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dexar [7]
3 years ago
14

An astronaut on the moon throws a baseball upward. the astronaut is 6ft, 6 in. tall and the initial velocity of the ball is 30 f

t per second. the height s of the ball in feet is given by the equation 2.7 30 6.5, 2 s ï½ ï­ t ï« t ï« where t is the number of seconds after the ball is thrown. (note: see where the initial height and initial velocity values play a role in the equation).

Physics
2 answers:
Ainat [17]3 years ago
5 0
The equation is garbled and the question is missing.

I found this equation for the same statement:

S = - 2.7t ^2 + 30t + 6.5

And one question is: after how many seconds is the ball 12 feet above the moon's surface?

Given that S is the height of the ball, you just have to replace S with 12 and solve for t.

=> 12 = - 2.7 t^2 + 30t + 6.5

=> 2.7t^2 - 30t - 6.5 + 12 = 0

=> 2.7t^2 - 30t + 5.5 = 0

Now you can use the quadratic equation fo find t:

t = { 30 +/- √ [30^2) - 4(2.7)(5.5)] } / (2*2.7)

=> t = 0.186s and t = 10.925 s

Answer: after 0.186 s the ball is at 12 feet over the surface, and again 10.925 s

katovenus [111]3 years ago
3 0

The maximum height of the baseball is about 90 feet

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

The height s of the ball in feet is given by the equation:

s = -2.7t^2 + 30t + 6.5

To find the velocity function, the above equation will be derived as follows:

v = \frac{ds}{dt}

v = -2(2.7)t^{2-1} + 30

\boxed {v = -5.4t + 30}

At the maximum height, the speed is 0 m/s , then:

v = -5.4t + 30

0 = -5.4t + 30

5.4t = 30

t = 30 \div 5.4

\boxed {t = \frac{50}{9} ~ s}

s = -2.7t^2 + 30t + 6.5

s_{max} = -2.7(\frac{50}{9})^2 + 30(\frac{50}{9}) + 6.5

s_{max} = \frac{539}{6}

\large {\boxed {s_{max} \approx 90 ~ feet} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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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.
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<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.
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<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.

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V = I/(n × A × Q)

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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.

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