Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
Answer:
1)
a) f = 1m × 2 × (5A / √2) × (5A / √2) / 0.003m = 0.00166... (66 is repeating)
b) The currents on two wires on a AC chord are always moving in opposite direction and so they are always replusing.
c) There needs to be a sheath to dampen the replusing, fluctuating force of the wires.
2)
a) v = √( ( (-2)(-1.6 × 10^(-16))(3000V) ) / (2.84 × 10^(-20)kg) ) = 5.81227 × 10^3
b) Any ion transversing a chamber having a magnetic field will deflect.
c) The direction of the electric field is vertical because it's perpendicular to the plates. The electric field magnitude is independent from the magnitude of the magnetic field and charge. So it's not possible to find the magnitude of the electric field, without knowing the voltage on the plates, the distance between the plates, and the dielectric constant.
d) Assuming the mangetic field remained, the path of the negative ions will be deflected vertically given that the magnetic field is horizontally perpendicular to the negative charged ions movement.
Sorry it took so long :) If anything is incorrect please let me know.
Answer:
C. 10kg to 10kg
Explanation:
You have to picture to it I think
Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=
Where 1 kg=1000 g
Frequency of oscillation=
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=
Where
=Angular frequency
A=Amplitude

Using the formula



Hence, the amplitude of oscillation=A=0.199
8.15m. La altura de la casa de la cual se deja caer un objeto que tarda en caer
es de 8.15m.
La clave para resolver este problema es mediante caída libre, la velocidad inicial del objeto es 0 por lo que podemos calcular la altura h de la casa mediante la relación
.
Conocemos la gravedad
y el tiempo en que tarda en caer el objeto
, sustituyendo los valores tenemos que:
