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3 years ago

Will increasing length and cross-sectional area increase the resistance of an object?

Physics

1 answer:

Klio2033 [76]3 years ago

5 0

Answer:

Increasing length increases resistance

increasing cross sectional area reduces the resistance

Explanation:

The formula for resistance of an object is

$r = d \frac{l}{a}$

where r is resistance, d is resistivity of the material, l is length of material and a is cross sectional area of the object. This equation shows us that resistance is directly proportional to length and inversely proportional to cross sectional area. Hence, increasing length increases resistance while increasing cross sectional area reduces the resistance.

If these 2 variables are varied to the same extent, the net effect can be zero on the resistance.

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Suppose a point charge is located at the center of a spherical surface. The electric field at the surface of the sphere and the

emmasim [6.3K]

Answer:

<em>The flux through the sphere will remain the same, and the magnitude of the electric field will increase by four times.</em>

Explanation:

The electric flux is the number of electric field, passing through a given area. It is proportional to the electric field strength and the area through which this field passes.

If the radius of the sphere is halved, the area of the sphere will reduce by square of the reduction, which will be four times. The electric field lines will become closer together, or technically increase by a fourth of its initial value. The resultant effect is that the electric flux will remain the same.

If originally,

Φ = EA cos∅

where Φ is the electric flux through the sphere

E is the electric field on the sphere

A is the area of the sphere.

If the area of the sphere is reduced to half, then,

the area reduces to A/4,

and the electric field increases to be 4E on the sphere.

The flux now becomes

Φ = 4E x A/4 cos∅

which reduces to

Φ = EA cos∅

which is the initial electric flux on the sphere.

8 0

3 years ago

*PLEASE HELP WILL GIVE BRAINLIEST*

levacccp [35]

Answer: The order is 2,1,5,4,3

Explanation: Hope This helps.

3 0

3 years ago

Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(

zheka24 [161]

Answer:

Explanation:

=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!

Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.

Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From

Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?

a.There is no energy stored in the string: The string will remain straight for all subsequent times.

b.Energy will flow into the string, causing the standing wave to form at a later time.

c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.

d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.

3 0

3 years ago

A model rocket fired vertically from the ground ascends with a constant vertical acceleration of 52.7 m/s2 for 1.41 s. its fuel

olya-2409 [2.1K]

For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is

a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.

-52.7 = (0-v1)/1.41
v1 = 74.31 m/s

We can use v1 for the formula of the maximum height attained by an object thrown upwards:

Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m

The maximum height attained by the model rocket is 281.42 m.

For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.

Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.

2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
x =52.4 m (distance it covered at constant acceleration)
Then. when it travels upwards only by a force of gravity,
d = v1(t) + 1/2*a*t^2
281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
t = 6.83 s (time when it lost fuel and reached its maximum height)

Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s

Therefore, the total time= 1.41+6.83+7.57 = 15.81 s

6 0

3 years ago

A commuter train travels from the airport for half an hour at 110 km/hr until it reaches the

nikklg [1K]

Answer:

it would be 10km

Explanation:

because that reaches max height before it falls stupid

7 0

3 years ago