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4 years ago

A main sequence star cools and expands. It must be entering which stage?

1 answer:

4 0

If a main sequence star is cooling and expanding, it is entering the red giant stage. This means that the star has burned up all of it's hydrogen and is now starting to burn its helium making it cooler.

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two plane mirrors inclined at an angle of 50 degree to each othe r a ray is incident on mirror1 then thedeviation x of ray after

Kipish [7]

The collapsed answer of Penchalreddy Badepalli is correct. The composition of two reflections via two mirror making an angle \alpha is equivalent to a single rotation by an angle 2\alpha, hence 2 * 60 deg = 120 deg. And turns is independent of the absolute orientation of the two mirrors in space and/or the direction of incidence of the incoming ray.

One could use elementary geometry to prove this (if you presume the direction of incidence is irrelevant imagine hitting the first mirror at 90 deg, then going retro right back along the normal to the first mirror, and follow the directions).

6 0

4 years ago

A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin

Greeley [361]

Answer:

The equivalent resistance of the combination is R/100

Explanation:

Electric Resistance

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

$\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}$

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

$\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}$

This sum is repeated 10 times. Operating each term:

$\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}$

All the terms have the same denominator, thus:

$\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}$

Taking the reciprocals:

$R_e=R/100$

The equivalent resistance of the combination is R/100

6 0

3 years ago

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri

igor_vitrenko [27]

Answer:

a) the particles are 0.217 m apart

b) the particles are moving in the same direction.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

= -0.155 A + 0.372 A

= 0.217 A

since A = 1 m

Thus,

Δx = 0.217 m

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

= (-πA / T) sin(2πt / T)

= -(π(1) / 1.5) sin(2π(0.45) / 1.5)

= -1.99

and,

v₂ = dx₂ / dt

= (-πA / T) sin((2πt / T) + π/6)

= -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

= -1.40

Since both v₁ and v₂ are negative, this shows that the particles are moving in the same direction.

6 0

3 years ago

Two objects, with different sizes, masses, and temperatures, are placed in thermal contact. in which direction does the energy t

s344n2d4d5 [400]

(c) energy travels from the object at higher temperature
to the object at lower temperature.

Size and mass have no effect.

3 0

3 years ago

: A honeybucket man is carrying his load. He has a pole 2 m long with a bucket hanging from each end. The buckets have a mass of

nasty-shy [4]

Answer:

Explanation:

Using the principle of moment, assuming the rod is uniform rod of mass 1 kg

the center of mass of the rod will be at 1 m

assuming the system is in equilibrium,

clockwise moment = anticlockwise moment

let the distance of the man shoulder be x from the center of gravity and also is the pivot point

total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and

2kg + 5 kg = 7 kg for front bucket

( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)

5 + 5 x + x = 7 - 7x

5 + 6x = 7 - 7x

6x + 7x = 7 - 5

13x = 2

x = 2 / 13 = 0.154 m

the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).

8 0

4 years ago