Answer:
Explanation:
We name M the larger mas, and m the smaller mass, so base on this we write the following conservation of momentum equation for the collision:
We can write this in terms of what we are looking for (the quotient of masses :
We use now the information about Kinetic Energy of the system being reduced in half after the collision:
We can combine this last equation with the previous one to obtain:
where solving for the quotient m/M gives:
The acceleration of the train at the give initial and final velocity is determined as 10.5 m/s².
<h3>Acceleration of the train</h3>
The acceleration of the train is calculated as follows;
a = Δv/t
where;
- u is initial velocity = 162 km/hr = 45 m/s
- v is final velocity of the car = 540 km/hr = 150 m/s
a = (150 - 45)/10
a = 10.5 m/s²
Thus, the acceleration of the train at the give initial and final velocity is determined as 10.5 m/s².
Learn more about acceleration here: brainly.com/question/605631
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Frequency of any wave = (the wave's speed) divided by (its wavelength)
Speed of light = about (3 x 10 to the 8) meters per second, in vacuum.
Frequency = (3 x 10 to the 8) / (4.1 x 10 to the -12) = 7.317 x 10 to the 19 Hz.
That's the same as 73,170,000,000 GHz.
Note:
That's the answer for the information given in the question, but the information in
the question is incorrect by 5 orders of magnitude. The wavelength of violet light,
or anything in the visible range, is a few hundred 'nanometers' ... a few hundreds
of 10 to the -9 . So the number in the question should be 4.10 x 10 to the -7,
not -12, and the frequency should be 731,700 GHz.
Answer:
120 km/hr
Explanation:
Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.
Now, at any instant, D² = d² + L²
= d² + 4²
= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.
differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have
d(D²)/dt = d(d² + 16)/dt
2DdD/dt = 2d[d(d)/dt]
dD/dt = 2d[d(d)/dt] ÷ 2D
Now d(d)/dt = vertical speed of rocket = 200 km/hr
dD/dt = 200d/D [D = √(d² + 16)]
dD/dt = 200d/[√d² + 16]
Now substituting d = 3 km, the distance the rocket has risen into the equation, we have
dD/dt = 200(3)/[√(3² + 16)]
dD/dt = 600/[√(9 + 16)]
dD/dt = 600/√25
dD/dt = 600/5
dD/dt = 120 km/hr
So, the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.