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Charra [1.4K]
2 years ago
15

Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp

ress your answer with the appropriate units. Enter positive value if the force is in the positive xx-direction and negative value if the force is in the negative xx-direction.

Physics
1 answer:
UkoKoshka [18]2 years ago
6 0

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

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The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done =  mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0
2 years ago
A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?
Shtirlitz [24]
<span>Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.</span>
8 0
2 years ago
Monochromatic light is incident on a grating that is 75 mm wide and ruled with 50,000 lines. the second-order maximum is seen at
DIA [1.3K]

Answer:

The wavelength of the incident light is \lambda = 400 nm

Explanation:

Given data

Distance between the sits

d = \frac{0.075}{50000}

d = 1.5 × 10^{-6} m

\theta = 32.5°

m = 2

We know that the wavelength of the incident light is given by

\lambda = \frac{d\sin \theta}{m}

Put all the value in above formula we get

\lambda = \frac{1.5 (\sin 32.5)}{2}×10^{-6}

\lambda = 4 × 10^{-7} m

\lambda = 400 nm

Therefore the wavelength of the incident light is \lambda = 400 nm

4 0
2 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
2 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
ollegr [7]
The answer:
the relationship between elementary charge, potential difference and electrical potential energy is given by 
   E= qV
E:  lectrical potential energy
q:   elementary charge
V:   potential difference

but we have  e=abs val(q)=3 
so we have E= qV=3ex4.5V=<span>13.5 eV
</span>
the answer is <span>(4)13.5 eV</span>

8 0
2 years ago
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