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2 years ago

10

A football has a mass of 2 kg and it accelerates at 20 m/s2. What is its force? (You can use a calculator)

2 answers:

professor190 [17]2 years ago

8 0

Answer:10mps

Explanation:I do FLVS

miskamm [114]2 years ago

7 0

Answer:

Kent is making a scale model of his favorite train. The actual train is 12 feet long and 4 feet wide. Kent wants his mdth of his model if he uses the same ratio?

Explanation:

Kent is making a scale model of his favorite train. The actual train is 12 feet long and 4 feet wide. Kent wants his model to be 6 inches in length. Which would be the width of his modelhe uses the same ratio?

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Latitude, elevation, ocean currents, topography, and prevailing winds. There's probably a few others but these are the most important.

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A 0.72-m long string has a mass of 4.2 g. The string is under a tension of 84.1 N. What is the speed of a wave on this string?

frosja888 [35]

The speed of the wave in the string is 83.4 m/s

Explanation:

For a standing wave in a string, the speed of the wave is given by the equation:

$v=\frac{1}{2L}\sqrt{\frac{T}{m/L}}$

where

L is the length of the string

T is the tension in the string

m is the mass of the string

In this problem, we have:

L = 0.72 m

m = 4.2 g = 0.0042 kg

T = 84.1 N

Solving the equation, we find the speed of the wave:

$v=\frac{1}{2(0.72)}\sqrt{\frac{84.1}{0.0042/0.72}}=83.4 m/s$

Learn more about waves:

brainly.com/question/5354733

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5 0

3 years ago

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C

Alex_Xolod [135]

Answer:

26.466cm³/min

Explanation:

Given:

Volume 'V'= 320cm³

P= 95kPa

dP/dt = -11 kPa/minute

pressure P and volume V are related by the equation

P $V^{1.4}$ =C

we need to find dV/dt, so we will differentiate the above equation

$V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt} = \frac{d[C]}{dt}$

$\frac{dP}{dt} V^{1.4} + P(1.4)V^{0.4} \frac{dV}{dt} =0$

lets solve for dV/dt, we will have

$\frac{dV}{dt} =\frac{-\frac{dP}{dt} V^{1.4} }{P(1.4)V^{0.4} } \\\frac{dV}{dt} =- \frac{-\frac{dP}{dt} V}{P(1.4)}$

$\frac{dV}{dt} = -\frac{(-11 ) 320}{95(1.4)}$ (plugged in all the values at the instant)

$\frac{dV}{dt}$ = 26.466

Therefore, the volume increasing at the rate of 26.466cm³/min at this instant

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3 years ago

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