Current what can do work in an electric device

When the distance between two stars decreases by one-third, the force between them

pashok25 [27]

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

$r'=\frac{2}{3}r$

So the new force will be

$F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F$

So, the force will be 2.25 times the previous value.

4 0

3 years ago

How is the frequency of an electromagnetic wave related to energy

aivan3 [116]

Answer: The energy carried by an electromagnetic wave is proportional to the frequency of the wave.

Explanation:

Electromagnetic waves that are of higher energy than visible light (higher frequency, shorter wavelength) include ultraviolet light, X-rays, and gamma rays.

7 0

3 years ago

A star with the same mass and diameter as the sun rotates about a central axis with a period of about 24.0 days. Suppose that th

Mila [183]

Answer:

a) w = 2.52 10⁷ rad / s, b) K / K₀ = 1.19 10⁴

Explanation:

a) We can solve this exercise using the conservation of angular momentum.

Initial instant. Before collapse

L₀ = I₀ w₀

Final moment. After the collapse

L_f = I w

angular momentum is conserved

L₀ = L_f

I₀ w₀ = I w (1)

The moment of inertia of a sphere is

I = 2/5 m r²

we take from the table the mass and diameter of the star

m = 1,991 10³⁰ kg

r₀ = 6.96 10⁸ m

r = 6.37 10⁶ m

to find the angular velocity let's use

w = L / T

where the length of a circle is

L = 2π r

T = 24 days (24 h / 1 day) (3600 s / 1h) = 2.0710⁶ s

we substitute

w = 2π r / T

wo = 2π 6.96 10⁸ / 2.07 10⁶

wo = 2.1126 10³ rad / s

we substitute in equation 1

w = $\frac{I_o}{I}$

w = 2/5 mr₀² / 2/5 m r² w₀

w = ( $\frac{r_o}{r}$ ) ² wo

w = (6.96 10⁸ / 6.37 10⁶) ² 2.1126 10³

w = 2.52 10⁷ rad / s

b) the kinetic energy ratio

K = ½ m w²

K₀ = ½ m w₀²

K = ½ m w²

K / K₀ = (w / wo) ²

K / K₀ = 2.52 10⁷ / 2.1126 10³

K / K₀ = 1.19 10⁴

7 0

3 years ago

The mass of Earth is 5.972 x 1024 kg and its orbital radius is an average of 1.496 x 1011 m. Calculate its linear momentum, give

Anna007 [38]

Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s

Explanation:

Given that;

mass of Earth m = 5.972 x 10²⁴ kg

radius r = 1.496 x 10¹¹ m

period t = 3.15 x 10⁷ s

now we know that Earth rotates in a circular path so the distance travelled per rotation is;

d = 2πr we substitute

d = 2π × 1.496 x 10¹¹ m

= 9.4 × 10¹¹ m

Now formula for speed v is;

v = d/t

we substitute

v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s

v = 2.98 × 10⁴ m/s

now we determine the linear momentum p

linear momentum p = mv

we substitute

p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)

p = 1.78 × 10²⁹ kg.m/s

Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s

8 0

3 years ago

Equipotentials are lines along which Select one: a. the electric field is constant in magnitude and direction. b. the electric c

soldi70 [24.7K]

Answer:

option D is correct

Explanation:

It is important to note that equipotential lines are always perpendicular to electric field lines. No work is required to move a charge along an equipotential, since ΔV = 0. Thus the work is :

W = −ΔPE = −qΔV = 0.

Work is zero if force is perpendicular to motion. Force is in the same direction as E, so that motion along an equipotential must be perpendicular to E. More precisely, work is related to the electric field by:

W = Fd cos θ = qEd cos θ = 0.

- The change in kinetic energy Δ K.E by conservation should be:

Δ K.E = W

Since, W = 0:

Δ K.E = 0

- If change in kinetic energy is zero it means that charge moves at a constant speed. Hence, option D is correct.

4 0

3 years ago