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Arada [10]
3 years ago
14

A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sle

d attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Physics
1 answer:
ycow [4]3 years ago
4 0

Answer:

\mu=0.185

Explanation:

From the question we are told that:

Mass m=17kg

Force F=33N

Velocity v=1.6m/s

Distance d= 9.8m

Generally the equation for Work done is mathematically given by

 W=\triangle K.E+\triangle P.E

Where

 \triangle K.E=(F-F_f)*2

 F_f=F+\frac{\triangle K.E}{d}

 F_f=33+\frac{0.5*17*1.6^2}{9.8}

 F_f=30.8N

Since

 f = \mu*m*g

 \mu= 30.8/(m*g)

 \mu= 30.8/(17*9.81)

 \mu=0.185

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lidiya [134]

Answer:

D

Explanation:

because 50.0/10.0 = 5.0

5 0
3 years ago
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Daniel [21]

The red box must way more. Gravitational potential energy is the product of a an objects mass times the acceleration due to gravity (which is constant on earth) times its height. Since the objects are on the same shelf they are at the same height, and since gravitational acceleration is constant as long as we stay on planet earth, then the mass is the only possible thing that could have changed. This means that the red box must weigh more than the blue box.

3 0
3 years ago
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

V_{2} = 98,44 cm^{3}

Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

For Temperature the ideal gases formula is:

T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

P_{3} = 1,317 * 10^{10} Pa

7 0
3 years ago
In the bohr model, what is the ratio of its kinetic energy to its potential energy?
stich3 [128]

The centrifugal force C = mv^2/r = kq^2/r^2 = P centripetal force. m is the electron mass, q are the proton and electron charges (opposites), and r is the Bohr radius.

Thus 1/2 mv^2/r = 1/2 kq^2/r^2 and KE = 1/2 mv^2 = 1/2 kq^2/r = 1/2 PE

Therefore KE/PE = 1/2, no matter what state the electron is in.

8 0
3 years ago
The atomic mass of an element is
inessss [21]

Here are the answers to the question. Make sure to give a valid reason, please.

A. the sum of the protons and neutrons in one atom of the element.

B. a ratio based on the mass of a carbon-12 atom.

C. a weighted average of the masses of an element's isotopes.

D. twice the number of protons in one atom of the element.

6 0
4 years ago
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