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3 years ago

What caused the big bang? WILL UPVOTE!!!

1 answer:

8 0

The expansion of the observable universe began <span>with the explosion of a single particle at a definite point in time.</span>

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Light traveling from water to a gemstone strikes the surface at an angle of 80.0º and has an angle of refraction of 15.2º . (a)

7nadin3 [17]

Answer:

The the speed of light in the gemstone is $v= 0.599*10^8 m/s$

The unreasonable thing about this is that the speed of ligth in the gemstone is too low

the speed is 19% of the speed of light which is very low

One main unreasonable or inconsistent factor is that the assumption that the differnce between the angle of incidence and angle of refraction is very large

Explanation:

From the question we are told that

The angle of incidence $i = 80^o$

The angle of refraction $r = 15.2^o$

From Snell's law we have ,

$n_1 sin \theta_1 = n_2 sin \theta_2$

Where $n_1$ is the refractive index of the first medium (water) with a constant value of $n_1 = 1..333$

$n_2$ is the refractive index of the second medium (gem stone)

$\theta_1$ is the angle between the beam and perpendicular surface of the first medium

$\theta_2$ is the angle between the beam an the perpendicular surface of the second medium

Making the $n_2$ the subject of the formula

$n_2 = n_1 \frac{sin \theta_1}{sin \theta_2}$

$= (1.333)(\frac{sin (80.0)}{sin 15.2} )$

$= 5.007$

Generally refractive index of a material is mathematically represented

$n = \frac{c}{v}$

Where c is the speed light

v is the speed of light observed in a medium

Making v the subject

$v = \frac{c}{n}$

substituting value for gem stone

$v = \frac{3.0*10^8}{5.007}$

$v= 0.599*10^8 m/s$

7 0

3 years ago

Anne pushed a cart by 12 N. How far did the cart move if the work done was 9 J?

telo118 [61]

Force=12N
Work=9J

$\\ \sf\longmapsto W=Fd$

$\\ \sf\longmapsto d=\dfrac{W}{F}$

$\\ \sf\longmapsto d=\dfrac{9}{12}=0.72m$

6 0

3 years ago

Read 2 more answers

UN BARCO NAVAL ENVÍA UNA SEÑAL A UN SUBMARINO QUE SE ENCUENTRA EN EL MAR DEBAJO DEL BARCO, SI ESTA SEÑAL TIENE UNA LONGITUD DE O

VLD [36.1K]

NOooOoIooOOi Hsushrhndndnrbrbr rgehrbhrhrbr r he rjrnd

4 0

3 years ago

A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per

FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0

3 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0

3 years ago