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solong [7]
3 years ago
5

Two point charges exert a 6.10 N force on each other. What will the force become if the distance between them is increased by a

factor of 8
Physics
1 answer:
anzhelika [568]3 years ago
5 0

Answer:

0.0953125 N

Explanation:

Applying,

F = kq'q/r²................. Equation 1

Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.

From the equation,

If both charges remain constant,

Therefore,

F = C/r²

C = Constant =  product of the two charge(q' and q) and k

Fr² = F'r'²................ Equation 2

From the question,

Given: F = 6.10 N

Assume: r = x m, r' = 8x

Substitute these value into equation 2

6.1(x²) = F'(8x)²

F' = 6.1/64

F' = 0.0953125 N

Hence the new force will become 0.0953125 N

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The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr
True [87]

Answer:

<em>mass of the ice is 254980463.8T kg</em>

<em>where T is the value of the thickness omitted in the question.</em>

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice.  Answer in kg.

<em>The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.</em>

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = \pi r^{2} = 3.142*297^{2} = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v =  277152.678T  ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

7 0
3 years ago
What term refers to the part of a spacecraft that is occupied by the crew for takeoff and landing?
Semenov [28]
Command module ✅

service module

lunar module

annum module
7 0
2 years ago
Can we call centre of gravity centre of weight? ​
algol [13]

Answer:

Centre of gravity is a theoretical point in the body where the total weight of the body is thought to be concentrated. In a uniform gravitational field, the centre of gravity is identical to the centre of mass. Yet, the two points do not always coincide.

3 0
3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
A block is at rest on a incline plane as shown in the diagram.As angle is increased, the componet of the blocks weight parallel
Elodia [21]

Answer:

The component of block weight parallel to the plane, Wₓ = W cosθ

Explanation:

Let the weight of the block due to gravitation is W

The direction of the weight is vertically down

Let θ be the angle formed with the vertical weight of the block and the incline.

Taking two components of weight one along the vertical weight and another component perpendicular to it.

Then the component `of weight long the parallel of the plane is

                                   Wₓ = W cosθ        

6 0
3 years ago
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