Answer:
Maximum acceleration in the simple harmonic motion will be
Explanation:
We have given amplitude of simple harmonic motion is A = 0.43 m
Time period of the oscillation is T = 3.9 sec
We have to find the maximum acceleration
For this we have to find the angular frequency
Angular frequency will be equal to 
Maximum acceleration is given by 
So maximum acceleration in the simple harmonic motion will be
Answer:
Frequencia = 1.25 Hz
Explanation:
<u>Dados los siguientes datos;</u>
- Velocidad = 50 m/s
- Longitud de onda = 40 metros
Para encontrar la frecuencia de la onda;
Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Haciendo de la frecuencia el tema de la fórmula, tenemos;

Sustituyendo en la fórmula, tenemos;

<em>Frequencia = 1.25 Hz</em>
Answer:
520000 or 520000 pa
Force = 520N
Area of contact = 0.001
Pressure: 520000 or 520000
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:


DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation

according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero

solving for x we get

x = 0.200 m
b)electric field at half way mean x =0.25

E = 3.84*10^{-4} N/C
Answer:
14.0 cm
Explanation:
Draw a free body diagram of the block. There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.
Sum of forces in the y direction:
∑F = ma
ρVg − mg − k∆L = 0
(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0
∆L = 0.140 m
∆L = 14.0 cm