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2 years ago

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An elevator is pulled up by a cable with a force of 65,000 N. The upward acceleration of the elevator is 1.8 m/s/s. What is the

mass of the elevator?
a. 1.8 kg.
b. 1700 kg.
c. 54880 kg.
d. 5600 kg.
e. 36111 kg.

1 answer:

goblinko [34]2 years ago

6 0

Answer:

36111 kg

Explanation:

Given

Force = 65000N

Acceleration = 1.8m/s²

Required

Determine the mass of the elevator

This question will be answered using the following Force formula.

Force = Mass * Acceleration

Substitute values for Force and Acceleration

65000N = Mass * 1.8m/s²

Make Mass the subject

Mass = 65000N/1.8m/s²

Mass = 36111.11 kg

From the list of given options, option E answers the question.

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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

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2 years ago

A car weighing 14,700 N is speeding down a highway with a velocity of 99 km/h. What is the

tankabanditka [31]

Answer: 148348.6239 kg•m/s

Explanation: Firstly, we need to convert the 14700 N into kilograms, and to do so, use the formula net force is equal to mass times acceleration and rearrange the formula to find mass like shown below...

F = ma

F/a = m

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Now that we convert it into kilograms, plug all the numbers into the variable of the momentum formula.

Momentum formula is P = mass x velocity

Like this:

P = 1498.470948 x 99

p = 148348.6239 kg•m/s.

I believe that is your answer, hope that helps you even a bit out.

Thanks.

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