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3 years ago

7

An 8 Newton wooden block slides across a horizontal wooden floor at constant velocity. What is the magi notice of the force of k

inetic friction between the block and the floor

1 answer:

kondaur [170]3 years ago

8 0

Answer:

The force of kinetic friction between the block and the floor is 2.4 N

Explanation:

The given parameters are;

The weight of the block = 8 Newtons

The velocity of the block = Constant velocity

Taking the kinetic friction for wood, $\mu _k$ = 0.3, we have;

The normal reaction of the block on horizontal ground, N = The weight of the block = 8 N

The force of kinetic friction between , $F_k$ = $\mu _k$ × N

Therefore, we have;

The force of kinetic friction between the block and the floor, $F_k$ = 0.3 × 8 N = 2.4 N.

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Which of the following in the list is the best conductor: plastic, metal, styrofoam, or glass?

tensa zangetsu [6.8K]

Answer:

Metal

Explanation:

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A container of gas molecules is at a pressure of 2 atm and has amass density of 1.7 grams per liter. All of the molecules in the

Tpy6a [65]

Answer:

The speed of nitrogen molecule is 1.87 m/s.

Explanation:

Given that,

Pressure = 2 atm

Density = 1.7 grams/liter

Atomic weight = 28 grams

We need to calculate the temperature

Using formula of idea gas

$PV=nRT$

$P=\dfrac{WRT}{VM}$

$P=\dfrac{\rho RT}{M}$

$T=\dfrac{PM}{\rho R}$

Put the value into the formula

$T=\dfrac{2\times28}{1.7\times0.0821}$

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We need to calculate the speed of nitrogen molecule

Using formula of RMS speed

$V_{rms}=\sqrt{\dfrac{3RT}{M}}$

$V_{rms}=\sqrt{\dfrac{3\times0.0821\times401.2}{28}}$

$V_{rms}=1.87\ m/s$

Hence, The speed of nitrogen molecule is 1.87 m/s.

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3 years ago

Saul convert 2.392 hectoliters to liters what should his new number be?

murzikaleks [220]

His new number should be: 239.2 Liters

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4 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

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4 years ago

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Marina CMI [18]

Answer:

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This is the initial velocity in the negative y-direction. Impulse is given by:

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3 years ago