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3 years ago

2. A science book is being pushed across a table. Can the book have a constant speed and a

1 answer:

4 0

Yes, it can have both cuz of the book is pushed straight it has a constant speed moving forward and if the book curves then it is changing velocity

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Why does sound travel slower at a cold temperature?

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<h2>Answer:</h2><h2>It is due to a refractment of light.</h2>

Sound moves faster in warmer air than colder air the way bends away from the warm air and back towards of air.

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3 years ago

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A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the

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25 kg and 299N are the answers

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4 years ago

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You toss a coin into a wishing well full of a liquid denser than the coin. Which of the following could be true?

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An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

bulgar [2K]

To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.

The perimeter of a circle is defined by

$P = 2\pi r$

Where,

r= radius

While centripetal acceleration is defined by

$a=\frac{v^2}{r}$

Where,

v= velocity

r= radius

PART A)

The distance of a body can be defined based on the speed and the time traveled, that is

x = v*t

For our values the distance is equal to

x = 15*115=1725m

The plane when going to make the turn from east to south makes a quarter of the circumference that is

$\frac{P}{4} = \frac{2\pi r}{4}$

The same route you take is the distance traveled, that is

$x = \frac{P}{4}$

$x = \frac{2\pi r}{4}$

$1725 = \frac{2\pi r}{4}$

$r = 1098.17m$

PART B)

With the radius is possible calculate he centripetal acceleration,

$a = \frac{v^2}{r}$

$a = \frac{115^2}{1098.17}$

$a = 12.04m/s^2$

Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of $12.04m/s^2$

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4 years ago

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

<em>radius of the curve, r = 90 m</em>
<em>angle of inclination, θ = 10.8⁰</em>
<em>speed of the car, v = 75 km/h = 20.83 m/s</em>
<em>mass of the car, m = 1100 kg</em>

The normal force on the car is calculated as follows;

$F_n = mgcos(\theta)$

The frictional force between the car and the road is calculated as;

$F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)$

The net force on the car is calculated as follows;

$mgsin(\theta) + \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \ + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) + \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\$

$\mu_s = 0.69$

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

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3 years ago