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3 years ago

6

Estimate how long it would take one person to mow a football field using an ordinary home lawn mower. Assume the mower moves wit

h a 1 km/h speed, and has a 0.6 m width

1 answer:

bekas [8.4K]3 years ago

8 0

Answer:

t = 11.9 h = 0.49 day

Explanation:

First of all, we will find the area covered by the mower per unit time:

$Rate\ of\ Area\ Covered\ by\ Mower = vw$

where,

v = speed of mower = 1 km/h = 1000 m/h

w = width of mower = 0.6 m

Therefore,

$Rate\ of\ Area\ Covered\ by\ Mower = (1000\ m/h)(0.6\ m)\\Rate\ of\ Area\ Covered\ by\ Mower = 600\ m^{2}/h$

Now, the standard football ground has an area of:

A = 105 m x 68 m = 7140 m²

Therefore, The time required to mow the lawn will be:

$Time\ to\ Mow\ the\ Lawn = t = \frac{A}{Rate\ of\ Area\ Covered\ by\ Mower}\\t = \frac{7140\ m^{2}}{600\ m^{2}/h}$

<u>t = 11.9 h = 0.49 day</u>

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It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used

stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

$a_{c}$ = rω²

ω² = $a_{c}$ / r

ω = √( $a_{c}$ / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a $a_{c}$ = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²

ω = 0.0500647 ≈ 0.05 rad/s

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM

1 rpm = 60 rph

so

ω = 0.478082 × 60

ω = 28.6849 revolutions per hour

Therefore, the required revolution per hour is 28.6849

7 0

3 years ago

6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory

Svetllana [295]

Answer:

Engular velocity: $w=1,24*10^{-3}/s$

Linear velocity: $V=7905 m/s$

The time it takes:

$t=5060s=84min$

Explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1) $a=r*w^{2}$

Solving for w:

(2) $w=\sqrt{\frac{a}{r} }$

Replacing a=9,8m/s2 and r=6,375,000m:

(3) $w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s$

And the angular velocity relates to the linear velocity:

$V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s$

The perimeter of the orbit is:

$P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m$

The time it takes:

$t=\frac{P}{V} =\frac{40.05*10^{6}m}{7905 m/s}=5060s=84min$

5 0

3 years ago

A spring vibrates 120 times in 2 mins find its frequency and time period

german

2 minutes is 120 seconds, so if you were finding vibrations per minute, it would be 60 times a minute.

3 0

2 years ago

4. Uncle Harry weighs 180 pounds. What is his mass in kilograms?

MatroZZZ [7]

180 pounds (lb) converts to 81.647 kilograms (kg).

5 0

2 years ago

A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t

vampirchik [111]

The new period is D) √2 T

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum 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>

$\texttt{ }$

$\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}$

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

$T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}$

$T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}$

$T : T_2 = \sqrt{L} : \sqrt{2L}$

$T : T_2 = 1 : \sqrt{2}$

$\boxed {T_2 = \sqrt{2}\ T}$

$\texttt{ }$

<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

3 0

3 years ago

Read 2 more answers