Answer:
45m/s
Explanation:
(here I'm taking gravity due to acceleration as 10 m/s^2 and not 9.8 m/s^2 to match the options)
We know that when an object is dropped from a certain height it has initial velocity 0 m/s
so;
u= 0m/s
s= 100m
a= 10m/s^2
Using 4th law of motion;
v^2 - u^2 = 2× a× s
v^2 - 0^2 = 2 × 10 × 100
v^2 = 2000
v = √2000
v = 44.72 m/ s
Final velocity = 45 m/ s (rounding off to 45)
Please feel free to tell me if you have any confusion.
The velocity of the combination of Jackie and the bicycle is 3.328 m/s.
Explanation:
From the given data the constant kinetic energy is 3.6 J. The mass of combination is 0.65 kg. To find the velocity of the combination of Jackie and the bicycle the formula is
KE = 0.5 x mv2.
To find velocity,
V2=ke/(0.5×m)
V=![\sqrt{(ke/(0.5*m)](https://tex.z-dn.net/?f=%5Csqrt%7B%28ke%2F%280.5%2Am%29)
v= 3.6/(0.5×0.65)
=![\sqrt (11077/10)/10](https://tex.z-dn.net/?f=%5Csqrt%20%2811077%2F10%29%2F10)
v= 3.328 m/s
Hence, the velocity of the combination of Jackie and the bicycle is 3.328m/s.
I think that when work is done and a force istransferred an object must move in the direction of the force.
A distance of d is covered with 53 mile/hr initially.
Time taken to cover this distance t1 = d/53 hour
Next distance of d is covered with x mile hours.
Time taken to cover this distance t2 = d/x hours.
We have average speed = 26.5 mile / hour
= Total distance traveled/ total time taken
= ![\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}](https://tex.z-dn.net/?f=%5Cfrac%7B2d%7D%7B%5Cfrac%7Bd%7D%7B53%7D%2B%5Cfrac%7Bd%7D%7Bx%7D%7D%20%3D%20%5Cfrac%7B2%7D%7B%5Cfrac%7B1%7D%7B53%7D%2B%5Cfrac%7B1%7D%7Bx%7D%20%7D%20%20%3D%20%5Cfrac%7B106x%7D%7Bx%2B53%7D)
![26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour](https://tex.z-dn.net/?f=26.5%20%3D%20%5Cfrac%7B106x%7D%7Bx%2B53%7D%20%5C%5C%20%5C%5C%2079.5%20x%20%3D%201404.5%5C%5C%20%5C%5C%20x%20%3D%2017.67%20miles%2Fhour)
Answer: 0.56 m/s
Explanation:
hello, there is 25° inclination angle for the chute in the drawing. Thankfully, I know this problem. The conservation of momentum.
so there are X and Y components for the momentum in this problem. The Y component is not conserved as when the coal gets in the cart, the normal force exerted by the surface reduces it to 0.
Now, the X component is definitely conserved here.
so you have the momentum of the cart which is 440*0.5 added to the momentum of the chunk which is 150*0.8*cos(25°), that is the momentum before the coupling between the objects. Afterwards both objects will have the same velocity, so we write the equation like this:
![440*0.5 + 150*0.8*cos(25) = 440*v_{final} + 150*v_{final} \\ => 220+120*cos(25) = (440+150)v_{final} => v_{final} = \frac{220+120*cos(25)}{590} = 0.56 m/s](https://tex.z-dn.net/?f=440%2A0.5%20%2B%20150%2A0.8%2Acos%2825%29%20%3D%20440%2Av_%7Bfinal%7D%20%2B%20150%2Av_%7Bfinal%7D%20%5C%5C%20%3D%3E%20220%2B120%2Acos%2825%29%20%3D%20%28440%2B150%29v_%7Bfinal%7D%20%3D%3E%20v_%7Bfinal%7D%20%3D%20%5Cfrac%7B220%2B120%2Acos%2825%29%7D%7B590%7D%20%20%3D%200.56%20m%2Fs)