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Damm [24]
3 years ago
8

If a long distance runner with a weight of 596.82 newtons does 35,674.7 joules of work during a portion of a race, what distance

will she cover during that portion?
Physics
1 answer:
Tpy6a [65]3 years ago
4 0

Answer:

\boxed {\boxed {\sf About \ 59.77 \ meters}}

Explanation:

Work is equal to the product of force and distance.

W=F*d

We are solving for the distance, d. We know the force is 596.82 Newtons and the work is 35,674.7 Joules.

  • If we convert the units for Joules, the problem will be simpler later.
  • 1 Joule is equal to 1 Newton meter.
  • The work done by the runner, 35,674.7 J, equals 35,674.7 N*m

F= 596.82 \ N\\W= 35,674.7  \ N*m

Substitute the values into the formula.

35,674.7 \ N*m= 596.82 \ N * d

We want to solve for the distance, so we must isolate d on one side of the equation.

d is being multiplied by 596.82 Newtons. The inverse of multiplication is division. Divide both sides of the equation by 596.82 N

\frac{35,674.7 \ N*m}{596.82 \ N} =\frac{596.82 \ N*d}{596.82 \ N}

\frac{35,674.7 \ N*m}{596.82 \ N}=d

The Newtons will cancel, hence the prior unit conversion.

\frac{35,674.7 \ m}{596.82} =d

59.7746389 \ =d

Let's round to the nearest hundredth, so the answer is concise. The 4 in the thousandth place tells us to leave the 7 in the hundredth place.

59.77 \ m  \approx  d

The distance covered during the portion of the race is about <u>59.77 meters.</u>

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Answer:

11.3 m/s

Explanation:

KE₁ = KE₂

½m₁v₁² = ½m₂v₂²

½ (2 kg) v² = ½ (4 kg) (8 m/s)²

v ≈ 11.3 m/s

8 0
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The Rigidbody component adds collision to a GameObject
kifflom [539]

Rigidbodies are components that allow a GameObject<u> to react to real-time physics. </u>

Explanation:

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7 0
3 years ago
If the Sun were scaled down to the size of a grapefruit, about how far would you have to walk from our classroom to reach Alpha
Burka [1]

Answer:

2000 miles.

Explanation:

It's the Colorado scale model, in which sun is taken as the grapefruit and the distances are measured for different planets with respect to sun just for understanding.

6 0
3 years ago
An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a
Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = 3.1 m

Width of the window = 2.1 m

Area of the window = (3.1\times 2.1) = 6.51\ m^2

Difference in air pressure = Inside pressure - Outside pressure

                                           = (1.0-0.954) atm = 0.046 atm

Conversion of the pressure in its SI unit.

⇒  1 atm = 101325 Pa

⇒ 0.046 atm = 0.046\times 101325 =4660.95 Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ Pressure=\frac{Force }{Area}

⇒ Force =Pressure\times Area

⇒ Plugging the values.

⇒ Force =4660.95\times 6.51

⇒ Force=30342.78 Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
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