The boat is moving at 22 m/s while the man is moving at 23.1 m/s.
That means the man, relative to the boat, is moving at 23.1-22 = 1.1 m/s.
v =d/t, so t = d/v --> t = 3/1.1 = 2.7 s
Answer:
C.) A high velocity and Large mass.
Explanation:
Momentum of any object is defined by following formula
Here
: m = mass of object
v = velocity of object
now we know that since momentum is product of mass and velocity
So in order to have more momentum we need the value of this product to be more. So this product will me large is both the physical quantity will be more in magnitude. So if mass is large and velocity will be more then the product of them will be large and hence the momentum of object will be more. Btw I had that question too.
If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward. So since the current is going at an angle then it has a x and y component. So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.
Answer:
(a) 490 N on earth
(b) 80 N on earth
(c) 45.4545 kg on earth
(d) 270.27 kg on moon
Explanation:
We have given 1 kg = 9.8 N = 2.2 lbs on earth
And 1 kg = 1.6 N = 0.37 lbs on moon
(a) We have given mass of the person m = 50 kg
As it is given that 1 kg = 9.8 N
So 50 kg = 50×9.8 =490 N
(b) Mass of the person on moon = 50 kg
As it is given that on moon 1 kg = 1.6 N
So 50 kg = 50×1.6 = 80 N
(c) We have given that weight of the person on the earth = 100 lbs
As it is given that 1 kg = 2.2 lbs on earth
So 100 lbs = 45.4545 kg
(d) We have given weight of the person on moon = 100 lbs
As it is given that 1 kg = 0.37 lbs
So 100 lbs 
(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
Learn more about energy stored in capacitor here: brainly.com/question/14811408
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