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The orbital radius of the Earth (from Earth to Sun) is 1.496 x 10^11 m.

mrs_skeptik [129]

Explanation:

The orbital radius of the Earth is $r_1=1.496\times 10^{11}\ m$

The orbital radius of the Mercury is $r_2=5.79 \times 10^{10}\ m$

The orbital radius of the Pluto is $r_3=5.91 \times 10^{12}\ m$

We need to find the time required for light to travel from the Sun to each of the three planets.

(a) For Sun -Earth,

Kepler's third law :

$T_1^2=\dfrac{4\pi ^2}{GM}r_1^3$

M is mass of sun, $M=1.989\times 10^{30}\ kg$

So,

$T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s$

(b) For Sun -Mercury,

$T_2^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.79 \times 10^{10}\ m\\\\T_2=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.79 \times 10^{10}}\ m\\\\T_2=1.31\times 10^{-4}\ s$

(c) For Sun-Pluto,

$T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s$

8 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time

Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

4 0

3 years ago

Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the waterÂs surface to knoc

sergij07 [2.7K]

Answer:

.012

Explanation:

Take the mass of the fish and divide it by the mass of the water:

65/.30=216.667

Divide the given speed by the value we found above:

2.5/216.667=.0115

Answer can be rounded up to .012

3 0

2 years ago

Read 2 more answers

A device that has the capacity to receive and store electrical energy is a(n)

sergeinik [125]

The answer to your question is capacitor :)

6 0

2 years ago

Read 2 more answers

A blue ball is thrown upward with an initial speed of 21.8 m/s, from a height of 0.9 meters above the ground. 2.7 seconds after

worty [1.4K]

I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²

For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.

2. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds

Thus, the two balls would be at the same height after 2.25 seconds.

3 0

3 years ago

Here is it u can see it​

Here is it u can see it