Answer:
a)Amplitude ,A = 2 mm
b)f=95.49 Hz
c)V= 30 m/s ( + x direction )
d) λ = 0.31 m
e)Umax= 1.2 m/s
Explanation:
Given that
As we know that standard form of wave equation given as
A= Amplitude
ω=Frequency (rad /s)
t=Time
Φ = Phase difference
So from above equation we can say that
Amplitude ,A = 2 mm
Frequency ,ω= 600 rad/s (2πf=ω)
ω= 2πf
f= ω /2π
f= 300/π = 95.49 Hz
K= 20 rad/m
So velocity,V
V= ω /K
V= 600 /20 = 30 m/s ( + x direction )
V = f λ
30 = 95.49 x λ
λ = 0.31 m
We know that speed is the rate of displacement
The maximum velocity
Umax = 1200 mm/s
Umax= 1.2 m/s
Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;
where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;
<span>The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.</span>
<span>a = 25-13/6 = 12/6 = 2 m/s^2
Av speed: 25+13/2 = 38/2 = 19 m/sec
Dist = speed * time
19 * 6 = 114 meters</span>
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.