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Fynjy0 [20]
3 years ago
14

a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resi

stance r/9 what is the diameter of the second wire g
Physics
1 answer:
777dan777 [17]3 years ago
5 0

Answer:

<h2>d₂ = 3d</h2><h2>The diameter of the second wire is 3 times that of the initial wire.</h2>

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

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3.7. A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle 23.1 degrees north of east, and finally 15.0 m west.
andriy [413]

Answer:

Explanation:

We shall represent displacement of dog in vector form , in terms of i , j , i representing east  and  j representing north .

Dog travels 3.5 m south .

Displacement D₁ = - 3.5 j

then dog travels 8.2 m , 23.1 degree north of east

Displacement D₂ = 8.2 cos23.1 i + 8.2 sin23 j

D₂ = 8.2 cos23.1 i + 8.2 sin23.1  j

= 7.54 i + 3.22 j  

Third displacement

D₃ = - 15i

Total displacement = D₁ + D₂ + D₃

= - 3.5 j + 7.54 i + 3.22 j  -15i

= - 7.46 i - 0.28 j

Magnitude of displacement = √ ( 7.46² + .28²)

= √(55.65 + .08 )

= 7.46 m

b ) Direction of displacement

If Ф be angle , displacement makes with west direction

TanФ =  .08 / 55.65 = .00143

Ф = .082 degree south of west or almost west .

From east , this angle = 180 + .082 = 180.082 , counterclockwise .

5 0
2 years ago
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0
4 years ago
A stone is dropped from the upper observation deck of a tower, 250 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dista
Vitek1552 [10]

(a) y(t)=250 - 4.9 t^2

For an object in free-fall, the vertical position at time t is given by:

y(t) = h + ut - \frac{1}{2}gt^2

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

y(t) = h - \frac{1}{2}gt^2=0

So for the stone in the problem, we have

250 - 4.9 t^2 = 0

Solving for t, we find:

t=\sqrt{\frac{250}{4.9}}=7.14 s

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

v(t) = u - gt

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

6 0
3 years ago
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