Newton’s Third Law of Motion states that for every action there is an equal and opposite reaction. So look for a scenario in which something had force applied upon it and the reaction is a force in the opposite direction of the same size.
Answer:
B) 
Explanation:
The electric force between charges can be determined by;
F = 
Where: F is the force, k is the Coulomb's constant,
is the value of the first charge,
is the value of the second charge, r is the distance between the centers of the charges.
Let the original charge be represented by q, so that;
= 2q
= 
So that,
F = 
x 
= 2q x
x 
=
x 
=
x 
F =
x 
The electric force between the given charges would change by
.
Answer:
Kinetic energy = (1/2) (mass) (speed²)
Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules
Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules
Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules
Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts .
Explanation:
Dont report my answer please
Incomplete Question.The Complete question is
The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km
(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.
(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?
Answer:
(i) KE= 2.56e29 J
(ii) KE= 2.65e33 J
Explanation:
i) Treating the Earth as a solid sphere, its moment of inertia about its axis is
I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²
I = 9.69e37 kg·m²
About its axis,
ω = 2π rads/day * 1day/24h * 1h/3600s
ω= 7.27e-5 rad/s,
so its rotational kinetic energy
KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²
KE= 2.56e29 J
(ii) About the sun,
I = mR²
I= 5.97e24kg * (1.496e11m)²
I= 1.336e47 kg·m²
and the angular velocity
ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s
ω= 1.99e-7 rad/s
so
KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²
KE= 2.65e33 J
the focal length <span> is much more decent for a concave, and also worse</span><span> for a convex mirror. When the image that is given, distance is good and decent, images are always on the same area of the mirror as the object given , and it is not fake. images distance is </span>never positive <span>, the image is on the oppisite side of the mirror, so the image must be virtual.</span>