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3 years ago

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2 answers:

Nady [450]3 years ago

7 0

Answer: It's A) S C A N N I N G <em>V R O T H E R S</em>

Explanation: I know because I took the test and got it correct/FLVS, you're welcome.

Tanzania [10]3 years ago

4 0

I think it’s watching your mirrors so C

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Describe different types of force in nature at least five

sineoko [7]

Weathering, Erosion, deposition, acid rain, precipitation

Explanation:

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3 years ago

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a team of dogs is pulling a sled with a net force of 2000 N. The sled is accelerating a 2.2 m/s^2. What is the mass of the sled?

sweet-ann [11.9K]

Force, F = ma

Where m = mass in kg, a = acceleration in m/s², Force, F is in N.

F = ma

2000 = m*2.2

2.2m = 2000

m = 2000/2.2

m ≈ 909.09

Mass is ≈ 909.09 kg.

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3 years ago

An object has a velocity of 8 m/s and a kinetic energy of 480 j what is the mass of the object

dedylja [7]

We Know,

K.E. = 1/2 mv²

480 = 1/2 (m)(8)²

m = 960/64

m = 15 Kg

So, the mass of the object is 15 Kg

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3 years ago

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A dog running to the right at 4 m/s sees a ball and accelerates steadily to catch it. The dog accelerates to the right at a rate

antoniya [11.8K]

Answer:

D.-4.798m/s

Explanation:

Greetings !

Given values

$u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec$

Solve for V of the given expression

Firstly, recall the velocity-time equation

$v = u + at$

plug in known values to the equation

$v = (4) + (0.21)(3.8)$

solve for final velocity

$v = 4.792ms$

Hope it helps!

6 0

1 year 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

3 years ago