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1 month ago

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In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

he axis of rotation. The cables swing out at a constant angle of 45.0° when the ride is operating. What is the angular speed of rotation?

1 answer:

ValentinkaMS [3.4K]1 month ago

7 0

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

$T\cos\theta=mg$

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

Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Substituting the value into the equation

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Thus, the angular speed of rotation computes to 1.34 rad/s.

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A 1500 kg car enters a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h

Yuliya22 [3333]

Answer:

The question is incomplete; refer to the attachment for the diagram

Explanation:

Provided that,

The mass of the car

M = 1500kg

Entering the curve at Point A with a speed of

Va = 100km/hr = 100× 1000/3600

Va = 27.78m/s

The car slows down at a constant rate until it reaches point C at a speed of

Vc = 50km/hr = 50×1000/3600

Vc = 13.89m/s

Radius of curvature at point A

p = 400m

Radius of curvature at point B

p = 80m

The distance from point A to point B as indicated in the attachment is

S=200m

We need to determine the total horizontal forces at points A, B, and C applied by the road on the tire

The constant tangential acceleration can be computed using the equation of motion

Vc² = Va² + 2as

13.89² = 27.78² + 2 × a × 200

192.9 = 771.6 + 400a

400a = 192.9—771.6

400a = -578.7

a = -578.7 / 400

a = —1.45 m/s²

at = —1.45m/s²

The tangential acceleration is -1.45m/s² and it is negative due to the car decelerating

Since the vehicle is decreasing speed at a uniform rate, the tangential acceleration remains the same at every point

At point A

at = -1.45m/s²

At point B

at = -1.45m/s²

At point C

at = -1.45m/s²

Now,

We can compute the normal component of acceleration (centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration can be calculated using

ac = v²/ p

At point A ( p = 400)

an = Va²/p = 27.78² / 400

an = 1.93 m/s²

At point B (p = ∞), since point B is the inflection point

Then,

an = Vb²/p = Vb/∞ = 0

an = 0

At point C ( p = 80m)

an = Vc²/p = 13.89² / 80 = 2.41m/s²

an = 2.41 m/s²

Then,

The tangential force is

Ft = M•at

Ft = 1500 × 1.45

Ft = 2175 N.

Given that the tangential acceleration remains constant, this represents the tangential force at A, B, and C

Next, the normal force

At point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N.

Thus, the horizontal force acting at each point is

Using the right triangle vector method

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N

6 0

25 days ago

One day you are driving your friends around town and you drive quickly around a corner without slowing down. You are going at a

Ostrovityanka [3204]

Result:

1.60 g

Elucidation:

Based on the attached document:

we can infer that:

$v = v_x =v_y = 20 \ m/s \\t = 2s$

The distance covered in 2 seconds will be:

x = vt

x = 20 m/s × 2 s

x = 40 m

The segment corresponds to a quarter of a circle with radius r,

therefore, if 2 πr = 4 x

Then the radius (r) can be calculated as:

$r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}$

r = 25.5 m

Centripetal acceleration can be expressed as:

$a = \frac{v^2}{r}$

thus;

$a = \frac{(20 \ m/s^2)}{25.5 \ m}$

a = 15.7 m/s²

The acceleration magnitude suffered by your passengers in relation to the acceleration due to gravity can be calculated using:

$a' = \frac{a}{g} g$

$a' = \frac{15.7 \m/s ^2}{9.81 \ m/s^2}\ g$

$a' = 1.60 \ g$

∴ The acceleration magnitude experienced by your passengers while turning = 1.60 g

6 0

1 month ago

Read 2 more answers

A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic for

serg [3582]

Answer:

20.353125 V

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

q = Charge of proton = $1.6\times 10^{-19}\ C$

$v_A$ = Velocity of proton at point A = 50 km/s

$v_B$ = Velocity of proton at point B = 80 km/s

The relationship derived from energy conservation is as follows:

$\dfrac{1}{2}m(v_B^2-v_A^2)=q(V_B-V_A)\\\Rightarrow V_B-V_A=\dfrac{1}{2q}m(v_B^2-v_A^2)\\\Rightarrow V_B-V_A=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 1.67\times 10^{-27}(80000^2-50000^2)\\\Rightarrow V_B-V_A=20.353125\ V$

The determined potential difference is 20.353125 V

3 0

1 month ago

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

serg [3582]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Find out more about average velocity:

6 0

1 month ago

True or False: Molecules in a gas resist crowding and get as far apart as possible. Free electrons also resist crowding and get

ValentinkaMS [3465]

Answer:

This assertion is inaccurate.

Explanation:

The random nature of gas molecules results in their erratic motion and occasional collisions. While it is true that they tend to avoid being tightly packed, achieving the maximum separation from each other is not always feasible due to their lack of fixed positions. Consequently, gas molecules in a container cannot consistently maintain the furthest distance from their neighboring molecules.

In contrast, the separation among electrons is primarily influenced by repulsive forces, not random movement as in gases. Electrons maintain distance as a result of repulsion between similarly charged particles. Therefore, the arrangement of electrons on a charged copper sphere occurs not from a random distribution but rather due to repulsion, establishing a set distance between them.

4 0

1 month ago