It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
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 acceleration of the ball after leaving the hand is 
downward
Explanation:
In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.
After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is
where m is the mass of the ball and g is the acceleration of gravity (
), acting in the downward direction.
According to Newton's second law, the acceleration of the ball is given by
where
is the net force acting on the ball
After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:
This means that the acceleration of the ball remains downward for the entire motion, after leaving the hand.
Learn more about Newton's second law:
brainly.com/question/3820012
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Answer:
it ends when clouds above start to break apart. Some tornadoes only last seconds. Others can last much longer. They come in many shapes and sizes.
Answer:
Explanation:
= Length of wire = 65 m
= Initial current = 1.8 A
= Final current = 2.9 A
We know
and
so
The length of the wire remaining on the spool is .
