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Natasha2012 [34]
3 years ago
15

2. A 56N force is exerted on an 8kg object. What is the object's acceleration?

Physics
1 answer:
White raven [17]3 years ago
7 0

acceleration = 7 m/s²

Explanation:

To determine the object's acceleration we use the following formula:

force (N) = mass (kg) × acceleration (m/s²)

acceleration (m/s²) = force (N) / mass (kg)

acceleration = 56 N / 8 kg

acceleration = 7 m/s²

Learn more about:

acceleration

brainly.com/question/12097332

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The answer is C tension; stretching; thinning
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Explain why in the construction of a lightning conductor one end is made into a sharp point​
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Answer:

A lightning conductor is made up of a sharp pointed metal (usually copper metal, as it is a very good conductor) connected directly to the ground. ... The lightning conductor is placed higher than the roof so that if lightning strikes, it strikes the conductor before it can reach the house.

Explanation:

A lightning conductor is made up of a sharp pointed metal (usually copper metal, as it is a very good conductor) connected directly to the ground. ... The lightning conductor is placed higher than the roof so that if lightning strikes, it strikes the conductor before it can reach the house.

3 0
3 years ago
Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same len
Mekhanik [1.2K]

Answer:

  • The difference in length for steel is 2.46 x 10⁻⁴ m
  • The difference in length for invar is 1.845 x 10⁻⁵ m

Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, \gamma_{st.} =  3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, \gamma_{in.} =  2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂  = L₁ + L₁αθ

L₂  - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = \frac{\gamma}{3}

Difference in length, for steel at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC

ΔL  = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC

ΔL  = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

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Answer:

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Explanation:

Enter a curve slower than the posted speed if your vehicle has a high center of gravity or if surface Traction is less than ideal.

Traction is an act of drawing or pulling something over a surface specially a road or a track or it also defined as the grip of a tire on road or a wheel on rail. So speed when entering a curve should be slower when the tire has low traction to avoid accident.

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