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3 years ago

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Larry and Balky set up an experiment to analyze the motion of a marble as it rolled down an incline. They started the marble at

the top of the ramp and set up photogates to collect data as the marble rolled down the ramp.
Which graph format will allow Larry and Balky to calculate instantaneous accleration from the slope of the graph?
A) Velocity on the y-axis, time on the x-axis
B) Distance on the y-axis, time on the x-axis
C) Time on the x-axis, distance on the y-axis
D) Velocity on the y-axis, distance on the x-axis

2 answers:

MA_775_DIABLO [31]3 years ago

8 0

A) Velocity on the y-axis, time on the x-axis

erik [133]3 years ago

7 0

Answer:

Option A is the correct answer.

Explanation:

The instantaneous acceleration = Change in velocity in velocity/Time taken

The slope of the graph should give instantaneous acceleration.

Slope of a graph = Change in value of Y -axis / Change in values of X -axis

Comparing both the equations

Change in value of Y -axis = Change in velocity in velocity

Change in values of X -axis = Time taken

So velocity values should be on the Y axis and Time values should be on the X axis.

Option A is the correct answer.

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Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental const

ludmilkaskok [199]

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$ , Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$ , the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$ , so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$

Learn more about gravitational constant:

brainly.com/question/1724648

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3 years ago

An electron in a mercury atom jumps from level a to level g by absorbing a single

statuscvo [17]

Answer:

7.39ev

Explanation:

Energy levels are found inside the atom. Electrons occupy these energy levels depending on the energy they possess. Electrons can move from one energy level to another due to absorption or emission of a photon or other factors. As the electron, jumps from a higher energy level to a lower energy level emitting a photon of measurable frequency, the photon carries energy equal to the amount of energy between the gap of the levels. This idea was first proposed by Neils Bohr and became the forerunner of the wave mechanical model of the atom.

Hence the energy of a photon is the energy of the gap between the two energy levels. Since Ea= 2.48ev and Eg= 10.38 ev.

If an electron jumps from Ea to Eg, the energy of the photon absorbed is given by;

E=Eg-Ea

E= 10.38ev - 2.48ev

E= 7.39ev

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3 years ago

David is investigating the properties of soil using the sample shown.

charle [14.2K]

Answer:

IT IS A BC ION KNOW WHY

Explanation:

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2 years ago

How much heat is absorbed by 60 g of copper when it is heated from 20°C to 80°C?

ira [324]

Answer:

1,836J

Explanation:

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3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago