Answer:
s = vcos(x)t
50 = 25cos(45)t
cos(45)t = 2
t = 2/cos(45) = 2sqrt(2)
h = vsin(x)t + gt^2/2
h = 25sin(45)*2sqrt(2) - 4.9*8
h = 10.8 metres
Explanation:
Answer:
Explanation:
a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)
work done raised the potential energy
b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W
c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%
Not a very likely scenario.
Answer:
acceleration a = 1.04 m/s2
Explanation:
Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:
s = 75 + 180 = 255 m
We can use the following equation of motion to find out the distance traveled by the car:
where v = 23 m/s is the velocity of the car when it passes the worker,
= 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.



Answer:
The maximum speed will be 26.475 m/sec
Explanation:
We have given mass of the toy m = 0.50 kg
radius of the light string r = 1 m
Tension on the string T = 350 N
We have to find the maximum speed without breaking the string
For without breaking the string tension must be equal to the centripetal force
So 
So 

v = 26.475 m /sec
So the maximum speed will be 26.475 m/sec
Answer:
The workdone is
Explanation:
From the question we are told that
The potential difference is 
Generally the charge on
is 
Generally the workdone is mathematically represented as

=>
=>