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34kurt
2 years ago
12

The components of a 15 meters per second velocity at an angle of 60 degrees above the horizontal are?

Physics
1 answer:
mafiozo [28]2 years ago
8 0

Answer:

x-component of velocity: 7.5 m/s

y-component of velocity: 13 m/s

Explanation:

This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.

By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

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You multiply the mass by the acceleration 82*7.5=615; that's what I would do
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What are the two factors that determine a material's density?
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Mass and volume are the 2 factors to determine density
3 0
4 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
An iron wire has a cross-sectional area equal to 5.00×10⁻⁶ m² . Carry out the following steps to determine the drift speed of th
Doss [256]
  1. In mass, there are 55.85 × 10⁻³ kg/mol in in 1 mole of iron.
  2. The molar density of iron is equal to 1.41 × 10⁵ mol/m³.
  3. The density of iron atoms is equal to 8.49 × 10²⁸ atoms/m³.
  4. The number density of conduction electrons is equal to 1.70 × 10²⁹ conduction electrons/m³.
  5. The drift speed of conduction electrons is equal to 2.21 × 10⁻⁴ m/s.

<h3>How to calculate the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = (m × σ × V)/ρ × e × f × l)

V = I/(n × A × Q)

Where:

  • U represents the drift speed of the conduction electrons, in m/s.
  • m represents the molecular mass of the metal, in kg.
  • e represents the elementary charge, in C.
  • f represents the number of free electrons per atom.
  • σ represents the electric conductivity of the medium at a particular temperature in S/m.
  • ρ represents the density of the conductor, in kg/m³.
  • ℓ represents the length of the conductor, in m.
  • ΔV represents the voltage applied or potential difference across the conductor in V.

<h3>How many kilograms are there in 1 mole of iron? </h3>

Molar mass of iron = 55.85 g/mol.

In Kilograms, we have:

Mass = 55.85 × 1/1000

Mass = 55.85 × 10⁻³ kg/mol.

For the molar density of iron, we have:

Molar density = density/molar mass

Molar density = 7874/0.056

Molar density = 1.41 × 10⁵ mol/m³.

For the density of iron atoms, we have:

Density of iron atoms = Avogadro's constant × molar density

Density of iron atoms = 6.023 × 10²³ × 1.406 × 10⁵

Density of iron atoms = 8.49 × 10²⁸ atoms/m³.

For the number density of conduction electrons, we have:

Fe ---> Fe²⁺ + 2e⁻

Number density of conduction electrons = 2 conduction electrons/1 atom of iron

Number density of conduction electrons = 2 × 8.49 × 10²⁸

Number density of conduction electrons = 1.70 × 10²⁹ conduction electrons/m³.

For the drift speed of conduction electrons, we have:

V = I/(n × A × Q)

V = 30/(1.70 × 10²⁹ × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁶)

Drift speed, V = 2.21 × 10⁻⁴ m/s.

Read more on drift speed here: brainly.com/question/15219891

#SPJ4

Complete Question:

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire.

(a) How many kilograms are there in 1 mole of iron?

(b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter).

(c) Calculate the number density of iron atoms using Avogadro’s number.

(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom.

(e) If the wire carries a current of 30.0 A, calculate the drift speed of conduction electrons.

4 0
2 years ago
One of the fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974, was clocked at 100.8 mi/hr. If a pitch
lawyer [7]

<u>Answer:</u>

 The ball fall vertically 2.69 ft by the time it reached home plate 60.0 ft away.

<u>Explanation:</u>

    Fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974 = 100.8 mi/hr = 44.8 m/s

    The horizontal distance to home plate = 60.0 ft = 18.288 m

  We have the horizontal velocity = 44.8 m/s

    So time taken = 18.288/44.8 = 0.408 seconds.

  The distance traveled by baseball vertically is found out by equation s=ut+\frac{1}{2} at^2

  Here u =0m/s, a = 9.81 m/s^2 and t = 0.408 s

 Substituting

     S = 0*0.408+\frac{1}{2} *9.81*0.408^2 = 0.82 m\\ \\S = 0.82 m= 2.69 ft

So vertical distance traveled = 0.82 m = 2.69 ft

7 0
3 years ago
Read 2 more answers
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