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3 years ago

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A bicycle rider traveling east at 10 km/hr sees a blue car pass her appearing to travel west at 50 km/hr. What is the blue car's

apparent speed for an observer riding in a red car traveling 15 km/hr west relative to the Earth? g

1 answer:

BigorU [14]3 years ago

8 0

Answer:

$v_{bR} = 25 km/h$ towards west

Explanation:

As we know that the speed of the blue car as appear to the bicycle rider is given as

$v_{bc} = 50 km/h$ towards west

also it is given that bicycle is moving at speed of 10 km/h towards East

so here we have

$v_{bc} = v_b - v_c$

so we have

$v_b = v_{bc} + v_c$

$v_b = -50 + 10 = 40 km/h$ towards west

now speed of the red car is given as 15 km/h towards west

so here the relative speed of blue car with respect to red car is given as

$v_{bR} = v_b - v_R$

$v_{bR} = 40 - 15 = 25 km/h$ towards west

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3 years ago

If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average

Nina [5.8K]

Answer:

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Explanation:

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t= 0,25 s

So;

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3 years ago

To convert minutes per second into kilometre per hour we multiply the speed with

ankoles [38]

Answer:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

Explanation:

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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

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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.

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4 years ago

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tiny-mole [99]

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Where n is the number of spokes necessary to construct the wheel.

$I_{rim} = M*r^2 = 1.0 * 0.5^2$

$I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2$

Replacing the values at the general equation we have,

$0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )$

Solving for n,

$n = 36$

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

$M_w= M + n*m$

$M_w = 1.0 + 36* 10 * 10^{-3} Kg$

$M_w = 1.36 Kg$

Therefore the mass of the wheel must be of 1.36Kg

4 0

4 years ago