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Allisa [31]
4 years ago
6

Which statement best describes the density of the outer planets?

Physics
2 answers:
ICE Princess25 [194]4 years ago
7 0
Yes it is the only planet
gavmur [86]4 years ago
5 0

Answer:

Saturn is mainly composed of the lightest two gases known, hydrogen and helium. It is the only planet in our solar system whose density is less than water.

Explanation:

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How to determine my slope in physics graph​
Hatshy [7]

Answer:

Determine the coordinates of two points on the line. Calculate the difference between these two locations' y-coordinates (rise). Calculate the x-coordinate difference between these two places (run). Divide the y-coordinate difference by the x-coordinate difference (rise/run or slope).

5 0
2 years ago
A cannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and height of 12 m.
eimsori [14]

1) x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2

The initial data of the projectile are:

y_0 = 12 m is the initial height

v_0 = 32 m/s is the initial speed of the projectile, so its components along the x- and y- directions are

v_{0x} = v_0 cos \theta = (32 m/s)(cos 43^{\circ})=23.4 m/s\\v_{0y} = v_0 sin \theta = (32 m/s)(sin 43^{\circ})=21.8 m/s

The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:

x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2

2) 4.94 s

To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:

0=12+21.8t-4.9 t^2

Solving the equation with the formula, we have:

t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}

which has two solutions:

t = -0.50 s

t = 4.94 s

Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is

t = 4.94 s

3) 115.6 m

To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:

x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m

4) 2.22 s

The cannonball reaches its maximum height when the vertical velocity becaomes zero.

The vertical velocity at time t is given by

v_y(t)= v_{0y} -gt

where

g = 9.8 m/s^2 is the acceleration due to gravity

Substutiting v_y(t)=0 and solving for t, we find

t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s

5) 36.2 m

The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:

y(t)=y_0 + v_{0y}t-\frac{1}{2}gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m

3 0
4 years ago
Why doesn't the principle of mechanical energy conservation hold in situations when frictional forces are present?
vodomira [7]
When you are talking about the Principle of mechanical Energy Conservation, it is really only including the kinetic and potential energy in a total system. When frictional forces are present, although the conservation of energy law is still present, it does not work when it comes to the conservation of mechanical energy as there is another type of energy that is factored in. As friction acts on the object, that transition from potential to kinetic as it slide/falls will be completely different as some of that energy is being transformed into thermal energy. Which breaks the conservation of mechanical energy.
7 0
3 years ago
If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric wi
ahrayia [7]

Answer:

The max. Energy that can be stored in the neoprene rubber capacitor will be 0.304J

Explanation:

Detailed explanation and calculation is shown in the image below

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The bigger the starting force, the further the distance moved by the car?
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