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andrey2020 [161]

3 years ago

8

Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg

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2 answers:

Strike441 [17]3 years ago

7 0

Lincoln has a mass of 41 kilograms

GrogVix [38]3 years ago

4 0

Answer:

41 kg

Explanation:

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The most basic difference between elliptical galaxies and spiral galaxies is that ________.

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Which identification number is the most important component of your personal identity

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The answer is Social Security Number.

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3 years ago

A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction bet

Vlada [557]

Answer: $1.95 m/s^2$

Explanation:

Given

inclination $\theta =30.6^{\circ}$

coefficient of kinetic friction $\mu =0.36$

As crate is moving Down therefore friction will oppose the motion

using FBD

$mg\sin \theta -f_r=ma$

$f_r=\mu N$

$f_r=\mu mg\cos \theta$

$mg\sin \theta -\mu mg\cos \theta =ma$

$a=g\sin \theta -\mu g\cos \theta$

$a=g(\sin (30.6)-0.36\cdot \cos (30.6))$

$a=9.8\times 0.199$

$a=1.95 m/s^2$

6 0

3 years ago

How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t

QveST [7]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

$Q=mC\Delta T\\\\Q=\rho VC\Delta T$

$(m=\rho V)$

where,

Q = amount of heat required = ?

m = mass

$\rho$ = density of air = $1.20kg/m^3$

V = volume of air

C = specific heat of air = $1006J/kg^oC$

$\Delta T$ = change in temperature = $10.0^oC$

Now put all the given values in above formula, we get:

$Q=\rho VC\Delta T$

$Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)$

$Q=1.449\times 10^6J$

Now we have to calculate the time required.

Formula used :

$t=\frac{Q}{P}$

where,

t = time required = ?

Q = amount of heat required = $1.449\times 10^6J$

P = power = 1500 W

Now put all the given values in above formula, we get:

$t=\frac{1.449\times 10^6J}{1500W}$

$t=966s\times \frac{1min}{60s}=16.1min$

Thus, the time required is, 16.1 minutes.

3 0

3 years ago