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3 years ago

7

A soap box derby car at the top of a ramp has more _________ energy and at the finish line at the bottom of the ramp it has more

_________ energy.
A) kinetic, kinetic
B) potential, kinetic
C) kinetic, potential
D) potential, potential

1 answer:

astra-53 [7]3 years ago

5 0

Kinetic, potential because, at the top of the ramp it’s going faster. Potential at the bottom of the ramp is potential because, it’s not doing any motion.

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A long wire is known to have a radius greater than 4.0 mm and to carry a current uniformly distributed over its cross section. i

ollegr [7]

Magnetic field outside it due to long wire is given by

$B = \frac{u_o i}{2 \pi r}$

Magnetic field due to long wire inside wire at any point

$B = \frac{u_o i r}{2 \pi R^2}$

Now the ratio of two magnetic field is given by

$\frac{B_{in}}{B_{out}} = \frac{r_1/R^2}{1/r_2}$

$\frac{0.285}{0.200} = \frac{4*10}{R^2}$

$1.425 = \frac{40}{R^2}$

R = 5.3 mm

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3 years ago

Consider a Hydrogen atom with the electron in the n 8 shell. What is the energy of this system? (The magnitude of the ground sta

Shtirlitz [24]

Answer:

The energy of an electron in the 8th shell is given by: -0.2125 eV

The number of subshells is: 8

The number of orbitals is: 64

The number of electrons that fit on this shell is: 128

Explanation:

First, we find the energy of the electrons in the 8th shell. In order to do this, we recall that the energy of an electron (in the Hydrogen atom) whose principal number is n is given by:

$E_{n}=-13.6\frac{1}{n^{2}}$

Substituting n=8, we find that the energy is given by:

$E_{8} = -13.6\frac{1}{8^{2}}=-0.2125$

In order to find the number of subshells we recall that, for a given principal quantum number n, the possible values of the quantum number l, which corresponds to the number of subshells are:

0, 1, 2, ... , n-1

Since n = 8 in our problem, the possible values of l are: 0, 1, 2, 3, 4, 5, 6, 7. Therefore, the number of subshells are 8.

Now we continue with the number of orbitals. For every subshell l, we have 2l+1 possible values of m, which correspond to the orbitals. Since the possible values of l are: 0,1,2,3,4,5,6,7, therefore, we have to perform the sum:

$\sum_{l=0}^{7}(2l+1) = 8^2=64$

And we can conclude that the number of orbitals is equal to 64.

Finally, we know that we can fit two electrons per orbital, therefore we can have 64*2 = 128 electrons in the shell corresponding to n=8.

8 0

3 years ago

How many Sig Figs (Significant Figures) are in each number?<br><br> 5070.0<br><br> 870.064080

nlexa [21]

∑ Hey, Lethality ⊃

Answer:

5070.0 has 5 significant figures

870.064080 has 9 significant figures

Explanation:

<u><em>Given:</em></u>

<em>How many Sig Figs (Significant Figures) are in each number?</em>

<em>5070.0</em>

<em>870.064080</em>

<u><em>Solution:</em></u>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>5070.0</em>

<em>5070.0 has 5 significant figures ( 5 , 0 , 7 , 0 , and 0 )</em>

<em>Number of significant figures: 5</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>870.064080</em>

<em>870.064080 has 9 significant figures ( 8, 7 ,0,0, 6,4,0,8 and 0 )</em>

<em>Number of significant figures: 9</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em />

<u><em>xcookiex12</em></u>

<em>8/23/2022</em>

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1 year ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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3 years ago

Compared to a stationary galaxy, light from a galaxy that is moving away from earth will appear _____.

Vlad1618 [11]

"B" When an object moves away from us, the light is shifted to the red end of the spectrum, as its wavelengths get longer.

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3 years ago

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