Putting the moon back to its original size, if it were suddenly to move to twice the

If you neglect air resistance, the horizontal component of a projectile A. Is always changing B. Remains constant throughout the

Shtirlitz [24]

Answer:

B. Remains constant throughout the flight

Explanation:

If we completely neglect air resistance then the projectile will not have any horizontal resistances to deal with, therefore the horizontal component will remain constant throughout the flight. This would continue to be the case until it meets some form of resistance. Which assuming that everything else is normal would be the case since the force of gravity will push the projectile down (vertical component) until it hits the ground which in that instance would act as an opposing force to the horizontal component as well.

8 0

3 years ago

Does the sun always shine directly overhead at the equator at noon

Nikitich [7]

The sun always shines directly overhead at noon. This is because the equator always gets the equivalent amount of sunlight. The area always get 12 hours of sunlight, because it's 0 degrees north and south and it's at the center of the Earth.

4 0

3 years ago

1) Consider a source particle of charge qS=9 C located at (−9,5) [distances in meters]. Find the electric field vector at the ta

xeze [42]

Answer:

1) E = 2.25 i^+ 0.809j^) 10⁹ N / C , 2) E = 2.39 10⁹ N / C , 3) θ = 19.8º , 4) F = 19.12 10⁹ N , 5) E = (1.32 i^+ 3.56 j^) 109 N/C

Explanation:

1) The equation for the electric field is

E = k q / r²

Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point

Since the electric field is a vector magnitude, we can find its component

X axis

Ex = k q / x²

where the distance on the axis is

x = √ (X₂-x₁)²

x = √ (-15 + 9)² = 6 m

Eₓ = 8.99 10⁹ 9/6²

Eₓ = 2.25 10⁹ N /C

Y Axis

y = √ (y₂-y₁)² = √ (15-5)² = 10 m

$E_{y}$ = 8.99 10⁹ 9/10²

$E_{y}$ = 0.809 10⁹ N / C

E = Eₓ i^+ $E_{y}$ j^

E = 2.25 i^+ 0.809j^) 10⁹ N / C

2) the magnitude can be found using the Pythagorean triangle

E = √ (Eₓ² + $E_{y}$ ²)

E = √ (2.25² + 0.809²) 10⁹

E = 2.39 10⁹ N / C

3) to find the angle let's use trigonometry

tan θ = $E_{y}$ / Eₓ

θ = tan⁻¹ $E_{y}$ / Eₓ

θ = tan⁻¹ (0.809 / 2.25)

θ = 19.8º

Regarding the positive side of the x axis

4) a charge q2 = 8C is placed, let's calculate the force

F = q E

F = 8 2.39 10⁹

F = 19.12 10⁹ N

5) The total electric field at the origin, let's look for its components

q₁ = 9C

r₁ = -9 i ^ + 5 j ^

q₂ = 8 C

r₂ = -15 i ^ + 15 j ^

X axis

Eₓ = E₁ₓ + E₂ₓ

Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²

Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)

Eₓ = 1.32 109 N / A

Y Axis

$E_{y}$ = $E_{1y}$ + $E_{2y}$

$E_{y}$ = k q₁ / Δy₁² + k q₂ / Δy₂²

$E_{y}$ = 8.99 109 (9/5² + 8/15²)

$E_{y}$ = 3.56 109 N / A

E = (1.32 i^+ 3.56 j^) 109 N/C

7 0

4 years ago

Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc

Lapatulllka [165]

Answer:

a) d = 30.79 m , b) θ = -22.4° , θ = 22.4 South of East

Explanation:

The easiest way to solve problems with vectors is to use their components, for this the East-West direction coincides with the x-axis and the North-South direction coincides with the y-axis

Let's use the index for / Ricardo and the index for Jane, let's break down the displacements

Richard

X axis

x₁ = 26.0 sin (60)

x₁ = -22.52 m

Y Axis

y₁ = 26.0 cos 60

y₁ = 13 m / s

Jane

X axis

x₂ = 16.0 cos (180 +30)

x₂ = -13.85 m

Y Axis

y₂ = 16.0 sin (180 + 30)

y₂ = - 8.0 m

Now we can use Pythagoras' theorem to find the distance between them

d = √ [(x₂ -x₁)² + (y₂ -y₁)²]

d = √ [(-13.85 + 22.52)² + (-8 -13)²]

d = 30.79 m

Let's use trigonometry to enter the address

tan θ = Δy / Δx

θ = tan⁻¹ Δy / Δx

θ = tan⁻¹ (-13.85 + 22.52) / (-8 - 13)

θ = tan⁻¹ (-8.67 / 21)

θ = -22.4°

The negative sign indicates that the angle is measured from the axis clockwise.

In the form of cardinal s point is

θ = 22.4 South of East

4 0

3 years ago

With a force of 5 Newton's Amanda pushes the stacks of books to the right. At the same time Jeremih her little brother pushes th

Fynjy0 [20]

Alright, so solving for the net force is a rather simple and easy step. I want to briefly explain how you determine the net force, but before I do, let's examine the problem together. What has been provided to us? What direction is it going? Well, we can look at the information!

Amanda's Force: 5 N

Jeremiah's Force: 10 N

They're both pushing on the books, but in different directions. Left and right. If Amanda is pushing the books to the right 5 newtons, and Jeremiah is pushing the books to the left with 10 newtons, that means the net force is 5. The book is being pushed to the left! We can say that because Jeremiah's force is much larger than his sisters.

We can determine the net force from the following;

Net Force → 10 N - 5 N = 5 N

In conclusion, your answer should be a total of five (5) newtons

4 0

3 years ago

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