For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
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
Answer:
The heater power required is 2400 W. The power in the heater can be calculated as the product of the voltage line and the steady current:

Explanation:
This question is in complete.The question is
A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Answer:
distance=0.124 m
Explanation:

Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
(1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):

Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:

The number of electrons that trough the wire 3 is 2.18*10^18 e/s