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MA_775_DIABLO [31]
3 years ago
12

A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?

Physics
2 answers:
S_A_V [24]3 years ago
7 0

Answer:

Force, F = 57.134 N  

Explanation:

It is given that,

Mass of the sled, m = 53 kg

Coefficient of kinetic friction, \mu_k=0.11

We need to find the force of friction felt by sled. The force of friction is given by :

F=\mu mg

F=0.11\times 53\ kg\times 9.8\ m/s^2

F = 57.134 N

So, the force of friction felt by the sled is 57.134 N. Hence, this is the required solution.

Anon25 [30]3 years ago
6 0

Answer:

57N

Explanation:

F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }

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A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium a
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Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

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Temperature (T) = 28.0 °C

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Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

8 0
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A spring scale hung from the ceiling stretches by 5.2 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl
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The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha
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Answer:

H=41.3kJmol^{-1}

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}

if we insert values, we arrive at

ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}

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