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olchik [2.2K]
3 years ago
5

Light from the sun reaches the earth in 8.3 minutes. the speed of light is 3.0  108 m/s. in kilometers, how far is the earth fr

om the sun?
Physics
1 answer:
Flura [38]3 years ago
6 0
To determine the distance of the light that has traveled given the time it takes to travel that distance, we need a relation that would relate time with distance. In any case, it would be the speed of the motion or specifically the speed of light that is travelling which is given as 3x10^8 meters per second. So, we simply multiply the time to the speed. Before doing so, we need to remember that the units should be homogeneous. We do as follows:

distance = 3x10^8 m/s ( 8.3 min ) ( 60 s / 1 min ) = 1.494x10^11 m

Since we are asked for the distance to be in kilometers, we convert

distance = 1.494x10^11 m ( 1 km / 1000 m) = 149400000 km
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2 years ago
If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?
Alisiya [41]

Answer:

The spring constant is 3750 N/m  

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

=>

(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m  

4 0
3 years ago
a block measures 3.5 cm long 2.8 cm wide and 1.6 cm deep. the density of the block 2.5 g/cm. calculate the volume of the block.
luda_lava [24]
Volume of a block can be found by: length × width × height. So:

3.5cm × 2.8cm × 1.6cm = 15.68cm^3
6 0
3 years ago
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The answer is A. locations by the ocean typically do not get as cold in the winter or as hot in the summer as locations that are located inland.
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Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
Angelina_Jolie [31]

Answer:

a). \frac{\dot{W}}{m}= 311 kJ/kg

b). \frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0

\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})

\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})

\frac{\dot{W}}{m}= -30+1.1(980-670)

\frac{\dot{W}}{m}= 311 kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}

\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}

\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978

\frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

5 0
3 years ago
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