Ask question

Ask question

All categories

5 days ago

10

The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis

, start text, A, U, end text, right parenthesis.
Assuming Neptune's orbit is circular, what is the distance it travels in a single orbit around the Sun?

You might be interested in

(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the

Softa [3030]

3 is jxhuneuxndnzixbf

7 0

1 month ago

Read 2 more answers

Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.

Yuliya22 [3333]

The time required for the water balloon to reach the ground is given as

$h = \frac{1}{2} gt^2$

Here we understand that

$h = 18 m$

$g = 9.8 m/s^2$

Now applying the earlier mentioned formula

$18 = \frac{1}{2}*9.8* t^2$

$18 = 4.9 t^2$

$t = 1.92 s$

Now in the same time frame, we can conclude the distance covered will be

$d = v_x * t$

$d = 8.2 * 1.92 = 15.7 m$

Thus, it will land at a distance of 15.7 m from where it started

5 0

2 months ago

A skateboarder is attempting to make a circular arc of radius r = 16 m in a parking lot. The total mass of the skateboard and sk

kicyunya [3294]

To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

$r = 16m$

$m = 82kg$

$\mu_s = 0.63$

The maximum velocity can be determined using centripetal force,

$F_c = \frac{mv^2}{r}$

Should be equal to,

$\frac{mv^2}{r} = \mu_s mg$

$v = \sqrt{\mu_s gr}$

$v = \sqrt{(0.63)(9.8)(16)}$

$v = 9.93m/s$

As a result, the highest speed achievable through the arc without slipping is 9.93m/s

3 0

2 months ago

Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she

Yuliya22 [3333]

This is due to the fact that below 4°c, water behaves differently than other substances and decreases in density as its temperature drops further.

8 0

1 month ago

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th

Softa [3030]

Answer:

$H = 10.05 m$

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

$\Delta t = 4.1 - 0.5 = 3.6 s$

now we have

$\Delta t = \frac{2v}{g}$

$3.6 = \frac{2v}{9.8}$

$v = 17.64 m/s$

this indicates the speed at the flagpole's top

at this point we have

$v_f - v_i = at$

$17.64 - v_i = (-9.8)(0.5)$

$v_i = 22.5 m/s$

the height of the flagpole is stated as

$H = \frac{v_f + v_i}{2}t$

$H = \frac{22.5 + 17.64}{2} (0.5)$

$H = 10.05 m$

5 0

3 months ago