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Finger [1]
4 years ago
10

Ow does the light-collecting area of an 8-meter telescope compare to that of a 2-meter telescope?

Physics
1 answer:
fomenos4 years ago
6 0
<span>The area will be the square of the quotient of the amounts for each telescope. Comparing the two microscopes, the 8-meter model is 4 times larger than the 2-meter model. Squaring this value gives a ratio of 16 times the amount of visual area for the larger telescope.</span>
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3 years ago
Is "A firecracker that has not been lit kinetic, or potential energy?
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4 years ago
The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally
dalvyx [7]
(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
K= \frac{1}{2}mv^2
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 2.20 \cdot 10^{-15} J}{9.1 \cdot 10^{-31} kg} }=6.95 \cdot 10^7 m/s

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
qvB = m \frac{v^2}{r}
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
r= \frac{mv}{qB}= \frac{(9.1 \cdot 10^{-31} kg)(6.95 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(3.00 \cdot 10^{-5} T)}=13.18 m

And finally we can calculate the centripetal acceleration, given by:
a_c =  \frac{v^2}{r}= \frac{(6.95 \cdot 10^7 m/s)^2}{13.18 m}=3.66 \cdot 10^{14} m/s^2
5 0
3 years ago
in the same viewing window. Compare the magnitudes of f (x) and g(x) when x is close to 0. Use the comparison to write a short p
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6 0
3 years ago
Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
2aS=v_f^2 - v_i^2
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
v_f =0
So we can rewrite the equation as
2(-9.81 m/s^2) h=-v_i^2
from which we can isolate h
h= \frac{v_i^2}{19.62} (1)

Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
8 0
3 years ago
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