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fenix001 [56]
3 years ago
15

A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless

pulley, as shown in figure (a). The coefficient of kinetic friction between the block and the surface is 0.300. 
(a) Find the acceleration of the two objects and the tension in the string. 
(b) Check the answer for the acceleration by using the system approach. (Use the following as necessary: m1, m2, μk, and g.) 
(c) What if an additional mass is attached to the ball? How large must this mass be to increase the downward acceleration by 60%? 

For (a) I calculated acceleration to be 5.10 m/s^2 and tension to be 36.2N. 
For (b) I the equation is a = (m2g-ukm1g)/(m1+m2) 
It is (c) that I cannot figure out. Can anybody please help me?
Physics
1 answer:
Allisa [31]3 years ago
4 0
1.6a =  \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3}  \\  \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\  \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\  \\ m_3 =  \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\  \\ m_3 = 22.57 kg
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 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi }  = \frac{L}{g} \\
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7 0
3 years ago
A 35-gram stainless steel ball on a track is moving at a velocity of 9 m/s. On the same track, a 75-gram stainless steel ball is
elena-s [515]

Answer:

4.2 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v

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315 g m/s = (75 g) v

v = 4.2 m/s

5 0
3 years ago
Read 2 more answers
A bird is about 6.26.2 in.​ long, with a​ thin, dark bill and a​ wide, white wing stripe. If the bird can fly 9292 mi with the w
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Answer:

209 mph

Explanation:

V = Speed of bird in still air

v = Speed of wind = 44 mph

Consider the motion of the bird with the wind

D_{1} = distance traveled with the wind = 9292 mi

t_{1} = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

t_{1} = \frac{D_{1}}{V + v}

t_{1} = \frac{9292}{V + 44}                              eq-1

Consider the motion of the bird with the wind

D_{2} = distance traveled against the wind = 6060 mi

t_{2} = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

t_{2} = \frac{D_{2}}{V + v}

t_{2} = \frac{6060}{V - 44}                              eq-2

As per the question,

Time taken with the wind = Time taken against the wind

t_{1} = t_{2}

\frac{9292}{V + 44} = \frac{6060}{V - 44}

(9292) (V - 44) = (6060) (V + 44)

9292V - 408848 = 6060V + 266640

3232V = 675488

V = 209 mph

5 0
2 years ago
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Dahasolnce [82]

Answer:

15 J

Explanation:

Work = Force x Distance

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Have a blessed day!

7 0
3 years ago
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GenaCL600 [577]
Around 70-72% of earth’s surface is covered in water (most of it is salt water).
Hope this helps.
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