Answer:
All planets have an elliptical orbit
all planets have roughly the same SHAPE of orbit
Answer:

Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.

Now, substitute this into 'dF':

Now we can integrate dF over the rod.

According to funtriva.com, the piece that allows you to adjust the amount of light that's coming through the microscope is called the adjustable diaphragm. It is located under to stage (where what you are observing is placed on) and can be rotated to make the light<span> intensity change</span>
After impact velocity = 14.968 ft/s
Weight and mass of Bullet and wooden block:
Bullet: w = 1oz = 1/16 lb m = 0.001941 lb
wooden block : W = 5lb M = 0.15528 lb
velocity of block and bullet immediately after impact:
Σmv1 + ΣImp = mv2
Resolving vertical component
( m× v₀cos30⁰) + 0 = ( m+M) v'
v' = ( m× v₀cos30⁰)/ (m+M)
v' = 14.968 ft/s
Horizontal and vertical component of the impulse exerted by block on the bullet:
Here we will apply the principle of impulse and momentum.
Horizontal component:
-mv₀ cos30⁰ + RxΔt =0
RxΔt = mv₀sin30⁰
= 0.001941 × 1400sin30⁰
RxΔt = 1.3587 lb.s
Vertical component:
-mv₀cos30⁰ + RyΔt = -mv'
RyΔt = m( v₀cos30⁰-v')
RyΔt = 0.001941(1400cos30⁰ - 14.968)
= 2.32 lb.s
Learn more about impact here:
brainly.com/question/15008937
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Answer:
The translational kinetic energy is 225 J
The rotational kinetic energy is 225 J
Explanation:
Given;
mass of the wheel, m = 2-kg
linear speed of the wheel, v = 15 m/s
Transnational kinetic energy is calculated as;
E = ¹/₂MV²
where;
M is mass of the moving object
V is the velocity of the object
E = ¹/₂ x 2 x (15)²
E = 225 J
Rotational kinetic energy is calculated as;
E = ¹/₂Iω²
where;
I is moment of inertia
ω is angular velocity

E = ¹/₂ x 2 x (15)²
E = 225 J
Thus, the translational kinetic energy is equal to rotational kinetic energy