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levacccp [35]
3 years ago
9

A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

indow. How much time does it take for the package to reach the ground? (hint: the package initial velocity equals the helicopter velocity)
Physics
1 answer:
Luden [163]3 years ago
7 0

Here it is given that initial speed of the package will be same as speed of the helicopter

v_i = 5.10 m/s

displacement of the package as it is dropped on ground

d = -105 m

acceleration is due to gravity

a = -9.8 m/s^2

now by kinematics

y = v* t + \frac{1}{2}at^2

-105 = 5.1 * t - \frac{1}{2}*9.8*t^2

4.9 t^2 - 5.1 t - 105 = 0

by solving above equation we have

t = 5.2 s

so it will take 5.2 s to reach the ground

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Answer:

It is an example of negative feedback mechanism

Explanation:

Negative feedback mechanism is the process by which the output of a system, process or mechanism is fed back into the input in order to reduce the error in the system.

In the system described in this question, the reabsorption of water is the output, a feedback signal is sent to the control center, which is the input in order to control the release of the hormone responsible for reabsorption of water into the blood.

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If a dog walks north for 10 meters and then east for 10 meters, what is the direction of its displacement?
Katyanochek1 [597]

The direction of its displacement wil be

c.northeast

In fact, the dog walks north for 10 meters and east for another 10 meters. The path of the dog can be represented with two vectors, A pointing north (of magnitude 10 meters) and B pointing east (of magnitude 10 meters). The direction of the resultant vector (due to east) will be given by

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3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

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