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serious [3.7K]
2 years ago
15

Trained dolphins are capable of a vertical leap of 7.0m straight up from the surface of the water-an impressive feat.Suppose you

could train a dolphin to launch itself out of the water at this same speed but at an angle.What maximum horizontal range could the dolphin achieve?
Physics
1 answer:
anzhelika [568]2 years ago
5 0

Answer:

   θ = 45º

Explanation:

To find the solution, let's use the projectile launch equation

    R = vo² sin 2θ / g

Where vo is the initial speed of the dolphin, T is the angle of the jump and g the gravity acceleration.

To obtain a maximum range, the sine T function must be 1,

     sin 2θ = 1

     2θ = sin⁻¹ 1

     2θ = 90

     θ = 45º

Therefore, the dolphin should jump at an angle of 45º from the horizontal

You might be interested in
01 – (Valor – 2,0) O maior campo de testes de veículos da América Latina, localizado na cidade de Indaiatuba (SP), tem forma cir
Scilla [17]

Answer:

a) Calcule a frequência em RPM

= 0.6 RPM

b) a velocidade escalar do carro em m/s.

= 20m/s

Explanation:

a) Calcule a frequência em RPM

A fórmula para calcular a frequência é: 1/T

onde T= Tempo (seconds)

T = 100s

A frequência = 1/100s

A frequência = 0.01Hz

em RPM

A fórmula para calcular a frequência em RPM =

1 Hz = 60RPM

0.01Hz =

A frequência em RPM = 0.01Hz × 60

= 0.6 RPM

b) a velocidade escalar do carro em m/s.

A fórmula para calcular a velocidade escalar = diâmetro ou distância (m) ÷ tempo (s)

Diâmetro ou Distância = 2.0km

Converter 2.0km para m

1 km = 1000m

2km =

2 km × 1000m

= 2000m

A velocidade escalar = 2000m ÷ 100s

A velocidade escalar = 20m/s

Answer:

a) Frequency in RPM

= 0.6 RPM

b) Scalar Velocity

= 20m/s

Explanation:

a) Frequently in RPM

Formula : 1/T

Where T= Time (seconds)

T = 100s

= 1/100s

= 0.01Hz

Frequency in RPM =

1 Hz = 60RPM

0.01Hz = 0.01Hz × 60

= 0.6 RPM

b) Scalar velocity

The formula = Diameter or Distance ÷ Time

Diameter or Distance = 2.0km

Convert 2.0km to m

1 km = 1000m

2km =

2 km × 1000m

= 2000m

Scalar Velocity = 2000m ÷ 100s

Scalar Velocity = 20m/s

8 0
2 years ago
Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known
marissa [1.9K]

Answer:

Mendeleev predicted the atomic mass of each element along with compounds they each should form.

Explanation:

Based on other elements in the same group he predicted the existence of eka-aluminum and eka-silicon, later to be named gallium (Ga) and germanium (Ge).

6 0
2 years ago
A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed
Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

F = \dfrac{mv^2}{r}

m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

Hence, the tension in the string is quadrupled.

8 0
2 years ago
What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state?
Sindrei [870]

Answer:

9.12\cdot 10^{-8} m

Explanation:

The energy needed to ionize a hydrogen atom in the ground state is:

E=13.6 eV= 2.18\cdot 10^{-18}J

The energy of the photon is related to the wavelength by

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength

Solving the formula for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2.18\cdot 10^{-18}J}=9.12\cdot 10^{-8} m

7 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
2 years ago
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