The law of suggests that the orbit of planets is not circular but .
Answers
Answer:
Kepler's first law suggests that the orbit of planets is not circular but elliptical
Explanation:
The three Kepler's laws explain the motion of the planets orbiting the Sun:
- The first law tells that the orbits of the planets around the Sun are ellipses, with the Sun located at one of the two focii
- The second law tells that a line connecting the Sun with the planet sweeps out equal areas in equal amounts of time
- The third law tells that the square of the orbital periods of the planets is proportional to the cube of their average distance from the Sun
As we can read, the first law tells us that the orbit of the planets is not circular, but elliptical.
Related Questions
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. find the velocity and speed when t = 6. f(t) = 100 + 60t − 4.5t2
By adding mass to the car, will we increase the average speed of the car?
Draw a circuit of a kettle
One particular confound for the use of an adoption study of heritability is the phenomenon of _____ placement, in which a child is placed in a home that is similar to the home of the biological parents.A) unilateralB) forcedC) matchedD) selective
auditory
circulatory
respiratory
integumentary
muscular
reproductive
Answers
Answer: circulatory
respiratory
muscular
Explanation:
The physical fitness can be define as the state of well being and health. It indicates towards the ability of the organisms to perform various aspects of physical activities such as sports, daily routine work, and occupational tasks. It is achieved by moderate to vigorous physical exercises, sufficient amount of rests and proper nutrition.
The physical fitness involves the proper functioning of the vital systems of the body these includes the respiratory, circulatory and muscular. As the body performs exercise it requires more amount of oxygen also the muscles performing activity deprive of the oxygen, the rate of respiration increases, more amount of blood is pumped out by the heart to fulfill the need of oxygen.
Circulatory
Muscular
Answers
A) The temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W is; T_i = 38.52°C
B) The thickness of the layer contained within the fur so that the bear loses heat at a rate of 51.4 W is; t = 13.41 cm
We are given;
Diameter of sphere; d = 1.6 m
Radius of sphere; r = d/2
r = 1.6/2
r = 0.8 m
Thickness of bear; t = 3.9 cm cm = 0.039 m
Outer surface Temperature of fur; T_h = 2.8 ∘C
Inner surface Temperature of fat;T_f = 30.9 ∘C
Thermal conductivity of fat; K_f = 0.2 W/m⋅k
Thermal conductivity of air; K_a = 0.024 W/m⋅k
A) To find the temperature at the fat-inner fur boundary when heat loss is 51.4 W, we will use the heat current formula;
H = K_f•A(T_f - T_i)/t
Where;
A is area = 4πr²
A = 4π × 0.8²
A = 8.04 m²
T_i is the temperature we are looking for
H is heat loss = 51.4
t is thickness
Making T_i the subject gives;
T_i = (T_f × H × t)/(K_f × A)
T_i = (30.9 × 51.4 × 0.039)/(0.2 × 8.04)
T_i = 38.52°C
B) We want to find the thickness of the layer contained within the fur. Thus, we will use K_a instead of K_f. Let us make t the subject in the heat current formula to get;
t = (K_a•A(T_i - T_h)/H
t = (0.024 × 8.04 × (38.52 - 2.8))/51.4
t = 0.1341 m
t = 13.41 cm
Read more at; brainly.com/question/14548124
Answer:
Explanation:
Using the equation
H = Q/t = k A ( T hot - T cold) / L
where H is the rate of heat loss = 51.4 W, T cold be temperature of the outer surface, A is the surface area of the fat layer which is a model of sphere ( surface area of a sphere ) = 4πr² where diameter = 1.60 m
radius = 1.60 m / 2 = 0.80 m
A = 4 × 3.142 × ( 0.8²) = 8.04352 m²
making T cold subject of the formula
T cold = T hot - = 30.9° C - ( 51.4 W × 3.9 × 10⁻² m) / ( 0.2 W/mK × 8.04352 m² ) = 30.9° C - 1.25 ° C = 29.65° C
b) The thickness of air layer for the bear to lose heat t a rate of 51.4 W
thermal conductivity of air is 0.024 W/mK and rearranging the earlier formula
L = = (0.024 W/ m K × 8.04352 m²) ( 29.65° C - 2.8°C) / 51.4 W = 0.101 m = 10.1 m
Answers
Answer:
Displacement = 1725m
Explanation:
Final answer:
Using the equations of motion, we find that the plane covers a distance of 850 meters while it is accelerating from a velocity of 50 m/s to 65 m/s at the rate of 0.5 m/s².
Explanation:
You're asking about a motion problem in physics that requires using the equations of motion to solve. In this case, the formula we would use is v² = u² + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is acceleration, and 's' is the distance covered.
Given:
Initial velocity (u) = 50 m/s
Final velocity (v) = 65 m/s
Acceleration (a) = 0.5 m/s²
We are asked to find the distance covered (s). Rearranging the formula to find 's', we get: s = (v² - u²) / 2a
Substituting the given values, the distance covered (s) = (65² - 50²) / (2*0.5) = 850 meters. So, the plane covers a distance of 850 meters while it is accelerating.
Learn more about Equations of Motion here:
#SPJ2
Answers
velocity. It's still 40 m/s forward, and it WILL be until he hits the ground.
b. protons; neutrons
c. electrons; protons
d. protons; electrons
Answers
Answer:
C.Electrons,protons.
Explanation:
Cations:
When metal loos it electrons then it will form cations.
Ex:
Anion:
When metal gain electron then it will from anions.
Ex:
So from above we can say that ,the cations have fewer electrons than protons.Protons is also called as positive charged ions and electron are also called negatively charge ions.
SO the option is C is correct.
Therefore Cations have fewer electrons than protons.
So the answer is: c. electrons; protons.
the theory of Pangaea.
c.
sea-floor spreading.
b.
the age of the Earth.
d.
the theory of Gondwana.
help please
Answers
Magnetic reversal proves seafloor spreading because we can see the polarity of the Earth’s magnetic field in rocks. As magma cools, particles in it solidify in the direction of the magnetic field. By looking at these particles, scientists can see the polarity of the magnetic field at the time the rocks were created.
Answer: