Ask question

Ask question

All categories

1 year ago

14

The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1

, la, and the gravitational constant G.

1 answer:

Andrei [34K]1 year ago

4 0

Answer:

The expression of gravitational field due to mass $m_!$ at a distance $l_a$

Explanation:

We have given mass is $m_1$

Distance of the point where we have to find the gravitational field is $l_a$

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by $g=\frac{Gm_1}{l_a^2}$

This will be the expression of gravitational field due to mass $m_!$ at a distance $l_a$

You might be interested in

4. Susan observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered why changes

posledela

5,10,15,20,25,30, that's how much it should have been

3 0

2 years ago

Read 2 more answers

Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4

Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= $(m+ \frac{1}{2} \lambda)$

replacing ;

Δx = 2t ; we have:

2t = $(m+ \frac{1}{2} \lambda)$

Given that thickness t = 700 nm

Then

2× 700 = $(m+ \frac{1}{2} \lambda)$ --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = $(m+ \frac{1}{2} \lambda)$ ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus; the wavelength λ of the light when it is traveling in air = 560 nm

b)

For the smallest thickness $t_{min} ; \ \ \ m =0$

∴ $2t_{min} =\frac{\lambda}{2}$

$t_{min} =\frac{\lambda}{4}$

$t_{min} =\frac{560}{4}$

$t_{min} =140 \ \ nm$

Thus, the smallest thickness t of the air film = 140 nm

7 0

2 years ago

Read 2 more answers

Fred wants to summarize mitosis in the cell cycle. Which statement describes mitosis?

salantis [7]

The statement that most accurately describes mitosis simply is <span>that mitosis is a type of cell division in which one cell (the mother) divides to produce two new cells (the daughters) that are genetically identical to itself</span>. This is the most basic textbook definition, that is summary, of what the mitosis is.

6 0

2 years ago

Read 2 more answers

If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time.

olya-2409 [2.1K]

Answer:

$F=mkv$

Explanation:

Given that

$v = v_i - kx$

We know that acceleration a given as

$a=\dfrac{dv}{dt}$

$v = v_i - kx$

$\dfrac{dv}{dt}=\dfrac{dv_i}{dt}-k\dfrac{dx}{dt}$

$\dfrac{dv}{dt}=0-k\dfrac{dx}{dt}$

We know that

$F=m\dfrac{dv}{dt}$

$F=-mk\dfrac{dx}{dt}$

$F=-mkv$

So the magnitude of force F

$F=mkv$

5 0

1 year ago

The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o

MrMuchimi

Answer:

The total kinetic energy of both particles is $2.43\times10^{-13}$

Explanation:

Given that,

Kinetic energy of nucleus $K.E= 1.24\times10^{-13}\ J$

Kinetic energy of proton $K.E= 2.47\times10^{-13}\ J$

Radius of proton $r= 0.9\times10^{-15}\ m$

We need to calculate the final potential energy

Using formula of final potential energy

$U=\dfrac{kq^2}{r}$

Put the value into the formula

$U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}$

$U_{f}=1.28\times10^{-13}\ J$

We need to calculate the initial energy of both the particles

Using formula of energy

$E_{i}=(K.E_{n}+K.E_{p})+U_{i}$

$E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0$

$E_{i}=3.71\times10^{-13}\ J$

We need to calculate the total kinetic energy of both particles

Using conservation of energy

$E_{i}=E_{f}$

$E_{i}=K.E_{f}+U_{f}$

$3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}$

$K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}$

$K.E_{f}=2.43\times10^{-13}$

Hence, The total kinetic energy of both particles is $2.43\times10^{-13}$

3 0

1 year ago