Answer: 60m/s
Explanation:
From the diagram:
Θ = 30°
Vertical resolution (y-axis) :
Voy = VoSinΘ
g in the upward direction = negative (-) = - g
Vfinal = 0
Distance (H) traveled along y =
Time taken to reach maximum height :
From v = u + at
0 = usinΘ - gt
gt = usinΘ
t = usinΘ / g
Horizontal resolution:
S = ut + 1/2at^2
Substituting t = usinΘ / g ; Voy = usinΘ
S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2
S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)
S = (u^2sin^2Θ) / 2g
Now if S = maximum height = 45m
Then,
45 = [Vo^2sin^2(30°)] / 2(10)
45 =[ Vo^2 * (0.5)^2] / 20
45 =( Vo^2 * 0.25) / 20
20 * 45 = Vo^2 * 0.25
900 / 0.25 = Vo^2
3600 = Vo^2
Vo = sqrt(3600)
Vo = 60m/s
Answer:
(a) Workdone = -27601.9J
(b) Average required power = 1314.4W
Explanation:
Mass of hoop,m =40kg
Radius of hoop, r=0.810m
Initial angular velocity Winitial=438rev/min
Wfinal=0
t= 21.0s
Rotation inertia of the hoop around its central axis I= mr²
I= 40 ×0.810²
I=26.24kg.m²
The change in kinetic energy =K. E final - K. E initail
Change in K. E =1/2I(Wfinal² -Winitial²)
Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]
Change in K. E= -27601.9J
(a) Change in Kinetic energy = Workdone
W= 27601.9J( since work is done on hook)
(b) average required power = W/t
=27601.9/21 =1314.4W
Suppose that the cyclist begins his journey from the rest from the top of a wedge with a slope of a degree above the horizontal.
At point A (where it starts its journey), the energy is:
Ea = m * g * h
In other words, energy is only potential.
At point B (located at the bottom of the wedge), the energy is:
Eb = (1/2) * (m) * (v ^ 2)
In other words, the energy is only kinetic.
For energy conservation we have:
Ea = Eb
That is, we have that all potential energy is transformed into kinetic energy.
Which means that the cyclist has less kinetic energy at point A because that's where he has more potential energy.
answer:
the cyclist has less kinetic energy at point A because that's where he has more potential energy.
3. Kinetic energy
4. Potential energy
5. Kinetic energy because it’s moving towards the waterfall otherwise there wouldn’t be a waterfall.
6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
10. Kinetic energy
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
