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worty [1.4K]
3 years ago
14

During World War II, mass spectrometers were used to separate the radioactive uranium isotope U-235 from its far more common iso

tope, U-238. Estimate the radius of the circle traced out by a singly ionized lead atom moving at the same speed.
Physics
1 answer:
mariarad [96]3 years ago
6 0

Answer:

21.55 m

Explanation:

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2 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
a child is swinging on swing, describe what happens to both the kinectic energy and potential enegry of the child as she swings
Igoryamba

Answer:

The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.

Explanation:

  • A child is swinging on swing up and down has both kinetic and potential energy.
  • The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,

                                      E = K.E + P.E

  • The K.E is maximum when the child is at the vertical position.
  • The P.E is maximum at the extreme deviated position from the vertical.
  • And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.
5 0
3 years ago
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