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Naily [24]
1 year ago
12

A particle moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.

72 m/s at t=5.59s what is the particle's acceleration in m/s?
Physics
1 answer:
mrs_skeptik [129]1 year ago
4 0

The particle's acceleration is 5.1 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²

Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s

The given parameters are;

  • V1 = 2.35 m/s
  • V2 = - 8.72 m/s
  • T1 = 3.42s
  • T2 = 5.59s

Acceleration a = ΔV ÷ ΔT

a = (2.35 + 8.72) / (5.59 - 3.42)

a = 11.07 / 2.17

a = 5.1 m/s²

Therefore, the particle's acceleration is 5.1 m/s²

Learn more about Acceleration here: brainly.com/question/9069726

#SPJ1

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How is thermal equilibrium reached? Question 3 options: When both objects have the same temperature When objects have the same 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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