’re using energy which results in a force. Even though he was tired and sweaty, he did not move the boulder. So therefore he did not perform any work.

5 0

4 years ago

The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0

4 years ago

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such li

lozanna [386]

Answer: 4.50*10^-6T (0.00000450071T)

Explanation: A current carrying conductor has been knowing to generate a specific amount of magnetic field.

This is given by the Bio-savart law (mathematical).

The Bio-savart law is a mathematical equation that gives the value of strength of the magnetic field created by a current carrying conductor.

B=(Uo* I) /2πr

Where

B= strength of magnetic field

Uo = magnetic permeability in free space = 1.257 *10^-6

r = distance between current carrying conductor and any reference point.

By doing the neccesary algebra, we have

B=(1.257 *10^-6 * 180)/ (2 * 3.142 * 8)

B= 2.2626 *10^-4 / 50.2857

B=4.5 * 10^-6T (0.00000450071T)

6 0

3 years ago