Velocity, va2 = 10.5 ft/s
<u>Explanation:</u>
From the figure:
Length of the cable = Sa + 2Sb = l
∴ vₐ = -2vb
Applying the principle of Impulse and momentum in x-direction
Limit is t1 to t2
-(1)
Applying the principle of Impulse and momentum in y-direction
Limit is t1 to t2
-(2)
Solving equation (1) and (2), we obtain
T = 1.6lb
va2 = 10.5 ft/s
Using the formula for word
W=Fd
We can find work because we have the distance and force
d=1 km= 1000m
F= 34 N
Thus W= 34(1000)
W= 34,000 J
PS I'm super close to the next rank, I need one more "Brainiest" solution. If you could I would be very grateful
Answer : The maximum current will be,
Explanation :
The relationship between power, resistance and current is:
where,
P = power = 0.5 watts
I = current = ?
R = resistance = 6 kohms = 6000 ohms
Now put all the given values in the above formula, we get:
Thus, the maximum current will be,
1) G.mars= 9.8x(1/3)=3.26
Vi=10 m/s , T=12 sec
Dis. = Vi(t)+1/2(g)(t)^2
(10)(12)+1/2(3.26)(12)^2= 354.72
2)V=500 m/s
Hmax= v^/2(g) = 500^2/2(9.8) = 12755.1