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Vilka [71]
3 years ago
9

When the moon orbits the Earth, it has velocity. Inertia would make the moon continue in a straight line instead of orbiting at

that velocity. The fact that it orbits at the same distance t tells you that ________
Physics
1 answer:
Sholpan [36]3 years ago
8 0

Answer:

...there is an external force causing the moon not to continue in a straight line.

Explanation: According to Newton's first law of motion, every object will continue in its state of rest or uniform motion along a straight line unless acted upon by an external force.

The  external force causing the moon not to continue in a straight line is the force of gravitational attraction between the earth and the moon. This is in accordance to Newton's universal law of gravitation which states that every object attracts another object with a force that is directly proportional to the product of the masses of the objects and inversely proportional the the square of the distance between them. Also in this case, the gravitational pull between the moon and the earth is equivalent to the centripetal force that keeps the moon towards the earth, the earth being the center of motion for the moon. Recall that if the moon is orbiting the earth, then certainly the earth becomes the center of its motion.

Therefore the moon will only continue in a straight line in absence of the earth's gravitational pull.

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The carbon cycle relies on producers and consumers to survive. What is the role of each in the carbon cycle?
LiRa [457]
ANSWER: (ROLE of Producers in the cycle:totake in carbon dioxide from the atmosphere during photosynthesis and release carbon dioxide during respiration,ROLE of Consumers:to contribute to the carbon cycle when they perform respiration and expel carbon dioxide into the atmosphere)
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8 0
3 years ago
Un automovil transita por una curva en forma de U y recorre una distancia de 400m en 30s sin embargo su posición final está a so
pochemuha

Answer:

Definimos:

Rapidez media es igual al cociente entre la distancia recorrida y el tiempo que se tarda en recorrer esa distancia.

En este caso la distancia recorrida es 400m, y el tiempo que se tarda es 30s, entonces la rapidez media va a ser:

RM = 400m/30s = 13.33 m/s

La velocidad media por otro lado, es igual al cociente entre el desplazamiento y el tiempo necesario para desplazarse.

El desplazamiento es igual a la distancia entre la posición final y la posición inicial, que en este caso eso 40m, y el tiempo necesario sigue siendo 30s, entonces la velocidad media va a ser:

VM = 40m/30s = 1.33 m/s

8 0
3 years ago
A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h
babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

 f’= f₀  \frac{343}{343 + v_s}

Explanation:

This is a doppler effect exercise, where the sound source is moving

           f = fo \frac{v}{v-v)s}      when the source moves towards the observer

           f ’=f_o  \frac{v}{v+v_{sy}}  Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

          f’= f₀  \frac{343}{343 + v_s}

8 0
3 years ago
Describe the climate, landforms,and existing plant and animal life during the cretaceous period
Fiesta28 [93]
The climate<span> was generally warmer and more humid than today, probably because of very active volcanism associated with unusually high rates of seafloor spreading. 
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8 0
3 years ago
Read 2 more answers
A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra
bija089 [108]

Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is 1.28\frac{kg}{m^3}.

v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

7 0
3 years ago
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