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kondaur [170]
3 years ago
7

projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor

m electric fiedl of magnitude E.
Physics
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

Range, R = MV²/2QE

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

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How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b
Kay [80]

Answer:

333.3 m

Explanation:

Given

m =100g\ =\  0.1kg\\v = 100 m/s\\g = 10 m/s ^2

Potential energy =\frac{2}{3}\  of\  Kinetic\  energy......Equation(1)

We know that

Potential energy=mgh

Kinetic energy =\frac{1}{2} mv^{2}

Now From the Equation(1)

mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\  gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\  m

3 0
3 years ago
A basketball with a mass of 0.5 kilograms is accelerated at 2
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Answer: 1N

Explanation: its not 0N.

5 0
3 years ago
When the biker is at the top of the ramp, he has a speed of 10 m/s at
oksian1 [2.3K]

Answer:

25000

Explanation:

100 x 25 x 10 = 25000

8 0
3 years ago
What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element
Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0
3 years ago
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