Answer:
Datos:
q1 = -50 μC = 
q2 = +30 μC = 
F = 10 N
a) x si la <em>F = 10N</em>
Aplicando la Ley de Coulomb:
x =
=
= 1,162m
b) x si la <em>F = 20 N</em>
x=<em> </em>
<em> </em>= 0,822m
c)x si la <em>F = 50 N</em>
x =
= 0,520m
Answer:
The rate at which power is generated in the coil is 10.24 Watts
Explanation:
Given;
number of turns of the coil, N = 160
area of the coil, A = 0.2 m²
magnitude of the magnetic field, B = 0.4 T
time for field change = 2 s
resistance of the coil, R = 16 Ω
The induced emf in the coil is calculated as;
emf = dΦ/dt
where;
Φ is magnetic flux = BA
emf = N (BA/dt)
emf = 160 (0.4T x 0.2 m²)/dt
emf = 12.8 V/s
The rate power is generated in the coil is calculated as;
P = V²/ R
P = (12.8²) / 16
P = 10.24 Watts
Therefore, the rate at which power is generated in the coil is 10.24 Watts
Answer:
T = 4.905[N]
Explanation:
In order to solve this problem we must perform a sum of forces on the vertical axis.
∑Fy = 0
We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.
![T-m*g=0\\T=0.5*9.81\\T=4.905[N]](https://tex.z-dn.net/?f=T-m%2Ag%3D0%5C%5CT%3D0.5%2A9.81%5C%5CT%3D4.905%5BN%5D)
Good. You can do some very interesting experiments with that equipment.
Answer:
Approximately
(assuming that the projectile was launched at angle of
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing
is the same as the altitude
at which this projectile was launched:
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is
(upwards,) the vertical velocity right before landing would be
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be
. In other words,
.
Hence, the time it takes to achieve a (vertical) velocity change of
would be:
.
Hence, this projectile would be in the air for approximately
.