Which 2 types of element are joined in an ionic bond

the passengers in a roller coaster feel 50 heavier than their true weight as the car goes through a dip with a 30m radius of cur

NeTakaya

Answer:

v = 12.12 m/s

Explanation:

Given that,

Radius of the curvature, r = 30 m

To find,

The car's speed at the bottom of the dip.

Solution,

Let mg is the true weight of the passenger. When it is moving in the circular path, the centripetal force act on it. It is given by :

$F=\dfrac{mv^2}{r}$

The normal reaction of the passenger is given by :

$N=mg-50\%mg$

N = 1.5 mg

Let v is the car's speed at the bottom of the dip. It can be calculated as:

$1.5\ mg-mg=\dfrac{mv^2}{r}$

$v=\sqrt{0.5gr}$

$v=\sqrt{0.5\times 9.8\times 30}$

v = 12.12 m/s

So, the speed of the car at the bottom of the dip is 12.12 m/s. Hence, this is the required solution.

7 0

3 years ago

Select the correct answer. Physics is explicitly involved in studying which of these activities? A. the mixing of metals to form

geniusboy [140]

Answer: C. the motion of a spacecraft under gravitational influence.

Explanation:

A is Metallurgy, B is Biology, C is astro-physics, I am not sure what D is, but it's safe to say it's not physics, E, micro-biology, and the study of radiation. C is the only one involving physics.

8 0

4 years ago

If a freely falling object were equipped with a speedometer on a planet where the acceleration due to gravity is 20 m/s², then i

babymother [125]

Answer:

20 m/s

Explanation:

The acceleration of an object determines the increase in velocity per second.

In this case the acceleration is 20m/s^2.

$a=\frac{v}{t}=\frac{\frac{20m}{s}}{s}$

That is, the increase is 20m/s each second.

I hope this is useful for you

Regards

7 0

4 years ago

Read 2 more answers

The number of times that a wave vibrates in unit is called

kari74 [83]

I believe it's frequency.

6 0

4 years ago

As a moon follows its orbit around a planet, the maximum grav- itational force exerted on the moon by the planet exceeds the min

Mars2501 [29]

The gravitational force exerted on the moon by the planet when the moon is at maximum distance $r_{max}$ is
$F_{min}=G \frac{Mm}{r_{max}^2}$
where G is the gravitational constant, M and m are the planet and moon masses, respectively. This is the minimum force, because the planet and the moon are at maximum distance.

Similary, the gravitational force at minimum distance is
$F_{max}=G \frac{Mm}{r_{min}^2}$
And this is the maximum force, since the distance between planet and moon is minimum.

The problem says that $F_{max}$ exceeds $F_{min}$ by 11%. We can rewrite this as
$F_{max}=(1+0.11)F_{min}=1.11 F_{min}$

Substituing the formulas of Fmin and Fmax, this equation translates into
$\frac{1}{r_{min}^2}=1.11 \frac{1}{r_{max}^2}$
and so, the ratio between the maximum and the minimum distance is
$\frac{r_{max}}{r_{min}}= \sqrt{ 1.11 }=1.05$

8 0

4 years ago