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2 years ago

The rainbow is a result of a. Diffraction of the light.

1 answer:

5 0

Rainbows are caused by the dispersion of light, which itself consists of a combination of refraction and reflection of light around little droplets of water.

Choice C

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The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

$\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}$

Remove units to make the equation easier to read.

$\displaystyle 8.11045189 = \frac{v^2}{7008100}$

Multiply both sides by 7,008,100.

$56838857.89=v^2$

Take the square root of both sides.

$v=7539.154985$

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

5 0

2 years ago

The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the ele

Blizzard [7]

Answer:

The magnitude of the electric field intensity is $E = 7.89 *10^{6} \ V/m$

Explanation:

From the question we are told that

The voltage is $\epsilon = 72.7 \ mV = 72.7 *10^{-3} V$

The thickness of the membrane is $t = 9.22 \ nm = 9.22 *10^{-9} \ m$

Generally the electric field intensity is mathematically represented as

$E = \frac{\epsilon }{t}$

substituting values

$E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}$

$E = 7.89 *10^{6} \ V/m$

8 0

3 years ago

What name is used for a high-power variable resistor?

const2013 [10]

The answer would be C: Rheostat. :)

3 0

2 years ago

Read 2 more answers

On a windy day, the wind turbine transfers 78 W of power. Calculate the amount of energy the turbine transfers in 10s.

tangare [24]

Answer:

Energy = 780 Joules

Explanation:

Given the following data;

Power = 78 Watts

Time = 10 seconds

To find the energy transferred;

Energy refers to the amount or quantity of power which is being consumed by an individual, group of people or organization over a specific period of time.

Mathematically, energy is given by the formula;

Energy = power * time

Energy = 78 * 10

Energy = 780 Joules

Therefore, the amount of energy the turbine transferred is 780 Joules

4 0

2 years ago

An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.

bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

p = 1.7 10⁻⁹ 6.8 10⁻⁶

p = 1.16 10⁻¹⁴ C m

The potential energie dipole is described by the expression

U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

θ = 0º

U = 1.16 10⁻¹⁴ 1160 cos 0

U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

θ = 180º

cos 180 = -1

U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

ΔU = | U1 -U2 |

ΔU = 1.35 10-11 - (-1.35 10-11)

ΔU = 2.7 10 -11 J

6 0

2 years ago