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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
Answer:
<u>The correct answer is 0.556 Watts</u>
Explanation:
The computer monitor uses 200 Watts of power in an hour, that is the standard measure.
If we want to know, how much energy the computer monitor uses in one second, we will have to divide both sides of the equation into 3,600.
1 hour = 60 minutes = 3,600 seconds (60 x 60)
Energy per second = 200/3600
Energy per second = 0.0556 Watts
Therefore to calculate how much energy is used in 10 seconds, we do this:
Energy per second x 10
<u>0.0556 x 10 = 0.556 Watts</u>
<u>The computer monitor uses 0.556 Watts in 10 seconds</u>
We can use the equation E = k | Q | r 2 E = k | Q | r2 to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge,
<h3>What is electric and magnetic field ?</h3>
With the use of electricity and other types of artificial and natural illumination, invisible energy fields known as electric and magnetic fields (EMFs) and radiation are created.
- While the magnetic field is discernible by the force it exerts on other magnetic particles and moving electric charges, the electric field is actually the force per unit charge experienced by a non-moving point charge at any given location inside the field.
Learn more about Electromagnetic field here:
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