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3 years ago

The fact that we can define electric potential energy means that:

Physics

2 answers:

stealth61 [152]3 years ago

5 0

Answer: I'd go with Option B if you're allowed to pick One Option Only.

Option A is wrong. Electric Forces arent Non - Conservative. They're Conservative forces.

Option B: We Can Only Define Potential Energies for conservative Forces/fields✅

Option C is wrong. Since the Electric Force is Conservative... It doesn't depend on the path taken.

Option D... There's a Point where the electric Potential Energy is Zero but the separation Distance Would be large.

E is wrong because the Workdone around a closed path is Zero.

Marysya12 [62]3 years ago

4 0

Answer:

Explanation:

câuC

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Which property of a wave is labeled "x" on the diagram?

faust18 [17]

C wavelength due to the fact the a frequency does not have an x but a y

8 0

3 years ago

How many strings of length 10 over the alphabet (a, b, c, d] have at least one b somewhere in the string?

Lapatulllka [165]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is B

Explanation:

The number of alphabet is n= 4 (a , b , c , d )

Generally the total number of string of length 10 over the 4 alphabets is

$N = 4^{10}$

Gnerally the number of string of length 10 that does not include b is

$T = 3^{10}$

Generally the number of strings of length 10 over the 4 alphabets that have at least one alphabet b somewhere in the string is

$G = N - T$

=> $G = 4^{10} - 3^{10}$

8 0

3 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

8 0

3 years ago

An object that is in free fall seems to be ______

bekas [8.4K]

An object that is in free fall seems to be (D) weightless.

Objects which are in free fall are said to be weightless because they only have the force of gravity acting upon them. Objects in free fall do not experience air resistance.

3 0

3 years ago

A 75 cm straight wire moves straight up through a 0.53 T magnetic field with a velocity of 16 m/s. What is the induced emf in th

Luden [163]

Answer:

Induced emf of the wire is 6.36 Volts.

Explanation:

It is given that,

Length of the wire, l = 75 cm = 0.75 m

Magnetic field, B = 0.53 T

Velocity, v = 16 m/s

The wire is moving straight up in the magnetic field. So, an emf is induced in the wire. It is given by :