Answer: 14. 49 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
Where:
is the horizontal distance between the cannon and the ball
is the cannonball initial velocity
since the cannonball was shoot horizontally
is the time
is the final height of the cannonball
is the initial height of the cannonball
is the acceleration due gravity
Isolating from (2):
(3)
(4)
(5)
Substituting (5) in (1):
(6)
Finally:
Answer:
Explanation:
Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.
Let T be tension in the hanging string
T cosθ = mg ( for balancing in vertical direction )
for balancing in horizontal direction
Tsinθ = F ( F is force of repulsion between two charges sphere)
Dividing the two equations
Tanθ = F / mg
tan17 = F / (7.1 x 10⁻³ x 9.8)
F = 21.27 x 10⁻³ N
if q be charge on each sphere , force of repulsion between the two
F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17 = .41 m )
21.27 x 10⁻³ = (9 X 10⁹ x q²) / .41²
q² = .3973 x 10⁻¹²
q = .63 x 10⁻⁶ C
no of electrons required = q / charge on a single electron
= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .39375 x 10¹³
3.9375 x 10¹² .
Answer:
I = 8.75 kg m
Explanation:
This is a rotational movement exercise, let's start with kinetic energy
K = ½ I w²
They tell us that K = 330 J, let's find the angular velocity with kinematics
w² = w₀² + 2 α θ
as part of rest w₀ = 0
w = √ 2α θ
let's reduce the revolutions to the SI system
θ = 30.0 rev (2π rad / 1 rev) = 60π rad
let's calculate the angular velocity
w = √(2 0.200 60π)
w = 8.683 rad / s
we clear from the first equation
I = 2K / w²
let's calculate
I = 2 330 / 8,683²
I = 8.75 kg m
Explanation:
Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.
Hence, 
= 0. And, as = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.
Therefore,
Q = -(19000 J + 2000 J)
= -21000 J
or, |Q| = 21000 J
Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.
