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Slav-nsk [51]
3 years ago
8

A friend of yours who has not taken astronomy sees a meteor shower (she calls it a bunch of shooting stars). The next day she co

nfides in you that she was concerned that the stars in the Big Dipper (her favorite star pattern) might be the next ones to go. How would you put her mind at ease?
Physics
1 answer:
Oxana [17]3 years ago
3 0
Tell her meteors and stars aren't related a stars life time is extremely long and meteors are just rocks floating through space when they fly be the earth they can appear like falling stars but they aren't her fave constellation wont be going away any time soon
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A house burns down. On the house across the street, all of the vinyl siding is twisted, melted, and warped because of the heat.
ivann1987 [24]

Answer:

Radiation

Explanation:

The fire from the burning house is not directly touching the house. Also not convection because there is not water involved

7 0
3 years ago
Read 2 more answers
2) Um gás ideal sofre uma determinada transformação, conforme mostra o gráfico abaixo. Considere
Roman55 [17]

Answer:

yes

Explanation:

5 0
3 years ago
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa
Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

5 0
3 years ago
Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
2 years ago
A car moves round a circular track of radius 0.3m of two revolution per/sec find its angular velocity.
Pie

Answer:

the angular velocity of the car is 12.568 rad/s.

Explanation:

Given;

radius of the circular track, r = 0.3 m

number of revolutions  per second made by the car, ω = 2 rev/s

The angular velocity of the car in radian per second is calculated as;

From the given data, we convert the angular velocity in revolution per second to radian per second.

\omega = 2 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 4\pi \ rad/s = 12.568 \ rad/s

Therefore, the angular velocity of the car is 12.568 rad/s.

4 0
2 years ago
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