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3 years ago

2. An electron and a proton are separated by 5 cm:

Physics

1 answer:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

a) the charge of an electron is equivalent to the magnitude of the elementary charge but barring a negative sign since the side of the elementary charge is roughly 1.602 * 10 - 19 Columbus then the charge of the electronic is-1.602 * 10 - 19

b) b=2T on the electron moving in the magnetic field

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Which three elements have strong magnetic properties?

Elodia [21]

The answer is D. iron, cobalt, and nickel

8 0

3 years ago

Read 2 more answers

Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne

gavmur [86]

vector A has magnitude 12 m and direction +y

so we can say

$\vec A = 12 \hat j$

vector B has magnitude 33 m and direction - x

$\vec B = -33 \hat i$

Now the resultant of vector A and B is given as

$\vec A + \vec B = 12 \hat j - 33 \hat i$

now for direction of the two vectors resultant will be given as

$\theta = tan^{-1}\frac{12}{-33}$

$\theta = 160 degree$

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

$R = \sqrt{A^2 + B^2}$

$R = \sqrt{12^2 + 33^2} = 35.11 m$

so magnitude will be 35.11 m

6 0

3 years ago

Calculate the GPE of a rock with a mass of 55kg on a cliff that is 27m high

GaryK [48]

Answer:

P=mgh

P=mghm=55

P=mghm=55g=9.8 or ~10

P=mghm=55g=9.8 or ~10h=27

Explanation:

hope it will help you have a great day bye and Mark brainlist if the answer is correct

 $kai6417$ 

 #carry on learning

4 0

2 years ago

A 55 kg person falling with a velocity of 0.6

siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

3 0

3 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

4 years ago