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3 years ago

Two people must have the same speed and velocity if

2 answers:

7 0

If they both are moving with the same speed and direction
i.e. covering the same distance in the same time interval in the same direction

stich3 [128]3 years ago

5 0

Answer:

In general, 2 people have the same speed when their velocity vectors have the same magnitude and direction.

Explanation:

It is possible to have the same speed for 2 bodies in 2 cases:

1.Both move in the same direction with constant velocities of equal magnitude.

2. Both bodies are under the influence of the same force under the same conditions, i.e. the force pushes the bodies in the same direction and with the same magnitude. In this way the instantaneous velocity for both bodies is the same.

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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

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-

kirill [66]

Answer:0.478 c

Explanation:

Given

mass of lighter Particle $(m_1)=3\times 10^{-28} kg$

mass of heavier Particle $(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle $(v_1)=0.834 c$

Let speed of heavier particle $=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

$8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

$v_2=0.478 c$

3 0

3 years ago

What is measurement

HACTEHA [7]

Answer:

“Measurement” is the act of determining a target's size, length, weight, capacity, or other aspect. There are a number of terms similar to “measure” but which vary according to the purpose (such as “weight,” “calculate,” and “quantify.”) In general, measurement can be understood as one action within the term “instrumentation.”

Explanation:

• To show measured results using values and symbols. • To use measurement tools. Measuring a target can be done through either direct measurement or indirect measurement. An Indirect measurement is done, for example, by using a dial gauge to measure the height difference between a measurement target and a gauge block and using that height to indirectly determine the target's height. Because this type of measurement is based on a reference, indirect measurement is also referred to as “comparative measurement.

An ”Direct measurement is measurement done by bringing the target into contact with the measurement system to read the length, height, or other aspect directly. Although direct measurement allows measurement results to be known as they are, errors may occur depending on the skill of the person doing the measurement.

5 0

3 years ago

If wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:

GenaCL600 [577]

The frequency of the wheel is given by:
$f= \frac{N}{t}$
where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:
$f= \frac{100}{10 s}=10 s^{-1}$

And now we can find the angular speed of the wheel, which is related to the frequency by:
$\omega=2 \pi f=2 \pi (10 s^{-1})=62.8 s^{-1}$

6 0

4 years ago

The number of sound waves per unit time is called what

Andreyy89

Answer:

Frequency

Explanation:

Hoped this helped

3 0

3 years ago