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andrey2020 [161]
2 years ago
13

A marathon runner runs at a steady 15 km/hr. When the runner is 7.5 km from the finish, a bird begins flying from the runner to

the finish at 30 km/hr. When the bird reaches the finish line, it turns around and flies back to the runner, and then turns around again, repeating the back-andforth trips until the runner reaches the finish line. How many kilometers does the bird travel?
Physics
1 answer:
Whitepunk [10]2 years ago
6 0

Answer:

The value is D =  15 \  km

Explanation:

From the question we are told that

   The  speed of the marathon runner is  v  =  15 \  km /hr

   The distance from the distance from the finish is  d =  7.5 \  km

   The  speed of the bird is  v_b  =  30 \ km / hr

  Generally the time taken for the runner to reach the finish is mathematically represented as

       t =  \frac{d}{v}

       t =  \frac{7.5}{15}

        t =  \frac{1}{2}

So the distance covered by the bird is  

      D =  v_b  *  t

      D =  30  *  \frac{1}{2}

         D =  15 \  km

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Explica la relación entre momento de torsión y aceleración angular mencionando tres ejemplos Una varilla uniforme delgada mide 1
sukhopar [10]

Answer:

1) τ = I α    whereby the torque is provided by the angular acceleration

2)  L = 216 Kg m² / s

Explanation:

1) Let's start with Newton's second law

       F = m a

multiply by the arm or perpendicular distance

       F r = m a r

if the distance is not perpendicular a way of realizing the relations using the vector product

      τ = F r = F x r

the bold are vectors.  The angular and linear acceleration are related

       a = α r=

         

      τ = m (α r) r

      τ = (m r²) α

the inertia of the rotational motion is

        I = m r²

we substitute

       τ = I α

whereby the torque is provided by the angular acceleration.

As an example we have:

* a spinning disk

* a ball rotating in the air

* a pulley

2) The rotational momentum is

           L = I w

the moment of inertia of a rod that through its center

           I = m L²

we substitute

           L = m L² w

let's calculate

          L = 6 1.5 2 16

          L = 216 Kg m² / s

7 0
1 year ago
A paper clip is made of wire 0.5 mm in diameter. If the original material from which the wire is made is a rod 25 mm in diameter
gizmo_the_mogwai [7]

Answer:

longitudinal engineering strain = 624.16

true strain is 6.44

Explanation:

given data

diameter d1 = 0.5 mm

diameter d2 = 25 mm

to find out

longitudinal engineering and true strains

solution

we know both the volume is same

so

volume 1 = volume 2

A×L(1) = A×L(2)

( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)

( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)

0.1963 ×L(1) = 122.71 ×L(2)

L(1) / L(2) = 122.71 / 0.1963 = 625.16

and we know longitudinal engineering strain is

longitudinal engineering strain = L(1) / L(2)  - 1

longitudinal engineering strain = 625.16  - 1

longitudinal engineering strain = 624.16

and

true strain is

true strain = ln ( L(1) / L(2))

true strain = ln ( 625.16)

true strain is 6.44

3 0
2 years ago
A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

so that

Q = ba⁴ * ε₀

5 0
2 years ago
Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,
Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number                                   Symbol                Possible values

Principal Quantum Number  .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l  

Spin Quantum Number.................m_s.....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0
1 year ago
What is the instantaneous velocity of a freely falling object 9.0 s after it is released from a position of rest? Express your a
Umnica [9.8K]

Answer:

So instantaneous velocity after 9 sec will be 88.2 m/sec              

Explanation:

We have given time t = 9 sec

As the object is released from rest so its initial velocity u = 0 m/sec

We have to find its final velocity v

Acceleration due to gravity g=9.8m/sec^2

From first equation of motion we know that v=u+gt

v=0+9.8\times 9=88.2m/sec

So instantaneous velocity after 9 sec will be 88.2 m/sec  

5 0
2 years ago
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