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Furkat [3]
3 years ago
12

A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov

ing the charge? A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in moving the charge? Work is required in moving the positive charge from point A to point B. No work is performed or required in moving the positive charge from point A to point B. Work is performed in moving the positive charge from point A to point B. Work is both performed and required in moving the charge from point A to point B. SubmitRequest Answer Provide Feedback Next
Physics
1 answer:
Nitella [24]3 years ago
5 0

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

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Concerning the work done by a conservative force, which of the following statements, if any, are true? It can always be expresse
Vera_Pavlovna [14]

Answer:

It is independent of the path of the body and depends only on the starting and ending points.

Explanation:

In Physics we define a conservative force as a force that is independent of the path of the body and depends only on the starting and ending points.

For conservative forces we can write;

KEi + PEi = KEf +PEf

where;

KEi= initial kinetic energy

PEi= initial potential energy

KEf= final kinetic energy

PEf= final potential energy

This equation is known as the principle conservation of mechanical energy . It applies only to conservative forces where friction is negligible. The term KE + PE  is also known as the total mechanical energy of the system.

3 0
3 years ago
The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect
Fittoniya [83]

Answer:

Part a)

\rho = 0.0205 kg/m^3

Part b)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

PM = \rho RT

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

P = 900 Pa

T = 273 - 41 = 232 K

now we will have

(900)(0.044) = \rho (8.31)(232)

\rho = 0.0205 kg/m^3

Part b)

Now for the earth surface the density of air is given for

P = 101.6 kPa

T = 18 ^oC

so we will have

PM = \rho RT

(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

8 0
2 years ago
Light is an electromagnetic wave and travels at a speed of 3.00 × 108 m/s. The human eye is most sensitive to yellow-green light
Nimfa-mama [501]

Answer:

f = 5.95 \times 10^{14} Hz

Explanation:

As we know that the frequency of the wave is given as

f = \frac{c}{\lambda}

here we know that

c = 3 \times 10^8 m/s

also we know that

\lambda = 5.04 \times 10^{-7} m

now we have

f = \frac{3 \times 10^8}{5.04 \times 10^{-7}}

f = 5.95 \times 10^{14} Hz

6 0
3 years ago
A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring
topjm [15]

Here we can use the work energy theorem

W_f + W_s = K_f - K_i

here we know that

K_f = 0

as it come to rest finally

K_i = \frac{1}{2}mv_i^2

K_i = \frac{1}{2}\times 1\times 2^2

K_i = 2 J

now work done by friction force will be given as

W_f = - F_f \times d = -\mu mg d

W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu

Work done by spring force is given as

W_s = \frac{1}{2}k(x_i^2 - x_f^2)

W_s = \frac{1}{2}(10)( 0 - 0.10^2)

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so now plug in all data above

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4 0
3 years ago
Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advant
Lena [83]

Answer:

Impulse = Average force x time of contact

Explanation:

Impulsive force is a force which is very large but applied on a body for a very small duration of time.

Impulse is given by the change in momentum of the body.

Impulse = Average force x small time interval

When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.

So, when a person falls on the tile floor, there is no compression and thus, the time of contact is very small and thus the impulsive force is very large, due to  which the body may damage.

So, when a person falls on the carpeted floor, there is a compression and thus, the time of contact is comparatively large and thus the impulsive force is small, due to  which the body may safe.

3 0
3 years ago
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