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salantis [7]
4 years ago
10

Problem 1 Two positive point charges are released on a frictionless surface. Which of the following best describes their subsequ

ent motion? Select One of the Following: (a) The particles remain stationary. (b) The particles move apart with constant velocity. (c) The particles move apart with a velocity that decreases with time. (d) The particles move apart with a velocity that increases with time. (e) The particles move apart with a velocity that increases for a while and then becomes constant.
Physics
1 answer:
Tatiana [17]4 years ago
8 0

Answer:

(e) The particles move apart with a velocity that increases for a while and then becomes constant.

Explanation:

Each particle feels a repulsive (because they have same sign of charge) electric force from the each other:

F=\frac{kq_{1}q{2}}{d^{2}}

and

F=ma\\

So each particle feels a repulsive force proportional to the quadratic inverse of the distance.that means that the charges begin to move away, and the further away they are from each other, the force (and therefore the acceleration) decreases, at a rate inversely proportional to the square of the distance. Theoretically this acceleration will never be zero, but in practice it will at some point reach a value very close to zero. Then the speed will grow for a while and when the acceleration has reached almost zero, the speed will practically remain constant.

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A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced s
BARSIC [14]

Answer:

The wavelength is  \lambda_R  =  649 *10^{-9}\ m

Explanation:

From the question we are told that

   The wavelength of the red laser is  \lambda_r  =  632.8 \ nm =  632.8 *10^{-9}\ m

    The spacing between  the fringe is  y_r  =  6.00\ mm =  6.00*10^{-3}  \  m

   The spacing between  the fringe for smaller laser point  is  y_R   = 6.19 \ mm =  6.19 *10^{-3} \  m

      Generally the spacing between  the fringe is mathematically represented as

       y  =  \frac{D *  \lambda  }{d}

Here  D is the distance to the screen

    and  d is the distance of the slit separation

Now for both laser red light light and  small laser  point  D and  d are same for this experiment

So

         \frac{y_r}{\lambda_r}  =  \frac{D}{d}

=>      \frac{y_r}{\lambda_r}  = \frac{y_R}{\lambda_R}

Where \lambda_R  is the wavelength produced by the small laser pointer

  So

           \frac{6.0 *10^{-3}}{ 632.8*10^{-9}}  = \frac{ 6.15 *10^{-9}}{\lambda_R}

=>       \lambda_R  =  649 *10^{-9}\ m

7 0
3 years ago
A steel ball with mass m=5.21 g is moving horizontally with speed ????=412 m/s when it strikes a block of hardened steel with ma
docker41 [41]

Answer:

a) The speed of the block immediately after the collision is v_{2f}=(0.289\±0.002)m/s.

b) The impulse exerted on the block is p_{2}=(4.2772\±0.0296)kg*m/s.

Explanation:

Hi

a) As this is a perfectly elastic collision, we can use the formula  v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}+ (\frac{m_{2}-m_{1}}{m_{1}+m_{2}} )v_{2i}, due v_{2i}=0m/s, we obtain v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}. Then with the data that we know m_{1}=5.21g=0.00521kg, m_{2}=14.8kg and v_{1i}=412m/s, therefore v_{2f}=(\frac{2(0.00521kg) }{0.00521kg+14.8kg} ) 412m/s=0.289m/s or v_{2f}=(0.289\±0.002)m/s adding uncertainty.

b) Now that we know the speed we can use p_{2}=m_{2}*v_{f2} =14.8kg*(0.289\±0.002)m/s=(4.2772\±0.0296)kg*m/s.

8 0
4 years ago
Why are the stars that appear in the night sky not visible during the day?
xxTIMURxx [149]
Because the sun is not hitting the light on them and there not as bright like when we see them when it's dark.
6 0
3 years ago
Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop
Sonja [21]
For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
8 0
3 years ago
If you disconnect the wires from the battery and then reconnect them at the opposite ends of the battery, how does that change t
Likurg_2 [28]

Answer:

Explanation:

The current from the battery always flows from the positive terminal of the battery to the negative terminal of the battery.

If we connect a red coloured wire to the positive terminal of the battery and black coloured wire to the negative terminal of the battery, and then reverses the wire to their respective terminals, then there is no change in the direction of flow of current. It does not matter that which wire is connected to the particular terminal. The current always flow from positive to negative terminal of the battery outside the battery.

6 0
3 years ago
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