Define and measure the speed
choose this
heliocentric theory = Ptolomeu
radioactivity = Becquerel
reflecting telescope - Newton (not sure)
Answer:
It accelerate but will not spin.
Explanation:
If an irregular shaped object is dropped from rest without feeling any form of air resistance it will accelerate without spinning and this is due to the fact that there is no Torque around the center of gravity
Answer:
a = - 5 m/s²
Explanation:
The acceleration of the ball can be found by using the general formula of acceleration as follows:
where,
a = acceleration of ball = ?
ΔV = change in velocity = Final Velocity - Initial Velocity
ΔV = - 0.4 m/s - 0.6 m/s = -1 m/s
Δt = time of contact with the wall = 0.2 s
Therefore, using the values in equation, we get:
Therefore, the correct option is:
<u>a = - 5 m/s²</u>
<span>There are two possible arrangements of q1,q2,and q3 in this problem. They are:
q3, 2 cm gap, q1, 2 cm gap, q2
or
q1, 2/3 cm gap, q3, 4/3 cm gap, q2
We really don't care about the absolute magnitude of q, so the fact that it's 1.00 nano Coulombs is totally irrelevant to this problem. The only thing important is the relative charge and distances between the particles.
The force exerted between two particles is expressed as
F = q1*q2/r^2.
q1,q2 = charges on the particles.
r = distance between the particles.
Depending upon the relative charge (positive or negative) the force may be either attraction, or repulsion. But since the signs of all the charges mentioned are the same, I'll assume that the force will be repulsive.
For the distance between q1 and q3 I'll use the value "r". And since q1 and q2 are 2 cm apart, for the distance between q3 and q2, I'll use the value (2-r). So we have the following equations.
Force between q1 and q3
F = q1*q3/r^2
Force between q2 and q3
F = q2*q3/(2-r)^2
Set the 2 equations equal to each other
q1*q3/r^2 = q2*q3/(2-r)^2
Substitute the known values and solve for r.
q1*q3/r^2 = q2*q3/(2-r)^2
q*q/r^2 = 4q*q/(2-r)^2
q^2/r^2 = 4q^2/(4 - 4r + r^2)
1/r^2 = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4r^2/(r^2(4 - 4r + r^2))
0 = (4r^2-(4 - 4r + r^2))/(r^2(4 - 4r + r^2))
0 = (4r^2 - 4 + 4r - r^2)/(r^2(4 - 4r + r^2))
0 = (3r^2 - 4 + 4r)/(r^2(r-2)(r-2))
Now let's look at the numerator and denominator of the expression and see where we can get a value of 0. The denominator is allowed to have any value EXCEPT 0 and that will occur at r = 2, or r=0. And nothing else in the denominator will help the expression become 0. But if the numerator is 0, then the expression is 0. So let's see at what values the numerator is 0. Using the quadratic formula with A=3, B = 4 and C=-4, we get zeros at r = -2 and r = 2/3. Both of those values make sense. If r = -2, that means that the charges are arranges q3, q1, q2 with q1 being 2 cm from q1 and 4 cm from q2. And for r = 2/3, that also makes sense with the charges being arranged q1, q3, q2 and q3 is 2/3cm from q1 and 4/3cm from q2. In both cases, q3 is twice as from from q2 as it is from q1.</span>
Answer:
nearly 2 days or less hes fast but not that fast
Explanation:
but maybe he can run it in five minutes