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spin [16.1K]
3 years ago
5

Neylin engineered (drove) a horse train from Miami and traveled toward New York. 14

Physics
1 answer:
kirza4 [7]3 years ago
6 0

Answer:

The answer would come out to be 27km/h

Explanation:

I took the test.

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Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o
Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.  

\lambda_{proton} = 6.05x10^{-14}m < \lambda_{electron} = 1.10x10^{-10}m

Part B:

The proton has a smaller wavelength than the electron.

\lambda_{proton} = 1.29x10^{-13}m < \lambda_{electron} = 5.525x10^{-12}m

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

\lambda = \frac{h}{p} (1)

Where h is the Planck's constant and p is the momentum.

\lambda = \frac{h}{mv} (2)

Part A

Case for the electron:

\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

But J = Kg.m^{2}/s^{2}

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

\lambda = 1.10x10^{-10}m

Case for the proton:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}

\lambda = 6.05x10^{-14}m

Hence, the proton has a smaller wavelength than the electron.  

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

KE = \frac{1}{2}mv^{2}

2KE = mv^{2}

v^{2} = \frac{2KE}{m}

v = \sqrt{\frac{2KE}{m}}  (3)

Case for the electron:

v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}

but 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}

v = 1.316x10^{8}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}    

\lambda = 5.525x10^{-12}m

Case for the proton :

v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}

But 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}

v = 3.07x10^{6}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}

\lambda = 1.29x10^{-13}m    

Hence, the proton has a smaller wavelength than the electron.

7 0
3 years ago
Matter is made up of tiny particles which can be atoms,molecules, or electrically charged particles called?.
yuradex [85]

Answer: alpha particle. i think

Explanation:

6 0
3 years ago
Which measurements are equal to 321 decimeters
vlabodo [156]
You are currently converting Distance and Length units from Centimeters to Feet 321 Centimeters (cm) = 10.5315 Feet (ft) This is a hard one but see if this helps if not let me now and i can try again..
7 0
4 years ago
In a uniform circular motion map, what is always true? Check all that apply.
grandymaker [24]

Answer:

Velocity vectors are always perpendicular to the circle.

Acceleration vectors point toward the center of the circle.

Velocity vectors are the same length.

Explanation:

(THESE ARE NOT MY WORDS BTW)

1) Acceleration and velocity are vectorial quantities, which means they have magnitude and direction.

2) In a circular motion velocity direction changes all the time, which means that it is accelerated.

3) In a uniform circular motiion, the velocity changes in a constant value. This is the rate of change of velocity, which is the magnitude of the acceleration, is constant (uniform).

4) The velocity is perpendicular to the path, i.e. the circle. You can see it if you think that if the object stopped changing the direction, then the object would follow a straight path (as per inertia principle). That is why this velocity is called tangential velocity (to differentiate it of the angular velocity).

This is what the option C says "Velocity vectors are always perpendicular to the circle". Then this is true.

5) The constant change of direction in a circular path, means that the object is been pushed, accelerated, toward the center of a circle. This is, all the time the object in motion tries to follow the perpendicular path but a push (a force) directed to the center of the circle changes its direction. Such force accelerates the object toward the center of the circle. So, the acceleration vectors point toward the center of the circle, which is what the option D says. So, this is also true.

6) Since the motion is uniform, the magnitude or length of the velocity vectors are always the same, are constant. So, the option E. is also true.

6 0
3 years ago
Read 2 more answers
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
Nina [5.8K]

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

4 0
3 years ago
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