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3 years ago

14

When silk is rubbed with glass what silk is transferred?

1 answer:

serious [3.7K]3 years ago

5 0

Answer:

When glass rod is rubbed with silk clothes, glass quickly loses electrons, and silk takes electrons out of the glass rod

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Which units are used to measure both velocity and speed? Check all that apply. m/s d/t km/h mph lph

GaryK [48]

The units used to measure speed and velocity are as follows: m/s, Km/h, and mph.

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4 years ago

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SCIENCE whoever gets this first will get a brainlest

gulaghasi [49]

Answer:

im not 100% sure i should be doing this but i found a website that will help a ton.

https://geo.libretexts.org/Courses/Gettysburg_College/Book%3A_An_Introduction_to_Geology_(Johnson_Affolter_Inkenbrandt_and_Mosher)/02%3A_Plate_Tectonics/2.01%3A_Alfred_Wegener%E2%80%99s_Continental_Drift_Hypothesis

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3 years ago

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Pressure and volume changes at a constant temperature can be calculated using

V125BC [204]

Answer:

Option A, Boyle's law

Explanation:

The complete question is

Pressure and volume changes at a constant temperature can be calculated using

a. Boyle's law. c. Kelvin's law.

b. Charles's law. d. Dalton's law.

Solution

In Boyle’s law, the gas is assumed to be ideal gas and at constant temperature. With these two conditions fixed, Boyle’s established that volume of gas varies inversely with the absolute pressure.

The basic mathematical representation of this phenomenon is as follows -

$P \alpha \frac{1}{V}$

OR

$P = \frac{k}{V} \\PV = k$

Where P is the pressure of ideal gas, V is the volume and k is the constant of proportionality.

Hence, option A is correct

3 0

3 years ago

Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)

Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force, $(FS_p)s=-k\Delta s-q(\Delta s)^3$

$k=350N/m$

$q=750 N/m^3$

We have to find the work must you do to compress this spring 15 cm.

$\Delta s=15 cm=0.15 m$

Using 1 m=100 cm

Work done= $\int_{0}^{0.15}-Fd(\Delta s)$

W= $-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)$

$W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}$

$W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750$

$W=4.033 J$

Ideal spring work= $0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J$

Percentage increase in work= $\frac{4.033-3.938}{3.928}\times 100=2.4$ %

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4 years ago

How to solve for h using the formula eg=mgh?

SVEN [57.7K]

Divide each side of the equation by mg .

8 0

4 years ago