A diver having mass m climbs up the diving board.
We know that Gravitational potential energy is given as <span>P<span>EG</span>=mgΔh</span>
What changes is his Gravitational potential energy due to change of height <span>Δh</span> with reference to the ground/water level.
While standing on the diving board his velocity is zero. As such kinetic energy is also zero.
Once he jumps off the springboard we see he gets additional energy from the springboard and falls down under action of gravity g. Due to decrease of height above the ground level Gravitational potential energy decreases and gets converted in to his kinetic energy. <span>1/2m<span>v2</span></span>.
While in air he encounters air resistance. Some of his energy is spent in overcoming this resistance. Gets converted in to kinetic and thermal energy of surrounding air and his body.
Once diver reaches the water, we see water splashing and hear noise of splash. Thereafter the diver comes to rest. Now his potential energy becomes zero. And converted kinetic energy has been converted in to kinetic energy, heat energy and sound energy of water.
As such energy transformation equation looks like
<span><span>Gravitational PE+Elastic PE of springboard</span><span>→Kinetic energy of air and water+Sound energy of splash+thermal energy</span></span>
Answer:
a) 
, b) 
, c) D. The magnitud of the change in the ball's momentum.
Explanation:
a) The magnitude of the change in the ball's momentum is:
![\Delta p = (0.275\,kg)\cdot \left[\left(1.63\,\frac{m}{s} \right)-\left(-3.28\,\frac{m}{s} \right)\right]](https://tex.z-dn.net/?f=%5CDelta%20p%20%3D%20%280.275%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%281.63%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29-%5Cleft%28-3.28%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5Cright%5D)
b) The change in the magnitude of the ball's momentum:
![\Delta p' = (0.275\,kg)\cdot \left[(1.63\,\frac{m}{s} )-(3.28\,\frac{m}{s} ) \right]](https://tex.z-dn.net/?f=%5CDelta%20p%27%20%3D%20%280.275%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%281.63%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29-%283.28%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%20%5Cright%5D)
c) The magnitude of the change in the ball's momentum is more directly related to the net force acting on the ball, as it measures the effect of the force on change in ball's motion at measured time according to the Impact Theorem. So, the right answer is option D.
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
<h3>b) Time</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating :
(5)
(6)
(7)
<h3>a) Final velocity</h3>
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
<h3>c) Range</h3>
Substituting (7) in (2):
(11)
(12)
Answer:
a) v = 0.9167 m / s, b) A = 0.350 m, c) v = 0.9167 m / s, d) A = 0.250 m
Explanation:
a) to find the velocity of the wave let us use the relation
v = λ f
the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength
λ = x
λ = x
the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period
T / 2 = t
T = 2t
period and frequency are related
f = 1 / T
f = 1 / 2t
we substitute
v = x / 2t
v = 5.50 / 2 3
v = 0.9167 m / s
b) the amplitude is the distance from a maximum to zero
2A = y
A = y / 2
A = 0.700 / 2
A = 0.350 m
c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same
v = 0.9167 m / s
d) the amplitude is
A = 0.500 / 2
A = 0.250 m
Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank
, we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
