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3 years ago

PLEASE HELP!!!!!

Physics

1 answer:

Arte-miy333 [17]3 years ago

5 0

Answer:

it states that energy can either be gained or lost but it only changes its form.

Explanation:

for example:as a ball is still on the table it posses a potential energy of 100j and a k.e of 0j,as it falls it gains k.e so the midpoint the p.e is equal to the k.e (50j equally) as it approches the ground it completely gains k.e (100j) and the p.e is 0j.

total energy is 100j so it has been converted from p.e to k.e.

hope u have understood.

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A body of density 9.0cm appears to have mass 27.0g in a liquid of density 1.2gcm. What is the volume of the solid?

Alika [10]

Answer:

v * 9.0 = 27.0 + (v * 1.2)

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3 years ago

Is energy transformation from potential to kinetic 100%?

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3 years ago

The engine that moves the cables for the San Francisco cable cars delivers 390.3 kW of power for each line. How long does it tak

Musya8 [376]

Answer:

It would take 16.7 s for the work to be done by the engine.

Explanation:

From the question, given: Power = 390.3 kW

Work to be done = 6.5 x $10^{6}$ J

But, power and work done with respect to time, has a relationship of:

Power = $\frac{work done}{time}$

So that,

time = $\frac{work done}{Power}$

Thus,

time = $\frac{6.5*10^{6} }{390.3*10^{3} }$

= 16.6539

time = 16.7 s

Time required is 16.7 seconds.

Thus, it would take 16.7 s for the work to be done by the engine.

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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).

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

4 years ago

When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12Hz. When

anzhelika [568]

Answer:

Explanation:

Given

When $m_1$ mass is attached to the spring its frequency is $f_1=12\ Hz$

when another mass $m_2$ is attached to $m_1$ , frequency changes to $f_2=4\ Hz$

frequency of spring mass system is given by

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

for $m_1$

$f_1=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}$

$12=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}-----1$

for $m_1$ and $m_2$

$f_2=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}$

$4=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}----2$

divide 1 and 2

$3=\sqrt{\frac{m_1+m_2}{m_1}}$

squaring

$9=1+\frac{m_2}{m_1}$

$\frac{m_2}{m_1}=8$

8 0

3 years ago