The time it takes the plane to change its velocity is 9s.
<h3>What is time?</h3>
Time can be defined the measured or measurable period during which an action, process, or condition exists or continues.
To calculate the time it takes the airplane to change its velocity, we use the formula below.
Formula:
- t = (v-u)/a.......... Equation 1
Where:
- a = Acceleration
- v = Final velocity
- u = Initial velocity
- t = time
From the question,
- v = 40 m/s
- u = 22 m/s
- a = 2 m/s²
Substitute these values into equation 1
- t = (40-22)/2
- t = 18/2
- t = 9s
Hence, the time it takes the plane to change its velocity is 9s.
Learn more about time here: brainly.com/question/2854969
Answer:
Energy loss per minute will be 
Explanation:
We have given the star produces power of 
We know that 1 W = 1 J/sec
So 
Given time = 1 minute = 60 sec
So the energy loss per minute 
We multiply with 60 we have to calculate energy loss per minute
The best and most correct answer among the choices provided by your question is the fourth option or letter D. Trade winds blow towards the equator because t<span>he Equator receives the most heat energy.
</span>The surface air that flows from these subtropical high-pressure belts toward the Equator is deflected toward the west in both hemispheres by the Coriolis effect. These winds blow<span> predominantly from the northeast in the Northern Hemisphere and from the southeast in the Southern Hemisphere.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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Answer:
R1 = 5.13 Ω
Explanation:
From Ohm's law,
V = IR............... Equation 1
Where V = Voltage, I = current, R = resistance.
From the question,
I = 2 A, R = R1
Substitute into equation 1
V = 2R1................ Equation 2
When a resistance of 2.2Ω is added in series with R1,
assuming the voltage source remain constant
R = 2.2+R1, and I = 1.4 A
V = 1.4(2.2+R1)................. Equation 3
Substitute the value of V into equation 3
2R1 = 1.4(2.2+R1)
2R1 = 3.08+1.4R1
2R1-1.4R1 = 3.08
0.6R1 = 3.08
R1 = 3.08/0.6
R1 = 5.13 Ω
