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yarga [219]
3 years ago
5

What are 7 elements from the periodic table that are named after objects in outer space

Physics
1 answer:
suter [353]3 years ago
5 0
Helium... from the greek word helios... the sun 
<span>selenium... from the greek word selene... the moon </span>
<span>palladium.. after the asteroid pallas </span>
<span>tellurium...from the greek word tellus... the earth </span>
<span>mercury...after the planet mercury </span>
<span>cerium... after the asteroid ceres </span>
<span>uranium...after the planet uranus </span>
<span>neptunium.. after the planet neptune </span>
<span>plutonium.. after the planet pluto</span>
You might be interested in
This problem has been solved!
irakobra [83]

Answer:

The correct answer is option '5': The type of metal from which the plate is made.

Explanation:

According to the principle of photoelectric effect we know that electron's are only emitted from a surface of metal if the frequency of the light is larger than a threshold frequency that depends on the metal and is known as threshold  function of the metal. The ejection of the electrons is independent of intensity of the incident light meaning any light of frequency lower than work function will not eject electrons from the metal no matter whatever the intensity of the light, or the surface area or thermal conductivity, time of illumination.

5 0
3 years ago
A bullet of mass 6.20 10-3 kg, moving at 1320 m/s impacts a tree stump and penetrates 11.00 cm into the wood before coming to re
sladkih [1.3K]

Answer:

  F = -49.1   10³ N

Explanation:

Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant

   v_{f}² = v₀² + 2 a x

Since the bullet is at rest, the final speed is zero

    x = 11.00 cm (1 m / 100 cm) = 0.110 m

    0 = v₀² + 2 a x

   a = -v₀² / 2 x

   a = -1320²/(2 0.110)

   a = -7.92 10⁶ m / s²

With Newton's second law we find the force

   F = m a

   F = 6.20 10⁻³ (-7.92 10⁶)

   F = -49.1   10³ N

The sign means that it is the force that the tree exerts to stop the   bullet

8 0
3 years ago
Two very large parallel sheets are 5.00 cm apart. sheet a carries a uniform surface charge density of -9.70 μc/m2 , and sheet b,
MAVERICK [17]
Question is missing. Found on internet the complete text of the problem:

"<span>Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."

Solution:

(a) The electric field produced by a uniformly charged sheet at any distance is given by
</span>E= \frac{\sigma}{2 \epsilon _0}
where \sigma is the charge density and \epsilon _0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity.

First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are \sigma_A = -9.70 \mu C/m^2=-9.70\cdot 10^{-6}C/m^2 and \sigma_B = -11.5 \mu C/m^2 = -11.5\cdot 10^{-6} C/m^2.

Sheet a gives an electric field of
E_A= \frac{\sigma_A}{2\epsilon _0}= \frac{-9.7\cdot 10^{-6} C/m^2}{2\cdot 8.85\cdot 10^{-12} F/m} = -5.48\cdot 10^5 V/m
where the negative sign means the field points towards sheet A, in any point of the space.

The electric field produced by sheet B is given by:
E_B = \frac{\sigma_A}{2\epsilon _0}= \frac{-11.5\cdot 10^{-6} C/m^2} {2\cdot 8.85\cdot 10^{-12} F/m} =-6.50\cdot 10^{5} V/m
and again, the negative sign means that the field at any point of the space points towards sheet B.

The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
E=E_1-E_2= -5.48\cdot 10^5 V/m - (-6.50\cdot 10^{5} V/m)=1.02\cdot 10^5 V/m
And the direction is towards sheet B, since it has a field with stronger intensity.

(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
E=E_1+E_2=-11.98\cdot 10^5 V/m
towards right.

(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
E=E_1+E_2=11.98 \cdot 10^5 V/m
towards left.
7 0
4 years ago
AP PHYSICS SYSTEM OF EQUATIONS
Anon25 [30]

Walking at a speed of 2.1 m/s, in the first 2 s John would have walked

(2.1 m/s) (2 s) = 4.2 m

Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is

<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>

while Ryan's position is

<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>

where Ryan's velocity is negative because he is moving in the opposite direction.

(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :

4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>

(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m

(3.9 m/s) <em>t</em> = 95.8 m

<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s

(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).

<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m

8 0
3 years ago
If a rock falls for 5 seconds near the surface of the earth and with no air friction, it will reach a velocity of
sukhopar [10]

Answer: -49m/s.

Explanation:

As the rock only falls, we will assume that the initial vertical velocity is zero.

We neglect the air friction, so the only force acting on the rock is the gravitational force, this means that the acceleration is -g = -9.8m/s^2.

Then we can write:

a(t) = -9.8m/s^2

To write the velocity of the rock, we must ingrate over time and get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity, and as we said above, v0 = 0m/s

Then the vertical velocity as a function of time is:

v(t) = (-9.8m/s^2)*t

Now, the question is:

"...If a rock falls for 5 seconds near the surface of the earth and with no air friction, it will reach a velocity of..."

Then we need to evaluate the velocity equation in t = 5 seconds.

v(5s) = (-9.8m/s^2)*5s = -49m/s.

3 0
3 years ago
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