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Paladinen [302]
3 years ago
5

A child throws a baseball directly upward. what are the signs of the velocity and acceleration of the ball immediately after the

ball leaves the child's hand? enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. for example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . if you think that both are zero, then you would enter 0,0 .
Physics
1 answer:
navik [9.2K]3 years ago
3 0

(-,-)


when the ball is thrown upward, the ball experiences gravitational acceleration = 9.8 m/s² which is always downward.

So the velocity of ball starts decreasing and reaches 0 at highest point.

so, the velocity and acceleration are negative with respect to ball's direction of moving when the ball is going upward.

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Work = Force x Distance = 500 x 4 = 2000 Nm = 2000 J
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2 years ago
Explain in your own words how the Doppler Effect is also applicable in our study of light.
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so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

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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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