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3 years ago

A bend in a river shaped like a loop is called a(n) ________________

Physics

2 answers:

Llana [10]3 years ago

8 0

This loop-like bend in a river is called a meander

babymother [125]3 years ago

5 0

Meander

I think that would be it.

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Two small space probes have been slowed to 10m/s as they approach the moon from the same direction. Probe 1 has a mass of 86kg a

bulgar [2K]

Answer:

Explanation:

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2 years ago

Unlike the majority of psychologists, psychiatrists __________. A. can diagnose disorders B. use psychotherapy C. treat people w

ludmilkaskok [199]

D prescribe medication

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3 years ago

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A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa

son4ous [18]

Answer:

The answer is 0.83 seconds.

Explanation:

The formula of free fall is following:

$h=1/2*g*t^2$

Where g=9.8 m/s^2 and t=2 seconds, the rock takes:

$h=1/2*9.8*2^2=19.6$

19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:

$39.2=1/2*9.8*t^2\\t^2=8\\t=2.83$

2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds

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3 years ago

Which county in Florida is most in need of safe rooms and hurricane ties?

Murrr4er [49]

Probably Pasco. Lol.

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3 years ago

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago