What is an ideal diode? a. The ideal diode acts as an open circuit for forward currents and as a short circuit with reverse volt
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The orbital speed of an ice cube in the rings of Saturn is 11.2 Km/s. The correct answer is option C
<h3>What does Orbital speed depend on ?</h3>The speed of an object travelling around a circle depends on two quantities namely;
- Its angular velocity w
- Its distance from the center of the circle.
Given that an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 10^26 kg, and use an orbital radius of 3.00 x 105 km. (G= 6.67 × 10-11 N·m2/kg2)
The given parameters are:
- The mass of Saturn = 5.68 x 10^26 kg
- The orbital radius = 3.00 x 105 km
- G = 6.67 × 10-11 N·m2/kg2
Let us first calculate the gravitational field strength on the Saturn.
g = GM/r²
Substitute all the necessary parameters and convert km to m
g = (6.67 × × 5.68 ×
) ÷ (300000 × 1000)²
g = 3.79 × ÷ 9 ×
g = 0.421 m/s²
The orbital speed will be
V² = gr
V² = 0.4211 × 300000 × 1000
V² = 126333333.3
V = √126333333.3
V = 11239.8 m/s
Convert it to Km/s by dividing the answer by 1000
V = 11239.8/1000
V = 11.2 Km/s
Therefore, the orbital speed of an ice cube in the rings of Saturn is 11.2 Km/s
Learn more about Orbital Speed here: brainly.com/question/22247460
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The angular momentum is defined as,
Acording to this text we know for conservation of angular momentum that
Where is initial momentum
is the final momentum
How there is a difference between the stick mass and the bug mass, we define that
Mass of the bug= m
Mass of the stick=10m
At the point 0 we have that,
Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity
vector from the point of reference (O), and ve is the velocity
At the end with the collition we have
Substituting
Applying conservative energy equation we have
Replacing the values and solving
Substituting
l=\frac{13}{0.54(9.8)}
Answer:
Speed of second car will be 57.17 m/sec
Explanation:
We have given lead car travels = 44 laps
1 laps = 1.34 km = 1340 m
So total distance = 1340×44 = 58960 m
Speed of lead car = 55.9 m/sec
We know that
As the second car is 1 lap behind so distance traveled by second car = 45×1340 = 60300 m
So speed of second car will be