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Ber [7]
3 years ago
13

How can you define a solution to an equation?

Physics
1 answer:
sleet_krkn [62]3 years ago
8 0
A solution is a value or a collection of values.. when substituted for the unknowns, the equation become an equality.
Example : x + 2 = 7
When we out the 5 in place of x we get: 5 + 2 = 2
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Sound
Ann [662]
Longitudinal, because the sound can only travel at one direction
3 0
3 years ago
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PLS HELP I NEED IT
irinina [24]

Answer:

<h2>A. a spring & B. a well dried into an aquifer.</h2>

8 0
3 years ago
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A mover pushes a 46.0kg crate 10.3m across a rough floor without acceleration. How much work did the mover do (horizontally) pus
Alchen [17]

Answer:

<h3>2,321.62Joules</h3>

Explanation:

The formula for calculating workdone is expressed as;

Workdone = Force * Distance

Get the force

F = nR

n is the coefficient of friction = 0.5

R is the reaction = mg

R = 46 ( 9.8)

R = 450.8N

F = 0.5 * 450.8

F = 225.4N

Distance = 10.3m

Get the workdone

Workdone = 225.4 * 10.3

Workdone  = 2,321.62Joules

<em>Hence the amount of work done is 2,321.62Joules</em>

3 0
3 years ago
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
3 years ago
Newton’s second law of motion can also be described as _____.
anzhelika [568]

Answer:

F=ma

Explanation:

The Newton's laws of motion tries to explain the how bodies behave and the energy changes when theys are in motion. For the 3 of them to hold, the bodies must be moving in a straight line and with constant velocity.

The second one states that, "the change of momentum of a moving body is directly proportional to the force producing it and it takes place to the direction of force."

From the choices given, the appropriate answer is F=ma

3 0
3 years ago
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