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3 years ago

Astronomers initially had difficulty identifying the emission lines in quasar spectra at optical wavelengths because

Physics

1 answer:

Rus_ich [418]3 years ago

6 0

No one expected violet & ultraviolet spectral lines to be shifted towards the red.

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Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a

mamaluj [8]

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

6 0

2 years ago

Which of the following are examples of energy being conserved? Choose all that apply A light bulb constantly changes electrical

Degger [83]

A plant collects sunlight to form glucose, and your friend proposes an idea for a fan. Conserved = saving

4 0

3 years ago

Read 2 more answers

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago

A ball on a cart is moving at a rate of 2 m/s. The cart suddenly stops and the ball continues to travel in the same direction at

sukhopar [10]

Answer:

The first

Explanation:

Cause it will continue in motion till another force is applied

7 0

3 years ago

Read 2 more answers

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a

Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration= $88.6m/s^2$

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution= $2\pi rad$

2.59 rev= $2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s$

By using $\pi=3.14$

Angular velocity= $\omega=16.27rad/s$

Distance from axis=r=0.329 m

Tangential speed= $r\omega=16.27\times 0.329=5.4m/s$

Radial acceleration= $\frac{v^2}{r}$

Radial acceleration= $\frac{(5.4)^2}{0.329}=88.6m/s^2$

6 0

3 years ago