Answer:
a) P = 149140[w]; b) 1491400[J]; c) v = 63.06[m/s]
Explanation:
As the solution to the problem indicates, we must convert the power unit from horsepower to kilowatts.
P = 200 [hp]
![200[hp] * 745.7 [\frac{watt}{1 hp}]\\149140[watt]](https://tex.z-dn.net/?f=200%5Bhp%5D%20%2A%20745.7%20%5B%5Cfrac%7Bwatt%7D%7B1%20hp%7D%5D%5C%5C149140%5Bwatt%5D)
Now the power definition is known as the amount of work done in a given time
P = w / t
where:
w = work [J]
t = time [s]
We have the time, and the power therefore we can calculate the work done.
w = P * t
w = 149140 * 10 = 1491400 [J]
And finally, we can calculate the velocity using, the expression for kinetic energy
![E_{k}=w=0.5*m*v^{2}\\ where:\\v = velocity[m/s]\\m=mass=750[kg]\\w=work=1491400[J]\\](https://tex.z-dn.net/?f=E_%7Bk%7D%3Dw%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%20%20where%3A%5C%5Cv%20%3D%20velocity%5Bm%2Fs%5D%5C%5Cm%3Dmass%3D750%5Bkg%5D%5C%5Cw%3Dwork%3D1491400%5BJ%5D%5C%5C)
The key to solving this problem is to recognize that work equals kinetic energy
![v=\sqrt{\frac{w}{0.5*m}} \\v=\sqrt{\frac{1491400}{0.5*750}} \\v=63.06[m/s]](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7Bw%7D%7B0.5%2Am%7D%7D%20%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B1491400%7D%7B0.5%2A750%7D%7D%20%20%5C%5Cv%3D63.06%5Bm%2Fs%5D)
Answer:
Range, 
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
Answer:
26.67 m/s
Explanation:
From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.
p=mv where p is momentum, m is the mass of object and v is the speed of the object
Initial momentum
The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero
Final momentum
Differentiate the expression, to obtain expression for velocity. Set velocity to 0, this when max height is reached. Obtain the tmax from that expression.
<span>h(t) = –16t² + 32t + 6
</span><span>h'(t) = –32t² + 32
0 = </span>–32t² + 32
t max= 1
hmax = <span> –16(1)² + 32(1) + 6
hmax = 22
Therefore, first option is the correct answer.</span>
Answer:
Explanation:
Given
Object fall from a height of 
Considering initial velocity to be zero i.e. 
using
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
(b)Average acceleration
After falling 45 m, object strike the car and comes to rest after covering a distance of 0.5 m
again using
here final velocity will be zero i.e.
initial velocity 
(c)time taken by it to stop
