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3 years ago

Why a soccer ball that is quickly rolling across a grass field slows and finally stops rolling

Physics

2 answers:

Arturiano [62]3 years ago

5 0

Friction accord while the ball was rolling in the grass and slowed it down until it finally came to a stop

jenyasd209 [6]3 years ago

4 0

This is because of friction. If you want to add extra detail you can say that this ties in with Newton's First Law of Motion, which states that an object moving at a constant speed will stay moving at a constant speed unless there is an opposite force, and in this case, the opposite force against the ball is friction.

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In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho

Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

$R=u_xt=(ucos \theta)t$

Therefore,

$t=\frac{R}{ucos\theta} .......(1)$

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

$y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2$

Substitute the value of t from equation (1).

$y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )$

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

$y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s$

The shot put was thrown with a speed 15.02 m/s.

7 0

2 years ago

Describe metallurgy. Check all that apply.

mylen [45]

Answer:

A, C, D

Explanation:

got it right

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2 years ago

Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway

SashulF [63]

Answer:

v = 12.12 m/s

Explanation:

Given that,

The mass of the cart, m = 75 kg

The roller coaster begins 15 m above the ground.

We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s$

So, the velocity of the cart is 12.12 m/s.

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2 years ago

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1500 kg, which is required to travel upward 57 m in

Kazeer [188]

Answer:

Explanation:

it equals16 2211

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2 years ago

What is a plane mirror? state the characteristics of the image formed by a plane mirror

Nadusha1986 [10]

A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!

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