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OLga [1]
3 years ago
10

A 12.0 kg object is attached to a cord that is wrapped around a wheel of radius 10.0 cm. the acceleration of the object down the

frictionless incline is measured to be 2.00 m/s2. assuming the axis of the wheel to be frictionless, determine
a.the tension in the rope,
b.the moment of inertia of the wheel, and
c.the angular speed of the wheel 2.00 s after it begins rotating, starting from rest.
Physics
1 answer:
olganol [36]3 years ago
6 0

 a) We use the formula:

T = [g sinθ – a] * m

where g is gravity, θ is the angle, a is acceleration, m is mass

T = [9.81 sin37 - 2.00] * 12 = 46.85 N <span>

b) We use the formula for moment of interia:</span>

I = T * r² / a

I = 46.85 * 0.10² / 2.00 = .2343 kg∙m² <span>

c) The formula we can use here is:</span>

w = α * t = (a/r) * t

w = (2/.10) * 2

<span>w = 40 rad/sec</span>

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Reduce speed or slow down.

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7 0
3 years ago
If a man weighs 900 n on the earth, what would he weigh on jupiter, where the acceleration due to gravity is 25.9 m/s2?
balandron [24]
<span>So, if the man weight 900 newtons on Earth then that means, using F=ma, that the mass of the man is approximately 91.84 kg. This is because 900N=m(9.8m/s^2), and so it follows that 900/9.8=91.84. Using the man's found mass we then plug this into F=ma again. It follows that F=(91.84)(25.9)=2378.57N. This means that the man "weighs" 2378.57 Newtons on Jupiter, or about 2.5x as great as his weight on Earth. This makes sense, considering that 25.9/9.8 is approximately equal to 2.64.</span>
4 0
4 years ago
What is the free-fall acceleration at the surface of the jupiter?
Rufina [12.5K]

Answer:

24.79 m/s2

Explanation:

Let us assume that there is an object with a mass of 'm' on the

 

Jupiter. Jupiter will attract this object:

mg_j=G\frac{mM_j}{r_j^{2} }

g_j=G\frac{M_j}{r_j^{2} }

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4 0
3 years ago
Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷
victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

q_1 = -8.00 * 10^{-7} C

x_1 = 6.00 cm

q_2 = +8.00 * 10^{-7} C

x_2 = 21 cm

The distance between q_1 and q_2 is

21 - 6 = 15cm

Electric field is given as

E = \frac{kq}{r^2}

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to q_1 is:

E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}

E_1 = -1.28 * 10^6 N/C

The electric field at the midpoint due to q_2 is:

E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}

E_2 = 1.28 * 10^6 N/C

The net electric field will be:

E = E_1 + E_2

E = -1.28 * 10^6 + 1.28 * 10^6

E = 0 N/C

7 0
4 years ago
The gravitational force
vivado [14]
<span>b. weakens as 1/d, where d is the distance between objects.</span>
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