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andreev551 [17]
3 years ago
14

Two forces are acting on a wheelbarrow. One force is pushing to the right and an equal force is pushing to the left. What can yo

u say about the wheelbarrow's movement?
Physics
1 answer:
andreev551 [17]3 years ago
6 0

-- As far as we know, the forces on the wheelbarrow are balanced.

-- That tells us that the net force on the wheelbarrow is zero, just
as if there were no forces acting on it at all.

-- That tells us that the wheelbarrow's acceleration is zero ... its
speed and direction of motion are not changing.

-- That tells us that the wheelbarrow is moving in a straight line
at a constant speed.  It's very possible that relative to us, the speed
may be zero, but we can't tell that from the given information.

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The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete
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Answer:

The answer is below

Explanation:

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y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

3 0
3 years ago
Please help me with Part B & help me correct the rest of the questions
AleksAgata [21]

Answer:

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5 0
3 years ago
A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu
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Answer:

v = 10.89\ m/s

Explanation:

given,                          

radius of loop = 12.1 m                              

to find the minimum speed transverse by the rider to not to fall out upside down                                                                

centripetal force = \dfrac{mv^2}{r}

gravitational force  = m g

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mg = \dfrac{mv^2}{r}

v = \sqrt{rg}

v = \sqrt{12.1 \times 9.8}

v = \sqrt{118.58}

v = 10.89\ m/s

5 0
3 years ago
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