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3 years ago

Which of the following speeds is greatest? (2.54 cm = 1.00 in.)

Physics

1 answer:

Mama L [17]3 years ago

7 0

Answer:

Hi, There!

O 10 m/s

O 10 yd/s

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Have a great day!

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Which is the appropriate unit of measurement for pitch?

malfutka [58]

Hertz Pitch and frequency

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3 years ago

Experiment 1: A study is done to determine which of two fuel mixtures allows a rocket to travel farther over a period of time. R

leva [86]

Answer: D

Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.

This is the answer to the question because:

Both experiments do have a confounding variable.
Experiment 1 doesn't have to stay constant.
A double-blind experiment will not do anything to the placebo.
High blood pressure people will not make the results confusing.

The answer has to be the option D. Hope this helps you!

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3 years ago

I need to choose a theme for my physics assignment My experiment is finding g

Kobotan [32]

<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

$= \frac{GM}{ {R}^{2} }$

where, G = Gravitational constant

$= 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\$

M = Mass of the earth

$= 6 \times {10}^{24} \: kg$

R = Radius of the earth

$= 6.4 \times {10}^{6} m$

Putting these values of G, M and R in the above formula, we get

$g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2}$

So, the value of acceleration due to gravity is

$9.8m/s ^{2}$

Hope it helps.

Do comment if you have any query.

5 0

3 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

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3 years ago

Energy that flows from an objective with a higher temperature to an object with a lower temperature

dalvyx [7]

That's THERMAL energy, often referred to as "heat".

7 0

3 years ago