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Novosadov [1.4K]
3 years ago
12

Pls can anyone help me

Physics
1 answer:
Pie3 years ago
7 0

Answer:

the u.s

Explanation:

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From 0 seconds to 4 seconds is the acceleration positive or negative?
Alenkinab [10]

Answer:

From 0 -4 seconds the acceleration is positive. (The graph is going upwards.)

From 6-10 seconds the acceleration is negative. (The graph is going downwards.)

7 0
4 years ago
Which of these is something whose construction MOST LIKELY would be determined by the government? A) An airport B) A golf club C
vesna_86 [32]

The answer is; A

Airports are usually controlled by governments because they are the points of entry and exit into a country and therefore should have enough security and customs. The vulnerability of the industry also from factors such as terrorism and weather requires that the government regulates the industry to protect the country's interests.

5 0
3 years ago
If a net horizontal force of 7.5 N is applied to a baby carriage whose mass is 12.5 kg what acceleration is
11111nata11111 [884]

Answer:

<h2>0.6 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m} \\

f is the force

m is the mass

From the question we have

a =  \frac{7.5}{12.5}  \\

We have the final answer as

<h3>0.6 m/s²</h3>

Hope this helps you

8 0
3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
An early model of the atom, proposed by Rutherford after his discovery of the atomicnucleus, had a positive point charge +Ze(the
Amanda [17]

Answer:

a)  E = k Ze (1- r³ / R³)  1/r², b) E=0, c)   E = -6.62 10¹⁰  N / C

Explanation:

a) For this we can use the law of Gauus

         Ф = E- dA = q_{int} / ε₀

where we take a sphere as a Gaussian surface, so that the electric field lines and the radii of the sphere are parallel, consequently the dot product is reduced to the algebraic product

       E A =q_{int} / ε₀

the area of ​​a sphere  

      A = 4π r²

      E 4π r² = q_{int} / ε₀

      E = 1 / 4πε₀   q_{int} / r²

       k = 1 /4π ε₀

       E = k q_{int} / r²       (1)

       

let's analyze the charge inside the gaussian sphere,

let's use the concept of density for electrons, since they indicate that the charge is evenly distributed

     ρ = Q / V

where the volume of the sphere is

    V = 4/3 πr³

     Qe = ρ V

     Qe = ρ 4 / 3π r³

the density of the electrons is

     ρ = Ze 3 / (4π R³)

where R is the atomic radius

we substitute

       Qe = Ze r³/ R³

for protons they are in a very small space, the atomic nucleus, so we can superno that they are a point charge.

The net charge inside our Gaussian surface, the charge of the protoens plus the charge of the electroens (Qe)

     q_{int} = q_proton + Q_electron

     q_{int} = + Ze - Qe

     q_{int} = + Ze - Ze r³ / R³

     q_{int} = Ze (1- r³ / R³)

   

  we substitute in equation 1

     E = k Ze (1- r³ / R³)  1/r²

b) on the surface of the atom r = R

therefore the electric field is zero

      E = 0

c) Calculate the electric field for the Uranium for

       r = R / 2 = 0.10 10⁻⁹ / 2 = 0.05 10⁻⁹ = 5 10⁻¹¹ m

     

       E = 8.99 10⁹ 92 1.6 10⁻¹⁹ (1-1/2)³   1/ (5 10⁻¹¹)²

       E = -6.62 10¹⁰  N / C

6 0
4 years ago
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