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Roman55 [17]
3 years ago
10

The horse on a carousel is 3.5m from the central axis.A. If the carousel rotates at 0.13 rev/s , how long does it take the horse

to go around twice?B. How fast is a child on the horse going (in m/s)?
Physics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

a. 15.4 seconds

b. 0.455 m/s

Explanation:

a. The carousel rotates at 0.13 rev/s.

This means that it takes the carousel 1 sec to make 0.13 of an entire revolution.

This means that time it will take to make a complete revolution is:

1 / 0.13 = 7.7 seconds

Therefore, the time it will take to make 2 revolutions is:

2 * 7.7 = 15.4 seconds

b. Let us calculate the linear velocity. Angular velocity is given as:

\omega = v / r

where v = linear velocity and r = radius

The radius of the circle is 3.5 m and the angular velocity is 0.13 rev/s, therefore:

0.13 = v / 3.5

v = 3.5 * 0.13 = 0.455 m/s

Linear velocity is 0.455 m/s

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How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
two ends of an inextensible string of length 12m are attached to points Aand B 1.2m apart,in the same horizontal plane.a mass 20
AlekseyPX
All forces must add up to zero. See pictures below.

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What element from the periodic table rhymes with extreme
jonny [76]
Halite or sulfur or gold or silver
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3 years ago
Which are considered noble gases
sladkih [1.3K]

Answer:Argon

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4 0
2 years ago
Read 2 more answers
Help please<br> It’s kinda urgent
user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]

The negative sign of acceleration means that the ship slows down its velocity in order to land.

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2 years ago
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