If you are measuring how far a ball traveled in an experiment which example below could be an

The space shuttle travels at about 28,000 km per hour. Using that information, estimate how many hours it will take the shuttle

jasenka [17]

It depends what is the position of earth and saturn. Distance from earth to saturn varies depending on whether earth is between sun and saturn or sun is between earth and saturn. Obviously, the shortest distance will be if earth is between sun and saturn. we will take that the distance between earth and saturn is:

s = 1 275 000 000 km
The time required to travel that distance is:
t = s/v = 45535 hours or 1897.3 days

3 0

3 years ago

The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

3 years ago

A small hot-air balloon is filled with 1.01×106 l of air (d = 1.20 g/l). as the air in the balloon is heated, it expands to 1.10

lara31 [8.8K]

<span>Volume of air in the balloon 1.01 x 10^6 L
Density of air is 1.20 g/l
Mass = Density X Volume

So mass of the air in the Balloon= ( 1.01 x 10^6) X 1.20 = 1.212 x 10^6 g
As the air is heated, the volume of air in the balloon expands to 1.10x 10^6 L Density= Mass/ voume So the Density of heated air = 1.212 x 10^6/ 1.10x 10^6 = 1.101 g/l The answer is 1.101 g/l.</span>

5 0

3 years ago

A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about

Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity Vₓ is 0

The accelerationis given by =1000 m/s²