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3 years ago

If there are no unbalanced forces acting on an object, according to Newton's First Law:

Physics

1 answer:

Sergio039 [100]3 years ago

4 0

I think its b because when there is unbalanced forces it accelerates.

Hope this helps you out

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Example 7.3

marysya [2.9K]

Answer:

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3 years ago

A 50N girl pushes a 10,000 N car with force of 200N. What is the force the car pushes back at the girl? *

miss Akunina [59]

Answer:

200N

Explanation:

EVERY FORCE IS OPPOSED BY AN EQUAL FORCE, REGARDLESS OF THE WEIGHT OF THE OBJECTS APPLYING THE FORCE.

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3 years ago

Group 18 noble gases are inert because

hammer [34]

Answer:

Explanation:

have a full 8 electrons on the outermost ring

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3 years ago

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

jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

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

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

What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your

torisob [31]

<span>Radius distance from origin to particle = √ (2²+1²) = √5 m = R
I = MR² = (0.200)(5) = 1.00 kg-m²
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal
V = 3.0 m/s
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s
w = [V component / R] = 1.3433/√5 = 0.601 rad/s
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s</span><span>

i hope I'm right</span>

6 0

3 years ago