A black hole can be considered a star that has...

Convert the speed of light 3.0x10^8 m/s to km/day

Aleksandr [31]

Answer: $2.592 \times 10^{8}km/day$

$1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day$

7 0

3 years ago

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PLEASE HELP: Why are shooting stars bright?

mrs_skeptik [129]

the answer is c I hope this helps

4 0

3 years ago

A light ray is incident from air into glass (ng = 1.52) then onto water (nw= 1.33). The wavelength of light in air (na= 1) is ai

gregori [183]

The answer is is b hope this helped

3 0

3 years ago

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an

Lemur [1.5K]

Answer:

q=1.7346×10⁻⁶C

Explanation:

Since the electric field is perpendicular to the bottom and top of the cube,the total flux is equals the flux over the top of surface plus the flex over the lower surface

Ф(total)=Ф₃₀₀+Ф₂₃₀

But the flux is given by Ф=E.A=EACos(θ) where θ is the angle between Area vector and electric field

So

Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)

Ф(total)=A(E₃₀₀ - E₂₃₀)

The total flux is given by Gauss Law as:

Ф(total)=q/ε₀

q=ε₀Ф(total)

q=ε₀(A(E₃₀₀ - E₂₃₀))

Substitute the given values

q=(8.85×10⁻¹²){(70²)(100 - 60)}

q=1.7346×10⁻⁶C

7 0

3 years ago

Hola tengo un taller de física y nose como resolver este pregunta

Blizzard [7]

a) 0.26 h

b) 71.4 km

Explanation:

a)

In order to solve the problem, we have to know what is the final velocity of the car.

Here, we assume that the final velocity reached by the car is

$v=300 km/h$

Therefore, we can find the time taken by the car to reach this velocity by using the suvat equation:

$v=u+at$

where:

u = 250 km/h is the initial velocity

$a=190 km/h^2$ is the acceleration of the car

v = 300 km/h is the final velocity

t is the time

Solving for t, we find:

$t=\frac{v-u}{a}=\frac{300-250}{190}=0.26 h$

b)

In order to find the distance covered by the car, we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where:

s is the distance covered

u is the initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 250 km/h

t = 0.26 h (calculated in part a)

$a=190 km/h^2$

Therefore, the distance covered is

$s=(250)(0.26)+\frac{1}{2}(190)(0.26)^2=71.4 km$

8 0

3 years ago