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2 years ago

Take a look at a bicycle with gears. Using what you have just learned, answer the following questions:

Physics

1 answer:

Harlamova29_29 [7]2 years ago

4 0

Answer:

answer a: a large front gear with a small back gear

answer b: a small front gear with a large back gear

Explanation:

just simple gearing ratios

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Why do you see colors when you look at reflected light from a cd

Nady [450]

The colors that you see on the CD are created by white light reflecting from ridges in the metal. When light reflects off or passes through something with many small ridges or scratches, you often get rainbow colors and interesting patterns. They are called interference patterns.

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3 years ago

Give 10 examples of units you might use or see in any given day

Ierofanga [76]

Cups
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tablespoon
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3 0

3 years ago

A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the

REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0

3 years ago

Steel train rails are laid in 12.0-m-long segments placed end to end. The rails are laid on a winter day when their temperature

Yanka [14]

Answer:

a) Space = $6.05 x 10^{-3} m = 0.605 cm$

b) Stress= $-100.8 x 10^{6} Pa$

Explanation:

1) Data Given

$L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree$

2) Calculate the space using Linear thermal expansion formula

We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel. The formula is given by:

$\Delta L = L_i \alpha_{steel} \Delta T$

We have everything except the $\alpha_{steel}$ , so we look for this on a book and we find that $\alpha_{steel} = 1.2 x 10^{-5} C^{-1}$ , so we can replace.

$\Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm$

3) Calculate the stress of the steel

The Stress is the ratio of applied force F to a cross section area - defined as

$\sigma = \frac{F_n}{A}$

Since we don't have the force and the Area, we need to look for another way to find the stress.

For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:

$Y =\frac{F L}{A \Delta L}$ (1)