All categories

1 year ago

Note: Use g=9.8 N/kg in this assignment.

Physics

1 answer:

Karolina [17]1 year ago

5 0

Answer:

16.12

thistle seed oil benefits to be deposited on your way

You might be interested in

Acceleration Practice

Trava [24]

Answer:

0.85 m/s²

Explanation:

Acceleration is change in velocity over change in time. In mathematically, it can be expressed as:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{\vec v_2 - \vec v_1}{t_2-t_1}}$

Our final velocity is given to be 17 m/s in 20 seconds. Initial velocity is at starting point which is 0 m/s in 0 second. Therefore:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17-0}{20-0}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17}{20}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = 0.85 \ \, \sf{m/s^2}}$

Therefore, the acceleration of a horse from starting point to 17 m/s in 20 seconds is 0.85 m/s²

8 0

1 year ago

Read 2 more answers

What is a landform created by plate motion?

8090 [49]

Answer: Volcanoes and ridges are landforms that are created by the movement of tectonic plates.

Explanation:

5 0

2 years ago

Read 2 more answers

Which of the following is a subsurface event takes place during the rock cycle

alukav5142 [94]

Answer:

The answer is A. Cementing...

Explanation:

hope this helps

4 0

3 years ago

Read 2 more answers

What's forces are acting on the compass needle to cause it to point northeast

USPshnik [31]

Earth's magnetic field is the force acting on the compass needle to cause it to point northeast.

3 0

3 years ago

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):

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

7 0

2 years ago