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3 years ago

The track of a roller coaster is shown below. At which point will the riders experience centripetal acceleration?

Physics

2 answers:

bulgar [2K]3 years ago

8 0

<u>The answer to this question is Y</u>

Hope this helps :)

valentinak56 [21]3 years ago

4 0

Centripetal acceleration is applied in circular motion. Y is the point on the track that most closely resembles a circular path for the roller coaster.

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Light with energy equal to three times the work function of a given metal causes the metal to eject photoelectrons. What is the

Mrac [35]

The ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

<h3 /><h3>What is the photoelectric effect?</h3>

When a medium receives electromagnetic radiation, electrostatically charged particles are emitted from or inside it.

The emission of ions from a steel plate when light falls on it is a common definition of the effect. The substance could be a solid, liquid, or gas; and the released particles could be protons or electrons.

A particular metal emits photoelectrons when exposed to light with energy three times its work function:

$\rm KE=3 \phi$

The ratio of the maximum photoelectron kinetic energy to the work function will be;

$R=\frac{E}{\phi} \\\\ R=\frac{3 \phi}{\phi} \\\\ R= 3$

Hence, the ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

To learn more about the photoelectric effect refer to the link;

brainly.com/question/9260704

#SPJ1

5 0

1 year ago

A child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a

kirill [66]

Answer:

θ = 13.16 °

Explanation:

Lets take mass of child = m

Initial velocity ,u= 1.1 m/s

Final velocity ,v=3.7 m/s

d= 22.5 m

The force due to gravity along the incline plane = m g sinθ

The friction force = (m g)/5

Now from work power energy

We know that

work done by all forces = change in kinetic energy

( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²

(2 g sinθ - ( 2 g)/5 ) d = v² - u²

take g = 10 m/s²

(20 sinθ - ( 20)/5 ) 22.5 = 3.7² - 1.1²

20 sinθ - 4 =12.48/22.5

θ = 13.16 °

5 0

3 years ago

In 1977 off the coast of Australia, the fastest speed by a vessel on the water

fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$a_{ave}=1.80 m/s^{2}$

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0$ the initial velocity and $V_{f}$ the final velocity

$\Delta t=85.6 s$

Then:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}$

Since $V_{o}=0$ :

$a_{ave}=\frac{V_{f}}{\Delta t}}$

Finding $V_{f}$ :

$V_{f}=a_{ave} \Delta t$

$V_{f}=(1.80 m/s^{2})(85.6 s)$

Finally:

$V_{f}=154.08 m/s$

8 0

3 years ago

our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure

Dafna1 [17]

Answer:

The coefficient of static friction between your partner and the floor is 0.55

Explanation:

Given:

Mass $m = 59$ Kg

Frictional force $F_{s} = 318.3$ N

From the formula of frictional force,

$F_{s} = \mu_{s} mg$

Where $\mu _{s} =$ coefficient of static friction, $g = 9.8 \frac{m}{s^{2} }$

Put the above values and find the coefficient of static friction.

$318.3 = \mu_{s} \times 59 \times 9.8$

$\mu_{s} = 0.55$

Therefore, the coefficient of static friction between your partner and the floor is 0.55

4 0

3 years ago

Match these terms with the correct examples.

posledela

1. liquid solution to a. oceans
2. gaseous solution to b. clouds
Not sure about 3 and 4.
3 might be oxygen but I think that's 5. element.

Hope this helps, not sure about water and air though.

4 0

3 years ago

Read 2 more answers