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3 years ago

A divot is created _____.

Physics

2 answers:

STatiana [176]3 years ago

4 0

I think it is when shot scrapes off the top of the turf

insens350 [35]3 years ago

3 0

Answer: when shot scrapes off the top of the turf

Explanation: Divot is a term used in golf, where "Divot" is the piece of turf that is "cuted" out of the ground when the player tries to hit the ball.

Usually, the Iron or the Wedge are the ones that cause the divots, and while it may seem like a poor technique, it is actually pretty a common thing to see. At the point that in some cases, the divot itself is analyzed to see the technique of the player.

The correct option is the third one: "when shot scrapes off the top of the turf"

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Sunlight has its maximum intensity at a wavelength of 4.83 x 10'm; wha energy does this correspond to in e?

Lera25 [3.4K]

Answer:

E = 2,575 eV

Explanation:

For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength

E = h f

c = λ f

Where the Planck constant has a value of 6.63 10⁻³⁴ J s

Let's replace

E = h c / λ

Let's calculate for wavelengths

λ = 4.83 10-7 m (blue)

E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷

E = 4.12 10-19 J

The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J

E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E = 2,575 eV

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3 years ago

Which characteristic must a food have to receive a “natural” label from the FDA ?

adelina 88 [10]

No artificial ingredients
i only know that because my mom was vegan for two years

4 0

3 years ago

A ball rolls down a ramp. A small amount of friction heats up the ball. What energy transformations are taking place?

Natasha2012 [34]

Answer:kinetic energy converted to heat energy

Explanation:

As the ball rolls down kinetic energy is converted to heat energy

6 0

3 years ago

Read 2 more answers

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

A 0.45-kilogram football traveling at a speed of 22 meters per second is caught by an 84-kilogram stationary receiver. if the fo

Troyanec [42]

Impulse = Change in momentum.
The ball was moving with a momentum of 0.45 * 22 = 9.9
The ball comes to rest in the receivers arm; this means the ball's final velocity = 0. So mv2 = 0.45 * 0
The magnitude of the impact is just the change in momentum. 9.9 - (0.45 * 0) = 9.9

3 0

3 years ago