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Complete Question:
a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.
Answer in units of m/s2
b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under
these new conditions.
Answer in units of m/s2
Answer:
a. 2.875m/s²
b. 3.172m/s²
Explanation:
a. The formula for centripetal acceleration = (speed²) ÷ radius
Centripetal acceleration = (5.7m/s)²÷ 11.3m
Centripetal acceleration = 2.875m/s²
b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.
Centripetal acceleration ( acceleration x) = 2.875m/s²
Increase in the speed rate ( acceleration n) = 1.34m/s²
Magnitude of acceleration = √a²ₓ + a²ₙ
=√( 2.875m/s²)²+ (1.34m/s²)²
= √ 10.06m/s²
= 3.172m/s²
Explanation:
Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).
For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:
k is constant of proportionality
It is required to solve the above equation for k
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Answer:
40 meters. look for the dot above the 20 on the x-axis and follow it over to the left.
Explanation:
Answer:
Yes it would be different on Earth and the moon
