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beks73 [17]
3 years ago
14

A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm

2. What is the current in the wire
Physics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer:

5.95 A

Explanation:

From the question

R = ρL/A..................... Equation 1

Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.

Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m

Substitute these values into equation 1

R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶

R = 0.084 Ω.

Finally, using Ohm law,

V = IR

Where V = Voltage, I = current

Make I the subject of the equation

I = V/R............... Equation 2

I = 0.5/0.084

I = 5.95 A

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We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
a ball of mass 16 kg on the end of a string is spun at a constant speed of 2 m/s in a horizontal circle with a radius of 1m. Wha
miss Akunina [59]

The work done by the centripetal force during om complete revolution is 401.92 J.

<h3>What is centripetal force?</h3>

Centripetal force is a force that acts on a body undergoing a circular motion and is directed towards the center of the circle in which the body is moving.

To Calculate the work done by the centripetal force during one complete revolution, we use the formula below.

Formula:

  • W = (mv²/r)2πr
  • W = 2πmv²................... Equation 1

Where:

  • W = Work done by the centripetal force
  • m = mass of the ball
  • v = velocity of the ball
  • π = pie

From the question,

Given:

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Substitute these values into equation 1

  • W = 2×3.14×16×2²
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Hence, The work done by the centripetal force during om complete revolution is 401.92 J.
Learn more about centripetal force here: brainly.com/question/20905151

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