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Andrews [41]
3 years ago
15

4.

Physics
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:

3 0
3 years ago
Read 2 more answers
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.
5 0
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


7 0
3 years ago
Read 2 more answers
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
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