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prisoha [69]
3 years ago
8

~~~!Here's the question!~~~

Physics
2 answers:
vazorg [7]3 years ago
8 0
The correct answer is A) Gravity. Hope this helps.
gtnhenbr [62]3 years ago
6 0
The answer would most likely be A since obviously gravity weighs things down which helps the every other masses stay settled in place
You might be interested in
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

3 0
3 years ago
A new piece of exercise equipment has been added to your gym, and you try it out. To use this machine, you lie horizontal on a m
Shalnov [3]

Answer:

500 N

Explanation:

Since the work done on the spring W = Fx where F = force applied and x = compression length = 0.170 m (since the spring will be compressed its full length when the force is applied)

Since W = 85.0 J and we need to find F,

F = W/x

= 85.0 J/0.170 m

= 500 N

So, the  magnitude of force must you apply to hold the platform stationary at the final distance given above is 500 N.

6 0
3 years ago
Give reason : the action and reaction do not lead to equilibrium ?
Natalija [7]

the action and reaction do not lead equilibrium if action and reaction force react on different objects

4 0
3 years ago
An experiment is conducted such that an applied force is exerted on a 5kg object as it travels across a horizontal surface in wh
Katen [24]

If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.

The  graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.

  • A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m

For more information on work done, visit

brainly.com/subject/physics

5 0
2 years ago
Read 2 more answers
Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 m
Rufina [12.5K]
<h2>Answer:</h2>

143μH

<h2>Explanation:</h2>

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l                 --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

<em>From the question;</em>

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

<em>But;</em>

A = π d² / 4                     [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

<em>Substitute these values into equation (i) as follows;</em>

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209           [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

6 0
3 years ago
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