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16 days ago

Which characteristics of Earth’s orbit are in agreement with Kepler’s second law? Check all that apply.

Physics

2 answers:

serg [2.5K]16 days ago

6 0

Explanation:

The planet's angular velocity in an elliptical orbit changes, indicating that it moves slower when it's further from the sun and accelerates when it is nearer.

1. Yes, Earth faces the highest gravitational pull when nearest to the Sun.

2. Yes, Earth covers the maximum distance in a thirty-day period when closest to the Sun.

3. Yes, Earth travels with minimal tangential speed when it is at its farthest from the Sun.

4. Yes, Earth sweeps out identical areas over equal time intervals throughout its orbit.

Kepler's second law states that the line drawn from a planet to the sun covers the same area in equal durations of time.

Maru [2.3K]16 days ago

6 0

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At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th

serg [2593]

Answer:

$8616.7468 \ kg/m^3$

Explanation:

The measurement of pressure is indicated as $p=\rho gh$ where p denotes the pressure, $\rho$ signifies density, and h represents height

Given values include pressure $p=9.891\times 10^4\ Pa$ , gravity's acceleration $g=9.9870\ m/sec^2$ , and height =1.163 m

$\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3$

5 0

8 days ago

Read 2 more answers

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Yuliya22 [2438]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

It is known that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v are constant on both sides. Therefore we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let the final pressure be P and the temperature $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide equation (2) by equation (3)

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, the left side temperature equals 1.48 times the right side temperature

6 0

12 days ago

-. What is the acceleration of 4 kg trolling bag pulled by a girl with a<br> force of 3 N?

inna [2210]

Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

Force(F) = 3 N

Mass of object(m) = 4 kg

Find:

Acceleration(a)

Computation:

Force(F) = ma

3 = (4)(a)

Acceleration(a) = 3/4

Acceleration(a) = 0.75 m/s²

3 0

1 month ago

An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista

Sav [2230]

Answer:

h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

when the object descends, its weight matches the spring's force

weight = spring force

mg = ky... equation 1

potential energy stored in a stretched spring = work done by the spring

mgh = 0.5 x k x h^{2}....equation 2

Substituting from equation 1 into equation 2

kyh = 0.5 x k x h^{2}

y = 0.5 x h

2y = h

where y is 6.4, yielding the maximum elongation as

h = 2 x 6.4 = 12.8 cm

6 0

26 days ago

Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme

Ostrovityanka [2208]

Upon comparison, it is evident that the student with the highest percent error is A. Student 4, who measured 9.61 m/s². Four students recorded the acceleration of gravity, with the accepted local value being 9.78 m/s². Now let's find out which student's measurement exhibited the greatest percent error.

4 0

23 hours ago