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ANEK [815]
1 year ago
12

You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surfac

e for which the coefficient of static friction with the box is 0.8, what is the mass of the heaviest box you will be able to move?
Physics
1 answer:
RoseWind [281]1 year ago
8 0

Answer:

103.44g

Explanation:

R=f/u

=750/0.8

=937.5N

m = 937.5/10cos25

m = 103.44g

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B. To have an earthquake there must be a fault line (where two or more tectonic plates meet) so if San Fran. has earthquakes they’re on a fault line.
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A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
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Answer:

The thickness of the oil slick is 1.95\times10^{-7}\ m

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

2nt= m\lambda

Where, n = Index of refraction

t = thickness

\lambda = wavelength

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2\times1.28 \times t=1\times\times500\times10^{-9}

t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }

t=1.95\times10^{-7}\ m

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3 years ago
What is it called when a surface takes light in without reflecting it
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Calculate the force exerted on the wall assuming that force is horizontal and using the data in the schematic representation of
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The picture attached shows the calculation

8 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
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Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

7 0
3 years ago
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