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2 years ago

How to prove3 equations of motion graphically

1 answer:

6 0

Consider an object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time t = 0 and t = t are u and v respectively. During time t, let s be the total distance travelled by the object.

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Which nucleus completes the following equation?

IgorLugansk [536]

Answer: $^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr$

The chromium nucleus will complete the reaction.

Explanation:

The given reaction is a type of radioactive nuclei decay:

Positron emission: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as $\beta ^+$ -decay. In this the mass number remains same.

$_A^Z\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e$

Now according to the given reaction:

$^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr$

The chromium nucleus will complete the reaction.

4 0

3 years ago

Read 2 more answers

What is folk remedies

mario62 [17]

A "folk" remedy is something you choose to eat, take, or do,
in order to correct some health issue, because your friend,
your grandmother, your religion, your shaman, your astrologer,
or your medicine man says that it's the right thing to do.

It may or may not be effective, and it may or may not even be
safe. What distinguishes it from mainstream medicine is that
its practices are based on hearsay, and have not been subjected
to scientific methods of investigation and evaluation.

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3 0

3 years ago

What is the force required to produce an acceleration of 46.4

Brilliant_brown [7]

That depends on the mass of the object, and the unit of the '46.4' .

If the '46.4' is ' meters per second² ' , then the force required is

(mass of the object in kilograms) x (46.4) newtons .

7 0

3 years ago

Please help!!!! 70 points, i really need this! D:

jolli1 [7]

1. Frequency 2. measure from trough to trough

7 0

3 years ago

Read 2 more answers

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: