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Arlecino [84]
3 years ago
15

Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri

c force on one of the masses? What is the initial acceleration of each mass if they are released and allowed to move?
Physics
2 answers:
9966 [12]3 years ago
8 0

Force between two charges is given by

F =\frac{kq_1q_2}{r^2}

F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}

F = 5.625 * 10^{-4} N

Now in order to find the acceleration of each mass

we can use

F = ma

5.625 * 10^{-4} = 1 * a

a= 5.625 * 10^{-4} m/s^2

forsale [732]3 years ago
8 0
The force between two charged particles can be determined using the formula
F = k Q1 Q2 / d²

Using the given values
F = 9 x 10^9 (1.0 x 10^-9)(1.0 x 10^-9) / 4²
F = 5.625 x 10^-10 N

The acceleration is
ma = F
1 (a) = 5.625 x 10^-10
a = 5.625 x 10^-10 m/s²
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3 years ago
An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of
GaryK [48]

Explanation:

The given data is as follows.

     Velocity of bullet, c_{p} = 814.8 m/s

    Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

               t = \frac{24.7}{343}      (velocity in air = 343 m/s)

                 = 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

               d_{b} = c_{b} \times t

                          = 814.8 \times 0.072

                          = 58.66

                          = 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0
3 years ago
How does mass and material type effect thermal energy transfer
Vedmedyk [2.9K]
The mass contributes with the time of thermal energy transfer with respect to the material type  but most importantly the material type will determine rate at which the material absorbs the transfer of heat or thermal energy by either three types, conduction, convection and radiation.
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Problem A spring was at its resting position where it is attached to a wall at its left side and a block at its right side as sh
puteri [66]

Answer:

F = 19.1 N

Explanation:

To find the force exerted by the string on the block you use the following formula:

F=kx  (1)  

k: spring constant = 95.5 N/m

x: displacement of the block from its equilibrium position = 0.200 m

you replace the values of k and x in the equation (1):

F=(95.5N/m)(0.200m)=19.1N

Hence, the force exterted on the block is 19.1 N

 

8 0
3 years ago
A 3.04 kg particle is located on the x-axis at xm = −8 m, and a 5.61 kg particle is on the x-axis at xM = 3.56 m. Find the coord
Murljashka [212]

Answer:

center of mass = −0.50 m

Explanation:

given data

mass m1 = 3.04 kg

distance xm = -8 m

mass m2 = 5.61 kg

distance xM = 3.56 m

solution

we get here center of mass for n mass of system that is express as

center of mass = \frac{m_1x_1+m_2x_2......m_nx_n}{m_1+m_2...m_n}     ......................1

but we have only 2 particle system so we will get

center of mass = \frac{m1 \times xm+m2 \times xM}{m1+m2}      .................2

put here value and we will get

center of mass = \frac{3.04 \times (-8 )+5.61 \times 3.56}{3.04 + 5.61}

solve it we will get

center of mass = −0.50 m

8 0
3 years ago
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