Ask question

Ask question

All categories

2 years ago

10

If light travels 300,000 km/s how long does light reflected from mars take to reach earth when mars is 65,000,000 km away

1 answer:

azamat2 years ago

6 0

About <span>216.6 seconds if my math is correct.</span>

You might be interested in

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

2 years ago

Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)

Ostrovityanka [42]

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

3 0

2 years ago

In what state must matter exist for fusion reactions to take place?<br><br> ...?

valina [46]

It must exist in p<span>lasma state.</span>

4 0

2 years ago

Read 2 more answers

A space probe lands on a newly discovered planet. A small canister is released from the probe and falls a distance of 3 m in 0.5

ivanzaharov [21]

Answer:

24m/s²

Explanation:

Given

Distance S = 3m

Time of fall = 0.5sec

Required

Acceleration due to gravity

Using the equation of motion

S = ut+1/2gt²

Substitute the given values

3 = 0+1/2g(0.5)²

3 = 1/2(0.25)g

3 = 0.125g

g = 3/0.125

g = 24

Hence the value for the acceleration of gravity on this new planet is 24m/s²

3 0

2 years ago

1. Any push or pull is called

timama [110]

Answer:

b ok

Explanation:

7 0

2 years ago