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Valentin [98]
2 years ago
11

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

of the electric field at this location?
2.2 × 1026 NC
2.8 × 10−25 NC
1.5 × 1010 N/C
6.6 × 10−11 N/C
Physics
1 answer:
AveGali [126]2 years ago
7 0

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

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Why do raindrops fall with a constant speed during the later stages of their descents?
BartSMP [9]
Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)

Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
6 0
3 years ago
A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the
vitfil [10]

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}

v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}

\mathbf{v_{avg} =12.31 \ m/s}

8 0
2 years ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
3 years ago
What volume will be occupied by 20g of Copper if the density of Copper is 9.1g/ml?
Eva8 [605]

Answer:

The volume of copper is 2.198 ml

Explanation:

Given;

mass of copper, m = 20 g

density of copper, ρ = 9.1 g/ml

Density is given by;

Density = mass / volume

Volume = mass / density

Volume = (20 g) / (9.1 g/ml)

Volume = 2.198 ml

Therefore, the volume of copper is 2.198 ml

3 0
3 years ago
In which space, outdoors or in your classroom, would it be easier to hear a musician? Explain
bixtya [17]

Answer:

It is easier to hear a musician in the classroom than outdoors

Explanation:

It is easier to hear a musician in the classroom due to the improved acoustics provided by the walls of the classroom whereby along with the direct sound of the musician, which is the lead source of the sounds, there is an increased number of indirect sound reaching the ear in the classroom than outdoors and due to precedence effect, all the sound appear to come from the musician

In music played outside, along side the direct sound from the musician, the indirect sound that reach the ear is echoed from maybe by only the ground while the majority of the sound from the music wanders away with the wind and in other directions as well as being absorbed such that speakers will be required to improve the sound of the music outdoors.

7 0
3 years ago
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