Answer:
2943J
Explanation:
Ep = mgh
given that,
m = 15kg
h= 20m
and
g = 9.81 m/s^2
Ep= 15 × 9.81 × 20
Ep= 2943J
With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock?
Answer in units of $m / s$.
Answer:
To solve this problem, we can use the kinematic equation for horizontal motion, which relates the initial velocity ($v_{0}$), final velocity ($v_{f}$), acceleration ($a$), and displacement ($d$) of an object:
$d = v_{0} t + \frac{1}{2}at^{2}$
In this case, we want to find the minimum initial velocity ($v_{0}$) that the divers must have to clear the rock. To do this, we can assume that the divers just graze the rock at the start of their trajectory, so the displacement in the horizontal direction is equal to the distance from the cliff to the rock ($d = 9.34 m$). We also know that the acceleration in the horizontal direction is zero, so the only force acting on the divers is gravity in the vertical direction, which gives an acceleration of $a = 9.8 m/s^{2}$.
At the instant the divers leave the cliff, they have zero horizontal velocity, so $v_{0} = 0$. We can use the equation above to solve for the time it takes for the divers to fall from the cliff to the level of the rock:
$d = \frac{1}{2}at^{2} \Rightarrow t = \sqrt{\frac{2d}{a}}$
Plugging in the numbers, we get:
$t = \sqrt{\frac{2(9.34 m)}{9.8 m/s^{2}}} \approx 1.44 s$
Since the cliff divers want to clear the rock, they need to travel a horizontal distance of at least $9.34 m$ during this time. We can use the equation for horizontal motion again to solve for the minimum initial velocity:
$d = v_{0}t \Rightarrow v_{0} = \frac{d}{t} = \frac{9.34 m}{1.44 s} \approx 6.49 m/s$
Therefore, the minimum horizontal velocity that the cliff divers must have to clear the rock is approximately $6.49 m/s$.
Dogs are able to hear much higher frequencies than humans are capable of detecting. For this reason, dog whistles that are inaudible to the human ear can be heard easily by a dog. If a dog whistle has a frequency of 30000 Hz, what is the wavelength of the sound emitted?
The wavelength of the sound emitted is 3.33 x 10⁵ m.
The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.
Wavelength λ = 1/f
If a dog whistle has a frequency of 30000 Hz, then the wavelength will be
λ = 1/f
λ = 1/30000
λ = 3.33 x 10⁵ m
Thus, the wavelength of the sound of dog emitted is 3.33 x 10⁵ m
Learn more about wavelength.
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A) 30 m/s
The problem can be solved by using the law of conservation of momentum. In fact, the total momentum falcon+pigeon before the collision must be equal to the total momentum falcon+pigeon after the collision:
where
mf = 0.480 g is the mass of the falcon
uf = 45 m/s is the initial velocity of the falcon
mp = 0.240 g is the mass of the pigeon
up = 0 is the initial velocity of the pigeon
v is the final combined velocity of pigeon+falcon
Solving the equation for v, we find
B) 480 N
The average force on the pigeon during the impact is given by
where
is the change in momentum of the pigeon
is the duration of the collision
here we have:
- Change in momentum of the pigeon:
- Duration of the collision:
So the average force is
To determine the final speed of the falcon and pigeon, we need to use the principles of conservation of momentum. To calculate the average force on the pigeon during the impact, we can use the equation for impulse. The primary topic of this question is conservation of momentum and impulse.
To determine the final speed of the falcon and pigeon, we need to use the principles of conservation of momentum. Since the pigeon is assumed to be stationary, its initial momentum is zero. The final momentum of the falcon and pigeon combined must also be zero, according to the law of conservation of momentum. Using the equations for momentum and rearranging, we can solve for the final speed of both the falcon and pigeon.
To calculate the average force on the pigeon during the impact, we can use the equation for impulse, which is the change in momentum. Impulse is equal to force multiplied by the time of impact. Rearranging the equation, we can solve for force.
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