Hello,
I'd be glad to help.
Momentum = (mass) times (speed)
= (75 kg) x (18 m/s)
= 1,350 kg-m/s .
Hope this helps
D = 110 m, t = 5 s
v o = 110 cs : 5 m = 22 m/s
-------------------------------------
v = v o - a t
v = 0 m/s, v o = 22 m/s, t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
9.80 : 5.5 = 0.56
Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g.
Answer:
The sum of PE and KE remains constant
Explanation:
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:
<span>theta = arcsin(3/20) = approx. 8.63° </span>
<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>
<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>
<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
Answer: I think its 120
Explanation: thx for the free points :)
