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3 years ago

The metric system and metric conversions are?

1 answer:

8 0

The Metric System and Metric Conversions are both two different things , but still corresponds in a way. For example, the metric system is meter, liter, and gram as units of length. Metric conversions are when you use either meter, liter, or and gram and make a number to equal by moving a decimal.

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A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially

Alchen [17]

The new speed of car is 10.9 m/s

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According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

$m_1u_1 + m_2u_2 = (m_1 + m_2) v$

$v =\frac{m_1u_1}{m_1+m_2}$

$v = \frac{7950*15}{7950+2950}$

$v= 10.9 m/s$

Therefore, the new speed of care is 10.9 m/s

Learn more about momentum here:

brainly.com/question/22257327

#SPJ1

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2 years ago

Blank is the perceived frequency of a sound wave

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pitch is the perceived frequency of a sound wave.

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3 years ago

Why is newton's gravitational force called universal gravitational force

Lelechka [254]

Newton’s law of gravity is considered “universal” because it is believed to be applicable to the entire Universe (to a good approximation). Gravity does not just pull an apple from a tree, but also pulls the Moon to the Earth, the Earth to the Sun, and so on. Gravity is a fundamental force of nature.

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3 years ago

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Which of the following objects exerts a gravitational force?

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the earth exerts a gravitational force

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3 years ago

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A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L

GaryK [48]

Answer:

a) L₂ = 0.676 m

b) σ₁ = 2.28*10⁸ N/m²

σ₂ = 9.62*10⁸ N/m²

c) ε₁ = 0.00253678

ε₂ = 0.00457875

Explanation:

Given info

L₁ = 1.22 m

A₁ = 2.19 cm² = 2.19*10⁻⁴ m²

L₂ = ?

A₂ = 0.52 cm² = 0.52*10⁻⁴ m²

P = 5.00*10⁴ N

E₁ = 9*10¹⁰ N/m²

E₂ = 2.1*10¹¹ N/m²

In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that

ΔL₁ = ΔL₂ ⇒ P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)

⇒ L₂ = A₂*E₂*L₁ / (A₁*E₁)

⇒ L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)

⇒ L₂ = 0.676 m

The stress in the brass rod is obtained as follows

σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²

The stress in the niquel rod is obtained as follows

σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²

The strain in the brass rod is obtained as follows

σ₁ = E₁*ε₁ ⇒ ε₁ = σ₁ / E₁

⇒ ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678

The strain in the niquel rod is obtained as follows

σ₂ = E₂*ε₂ ⇒ ε₂ = σ₂ / E₂

⇒ ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875

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3 years ago