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2 years ago

If a car driving down the road doubles its speed what happens to its kinetic energy?

Physics

1 answer:

ivanzaharov [21]2 years ago

7 0

Answer:

Kinetic energy doubles

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In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

katrin2010 [14]

Time period of any moon of Jupiter is given by

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

from above formula we can say that mass of Jupiter is given by

$M = \frac{4 \pi^2 r^3}{GT^2}$

now for part a)

$r = 4.22 * 10^8 m$

T = 1.77 day = 152928 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}$

$M = 1.9* 10^{27} kg$

Part B)

$r = 6.71 * 10^8 m$

T = 3.55 day = 306720 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}$

$M = 1.9* 10^{27} kg$

Part c)

$r = 10.7 * 10^8 m$

T = 7.16 day = 618624 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}$

$M = 1.89* 10^{27} kg$

PART D)

$r = 18.8 * 10^8 m$

T = 16.7 day = 1442880 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}$

$M = 1.889* 10^{27} kg$

6 0

3 years ago

En una competencia automovilística, Germán tarda 38 s en dar una vuelta a la

attashe74 [19]

Explanation:

Análisis estadístico de resultados de ensayos de pavimentos asfálticos según la ... T38 Caracterizacin dinámica de suelos granulares ... Se retira y se da vuelta la probeta

3 0

3 years ago

Compared to the mass and charge of a proton, the electron has

maxonik [38]

B. opposite charge and smaller mass

8 0

4 years ago

Please help, and show steps. Thank you very much!

Vikentia [17]

V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)

6 0

4 years ago

A rock is thrown upward with an initial velocity of 16 ft/s from an initial height of 5 ft. write a quadratic function equation

Andrei [34K]

During upward projection the final velocity is zero, and the gravitational acceleration is -10 m/s² (against the gravity).
Therefore; using the equation;
S = 1/2gt² + ut
Where s is the height h, g is gravitational acceleration, and t is the time and u is the initial velocity u, is 16 ft/s.
Thus; h= 1/2(-10)t² + 16t
We get; h = -5t² + 16t
Therefore; the quadratic equation is 5t² - 16t + h =0

5 0

3 years ago