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cluponka [151]
3 years ago
10

A football player runs 20 meters North of a football field, and then 15 meters East. The total motion lasted 15 seconds. What wa

s his average speed
Physics
1 answer:
vlada-n [284]3 years ago
4 0
Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds

Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3  meters per second
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Doubling the voltage in this arrangement both doubles the voltage drop across the resistor and the current through it. The bulb will be much brighter.

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A sinusoidal sound wave moves through a medium and is described by the displacement wave function
ZanzabumX [31]

By applying the wave equation we know that the maximum speed of the element's oscillatory motion is 1716 micrometer / s.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

y = A cos (kx - ωt)

where y is y-axis displacement, A is amplitude, k is wave number, x is x-axis displacement, ω is angular speed and t is time.

the wavenumber and angular speed of the wave equation can be determined respectively by

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where k is the wavenumber, λ is wavelength and f is frequency.

From the question above, we know that:

y = 2.00cos (15.7x - 858t)

v = dy / dt

v = d(2.00cos (15.7x - 858t)) / dt

v = -858 x (-2.00sin(15.7x - 858t))

v = 1716 sin(15.7x - 858t) micrometer/s

maximum velocity can be reached when (sinθ = 1), hence

v = 1716 sin(15.7x - 858t)

v = 1716 x 1

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For more on wave equation on: brainly.com/question/25699025

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Force is a vector because it has both ___
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C. Size and Direction

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Iron ball sinks in water but not in mercury. Why?
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Read 2 more answers
Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
4 years ago
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