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Lady bird [3.3K]
2 years ago
7

Kepler’s first law states that:

Physics
1 answer:
Likurg_2 [28]2 years ago
6 0

Answer:

Kepler's first law means that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. ... It is zero for a perfect circle.

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Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in
Elanso [62]

Answer:

kinetic energy

Explanation:

a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.

4 0
2 years ago
Read 2 more answers
A constant force of 11.8 N in the positive x direction acts on a 4.7-kg object as it moves from the origin to the point (1.6i –
zhenek [66]

Answer:

W = 18.88 J

Explanation:

Given that,

Constant force, F = 11.8 N (in +x direction)

Mass of an object, m = 4.7 kg

The object moves from the origin to the point (1.6i – 4.6j) m

We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :

W=F{\cdot} d\\\\W=(11.8i){\cdot} (1.6i-4.6j)\\\\=11.8\times 1.6\\\\=18.88\ J

So, the work done by the given force is 18.88 J.

5 0
2 years ago
A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w
Zigmanuir [339]
Radial acceleration is given by

a_{rad}= \frac{v^2}{r}
where 

v=r \omega
then

a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2

Now

70\omega^2=3.90 \frac{m}{s^2}  \\  \\ \omega= \sqrt{ \frac{3.9}{70} }

Using the relation

\omega=2 \pi f

2 \pi f= \sqrt{ \frac{3.9}{70} }\\  \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz

Putting into rpm

\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm

8 0
3 years ago
As a system expands, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pres
Kaylis [27]

Answer:

Vi = 0.055 m³ = 55 L

Explanation:

From first Law of Thermodynamics, we know that:

ΔQ = ΔU + W

where,

ΔQ = Heat absorbed by the system = 52.5 J

ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

52.5 J = - 102.5 J + W

W = 52.5 J + 102.5 J

W = 155 J

Now, the work done in a constant pressure condition is given by:

W = PΔV

W = P(Vf - Vi)

where,

P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³

Vi = Initial Volume of System = ?

Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

Vi = 0.058 m³ - 155 J/50662.5 Pa

Vi = 0.058 m³ - 0.003 m³

<u>Vi = 0.055 m³ = 55 L</u>

7 0
3 years ago
A. 24.89<br> B. 25.89<br> C. 17.74<br> D. 19.73
Veronika [31]

Answer: D

Explanation:

Just did it got an 100

5 0
2 years ago
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