Which of the following forces best represents an equilibrant force in this system?

If you can throw a stone straight up to height h, what’s the maximum horizontal distance you could throw it over level ground?

Mariana [72]

To answer this question, first we take note that the maximum height that can be reached by an object thrown straight up at a certain speed is calculated through the equation,
Hmax = v²sin²θ/2g
where v is the velocity, θ is the angle (in this case, 90°) and g is the gravitational constant. Since all are known except for v, we can then solve for v whichi s the initial velocity of the projectile.

Once we have the value of v, we multiply this by the total time traveled by the projectile to solve for the value of the range (that is the total horizontal distance).

8 0

3 years ago

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An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e

frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

$Tsin\theta= \frac{mv^2}{r}$

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

$Tcos\theta = mg$

Matching both expressions with respect to the tension we will have to

$T = \frac{\frac{mv^2}{r}}{sin\theta}$

$T = \frac{mg}{cos\theta}$

Then we have that,

$\frac{\frac{mv^2}{r}}{sin\theta} = \frac{mg}{cos\theta}$

$\frac{mv^2}{r} = mg tan\theta$

Rearranging to find the velocity we have that

$v = \sqrt{grTan\theta}$

The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is

$r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}$

$r = \frac{11.1}{2}+2.41$

$r = 7.96m$

Replacing we have that

$v = \sqrt{(9.8)(7.96)tan(14.5\°)}$

$v = 4.492m/s$

Therefore the speed of each seat is 4.492m/s

6 0

3 years ago

A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui

raketka [301]

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

8 0

3 years ago

Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at

atroni [7]

Answer:

y = 52.44 10⁻⁶ m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

a sin θ= m λ

The first minimum occurs for m = 1

sin θ = λ / a

Now let's use trigonometry the object is a distance L = 0.205 m

tan θ = y / L

Since the angles are very small, let's approximate

tan θ = sin θ/cos θ = sin θ

sin θ = y / L

We substitute in the diffraction equation

y / L = λ / a

y = λ L / a

Let's calculate

y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

y = 52.44 10⁻⁶ m

7 0

4 years ago

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :