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3 years ago

8

What is the energy in space left over from the big bang called

1 answer:

Lilit [14]3 years ago

5 0

It is called the CMBR, which stands for cosmic microwave background radiation. It was discovered by Arno Penzias and Robert Wilson in 1964.

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Question 6. B) and c)

maks197457 [2]

The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds

4 0

4 years ago

Stages of global warming

ddd [48]

Answer:

- Glaciers melt

- The seas rise

- Temperature changes

- Humans add heat trapping greenhouse gasses

7 0

4 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

a

$\theta _2 = 13^o$

b

$\theta _1 =32.94^o$

c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

3 years ago

A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this

Hunter-Best [27]

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

$m = 3500lb = 1587.57kg$

The values of the velocities are,

$\text{Initial Velocity} = V_i = 0$

$\text{Final Velocity} = V_f = 60mph = 26.822m/s$

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

$a = \frac{v_f-v_i}{t}$

$a = \frac{26.822-0}{5}$

$a = 5.36m/s^2$

Now applying the Newton's second law we have,

$F= ma$

$F = (1587.57)(5.36)$

$F = 8516.36N$

Therefore the approximate magnitude is 8516.36N

5 0

4 years ago

A sleeping 68 kg man has a metabolic power of 79 w .

Lesechka [4]

<span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>

3 0

4 years ago