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3 years ago

This process releases a large amount of . Energy ,oxygen ,or water

Physics

1 answer:

EleoNora [17]3 years ago

4 0

Answer:

carbon dioxide i think please be right

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A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)

gavmur [86]

<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

5 0

3 years ago

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What is an odometer?

ra1l [238]

Answer:

It is a instrument used to measure the distance traveled by a vehicle.

Explanation:

This is what it looks like ↓

6 0

3 years ago

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A sportscar has a mass of 1500 kg and accelerates at 5 meters per second squared. What is the magnitude of the force acting on t

Hatshy [7]

Answer:

7500 Newtons

Explanation:

Mass of the sportscar= 1500 kg

Acceleration of the sportscar= 5m/s^2

Hence, let the Force acting on it be F

$We\ know\ that,\\Force=Mass*Acceleration\\F=ma\\\\Here,\\F=1500*5\\=7500 kg m/s^2\ or\ 7500\ Newtons$

4 0

3 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl

Furkat [3]

Answer:

a) $F = 78.606\,N$ , b) $F = 88.911\,N$

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m \cdot g \cdot \sin \theta$

$F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}$

$F = 78.606\,N$

b) The equations of equilibrium are the following:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m\cdot (a + g \cdot \sin \theta)$

$F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})$

$F = 88.911\,N$

3 0

3 years ago

An object start from rest with a constant acceleration of 8.00 m/s^2 along straight line.

san4es73 [151]

I’ll say c Bc it make more since to find the travel distance

7 0

3 years ago