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Crank
3 years ago
15

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t

ire at a distance of 0.379 m 0.379 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0

Explanation:

The given data is as follows.

       Angular velocity (\omega) = 2.23 rps

     Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

             = 2.23 revolutions per second

             = 2.23 \times 2 \times 3.14 rad/s

             = 14.01 rad/s

Now, tangential speed will be calculated as follows.

  Tangential speed, v = R \times \omega

                               = 0.379 x 14.01

                               = 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

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Answer:

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Explanation:

We have,

Mass of a squirrel is 0.765 kg.

He jumps off the tree and hits the ground with 125 joules of energy.

It is required to find the height up on the tree the squirrel was when it jumped.

The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

E=mgh

h is height up on the tree the squirrel was when it jumped

h=\dfrac{E}{mg}\\\\h=\dfrac{125\ J}{0.765\ kg\times 9.8\ m/s^2}\\\\h=16.67\ m

So, the squirrel will go to a height of 16.67 m.

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A 1.60 m cylindrical rod of diameter 0.450 cm is connected to a power supply that maintains a constant potential difference of 1
frosja888 [35]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

A 1.60 m cylindrical rod of diameter 0.450 cm is connected to a power supply that maintains a constant potential difference of 12.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.4 A , while at 92.0 ∘C it reads 17.4 A . You can ignore any thermal expansion of the rod.

a) Find the resistivity and for the material of the rod at 20° C .

b) Find the temperature coefficient of resistivity at 20° C for the material of the rod.

Given Information:

Room temperature = T₀ = 20° C

Temperature = T = 92° C

Current at 20° C = I₀ = 18.4 A

Current at 92° C = I = 17.4 A

Voltage = V = 12 V

Length = L = 1.60 m

Diameter = d = 0.450 cm = 0.0045 m

Required Information:

Resistivity of the material at 20° C = ρ = ?

Temperature coefficient of resistivity at 20° C = α = ?

Answer:

Resistivity of the material at 20° C = 2.062x10⁻⁶ Ω.m

Temperature coefficient of resistivity at 20° C = 7.986x10⁻⁴ per °C

Explanation:

a) We want to find out the resistivity of the material at 20° C

The resistivity of any material can be found using,

ρ = R₀A/L

Where R₀ is the resistance of the rod at 20° C, A is the area of rod and L is the length of the cylindrical rod.

We also know that area is given by

A = πr²

where r = d/2 = 0.0045/2 = 0.00225 m

A = π(0.00225)²  

A = 5.062⁻⁶ m²

We know that resistance of the material is given by

R₀ = V/I₀

R₀ = 12/18.4

R₀ = 0.6521 Ω

Therefore, the resistivity of the material is

ρ = R₀A/L

ρ = (0.6521*5.062⁻⁶)/1.60

ρ = 2.062x10⁻⁶ Ω.m

b) We want to find out the temperature coefficient of resistivity of the rod at 20° C

The temperature coefficient of resistivity is given by

α = R/R₀ - 1/(T - T₀)

Where R is the resistance of the rod at 90° C

R = V/I

R = 12/17.4

R = 0.6896 Ω

α = R/R₀ - 1/(T - T₀)

α = (0.6896/0.6521) - 1/(92° - 20°)

α = 0.0575/72°

α = 0.000798 per °C

α = 7.986x10⁻⁴ per °C

3 0
3 years ago
Read 2 more answers
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