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AnnyKZ [126]
2 years ago
11

How does nuclear fusion affect the life cycle of a star?.

Physics
1 answer:
Fittoniya [83]2 years ago
4 0

Answer:

As the main sequence star glows, hydrogen in its core is converted into helium by nuclear fusion. When the hydrogen supply in the core begins to run out, and the star is no longer generating heat by nuclear fusion, the core becomes unstable and contracts

Explanation:

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morpeh [17]
1)a 2)D 3)a. I think the answers are
4 0
3 years ago
What is the velocity of a wave with a frequency of 45 Hertz and a wavelength of 3 meters?
Juli2301 [7.4K]

Explanation:

By using v=( f )x( lambda )

v= 45 s^-1 x 3 m

Therefore v = 135 ms^-1

8 0
2 years ago
Please help ASAP please
Alex_Xolod [135]

Answer:

14

Explanation:

6 0
2 years ago
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a
frosja888 [35]

Answer:

a) Yes

b) 7 rad/s

c) 0.01034 J

Explanation:

a)

Yes the angular momentum of the block is conserved since the net torque on the block is zero.

b)

m = mass of the block = 0.0250 kg

w₀ = initial angular speed before puling the cord = 1.75 rad/s

r₀ = initial radius before puling the cord = 0.3 m

w = final angular speed after puling the cord = ?

r = final  radius after puling the cord = 0.15 m

Using conservation of angular momentum

m r₀² w₀ = m r² w

r₀² w₀ = r² w

(0.3)² (1.75) = (0.15)² w

w = 7 rad/s

c)

Change in kinetic energy is given as

ΔKE = (0.5) (m r² w² - m r₀² w₀²)

ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)

ΔKE = 0.01034 J

6 0
3 years ago
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
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