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Vesna [10]
3 years ago
7

Need help asap please and thank you​

Physics
1 answer:
nordsb [41]3 years ago
6 0
Distance = speed x time

distance = 116 x 10

distance = 1160 m
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What scientist was the first to propose the heliocentric model of the universe
Schach [20]
This theory was first proposed by Nicolaus Copernicus. Copernicus was a Polish astronomer. He first published the heliocentric system in his book: De revolutionibus <span>orbium </span>coelestium<span> , "On the revolutions of the heavenly bodies," which appeared in 1543.</span>
3 0
3 years ago
There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
3 years ago
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowin
leva [86]
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
 So as air speed indicator shows the negative speed in y component and adding it in
  air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
 v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
 solving
  v= -12.7 m/s x-40.7 m/s y
 if magnitude of velocity or speed is required then
  speed= sqrt(12.7^2 + 40.7^2)
 speed= 42.63 m/s
 if angle is asked
  angle = arctan (40.7/12.7)
 angle = 72.67 degrees south of west</span>
6 0
3 years ago
a man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s
alex41 [277]

Answer:

h = 90.10 m

Explanation:

Given that,

A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m

The initial speed of the stone, u = 10 m/s

The path followed by the projectile is given by :

h(t)=-4.9t^2+10t+85 ....(1)

For maximum height,

Put dh/dt = 0

So,

\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s

Put the value of t in equation (1).

h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m

So, the maximum height of the stone is equal to 90.10 m.

5 0
3 years ago
What happens to a sample of matter when it’s particles lose kinetic energy?
Mrrafil [7]
What will happen if the sample is the
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3 years ago
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