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2 years ago

10

The distance between earth and sun is 15000000km. Light takes 499 seconds to reach earth from sun. Calculate the speed of light

from the data provided. Mention the unit as well.

2 answers:

iVinArrow [24]2 years ago

7 0

To solve the problem we must know about the relationship between Speed, distance, and Time.

<h3>What is the relationship between Speed, distance, and Time?</h3>

We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

$\rm{Speed = \dfrac{Distance}{Time}$

The speed of the light is 30,060.12 km/sec.

Given to us

The distance between the earth and the sun is 15000000km
Light takes 499 seconds to reach earth from the sun.

We know that speed can be described as,

$\rm{Speed = \dfrac{Distance}{Time}$

Therefore,

<h3>What is the speed of the light?</h3>

$\text{Speed of light} = \dfrac{\text{Distance between the earth and the sun}}{\text{Time taken by the light to travel the distance}}$

Substitute the value,

$\text{Speed of light} = \dfrac{15,000,000\ km}{499\ seconds}$

$\text{Speed of light} = 30,060.12\ km/sec$

Hence, the speed of the light is 30,060.12 km/sec.

Learn more about Speed, distance, and Time:

brainly.com/question/15100898

fredd [130]2 years ago

6 0

Answer:

The speed of light is 30,060,120.24048 m/s.

Explanation:

15000000 km into m

1 km = 1000m

15000000 km = 15000000 × 1000 m

= 15000000000m

speed = distance travelled / time taken

or, speed = 15000000000/499 m/s

so, speed = 30,060,120.24048m/s

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Two solid balls (one larger, the other small) and a cylinder roll down a hill. Which has the greatest speed at the bottom and wh

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The answer is c hope this clarifies

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3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$

The frictional force on the block W₂, $F_{f2}$ ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ $F_{w2}$ = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + $\mathbf{F_{w2}}$

The frictional force from the ground, $\mathbf{F_{f1}}$ = N×μ + $\mathbf{F_{f2}}$ = P

Where;

P = The horizontal force applied to the block

P = (W₁ + $\mathbf{F_{w2}}$ ) × μ + $\mathbf{F_{f2}}$

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

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3 0

3 years ago

Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

5 0

3 years ago

The potential energy of a 35 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?

Anettt [7]

From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.

We need to find its height.

We will use the formula P.E = mgh

Therefore h = P.E / mg

where P.E is the potential energy,

m is mass in kg,

g is acceleration due to gravity (9.8 m/s²)

h is the height of the object's displacement in meters.

h = P.E. / mg

h = 14000 / 40 × 9.8

h = 14000 / 392

h = 35.7

Therefore the canon ball was 35.7 meters high.

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3 years ago

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