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2 years ago

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A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

r to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?

1 answer:

Marina86 [1]2 years ago

3 0

Answer: $\alpha =10.66 rad/s^2$

Explanation:

Given

mass of disk $m=5 kg$

diameter of disc $d=30 cm$

Force applied $F=4 N$

Now this force will Produce a torque of magnitude

$T=F\cdot r$

$T=4\dot 0.15$

$T=0.6 N-m$

And Torque is given Product of moment of inertia and angular acceleration $(\alpha )$

$T=I\cdot \alpha$

Moment of inertia for Disc $I= \frac{Mr^2}{2}$

$I=0.05625 kg-m^2$

$0.6=0.05625\cdot \alpha$

$\alpha =10.66 rad/s^2$

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nalin [4]

Answer:

This depends on what angle they are approaching each other before they collided.The two simple cases are if they are running in the same direction or opposite direction from each other. For either case, use the conservation of momentum equation to solve: M_total*V_result = M1*V1 + M2*V2

Explanation:

Here are two possible solutions.

Head-on collision: M1=78, V1=8.5, M2=72, V2=-7.5 (that's negative because he's running the other way), M_total = 78+72 = 150, so V_result = (78*8.5 - 72*7.5)/150 = 0.82 m/s. Sanity check, they weigh about the same and so most of their velocity should cancel out.

Running the same way: change the sign of V2 to positive so V_result = (78*8.5 + 72*7.5)/150 = 8.02 m/s. Sanity check, they weigh about the same and the resultant speed is between the two starting velocities.

<em>hope it helps:)</em>

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When is the next total solar eclipse in north america

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April 8, 2024

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2 years ago

A factory worker pushes a 30.0kg crate a distance of 4.4 m along a level floor at constant velocity by pushing downward at an an

andre [41]

Answer:

1) F = 94.58 N , 2) W1 = 360.4 J , 3) W2 = -360.4 J , 4) W3 =0

Explanation:

1- To find the value of the force, let's use Newton's second law, where in the x-axis we have the applied force and the friction force that opposes the movement and the axis and we have the normal and the weight, look at the attached to see A free body diagram.

We see that the applied force (F) must be decomposed

Cos θ = Fx / F

Fx = f cos θ

sinθ = Fy / F

Fy = F sin θ

As the box moves at constant speed the acceleration is zero

X axis

Fx-fr 0 = 0

Fx = fr

fr = μ N

F cos T = μ N

F cos T = μ N

Axis y

N -W- Fy = 0

N- F sin θ = mg

Let's write the equation system and solve it

F cos θ = μ N

N - F sin θ = mg

N = mg + F sinθ (1)

F cos θ = μ (mg + F sin θ)

F (cos θ - μ Sin θ) = μ mg

F (cos 30 - 0.24 sin 30) = 0.24 30.0 9.8

F = 70.56 / 0.746

F = 94.58 N

This is the value of the force applied

2- The work is defined as the scalar product of the force by the distance traveled, in this case as we have the components of the force the only one that performs work is the Enel X-axis component, because with the other the angle is 90º and the cosine of 90 is zero

W1 = Fx X

W1 = 94.58 cos 30 4.4

W1 = 360.4 J

3- Let's calculate the value of the friction force

fr = μ N

The value of normal can be found in equation 1

N = mg + F sin θ

N = 30.0 9.8 + 94.58 sin 30

N = 341.3 N

fr = 0.24 341.3

fr = 81.9 N

Now we can calculate the work, but let's look at the angle between the displacement and the friction force that is 180º so the cosine of 180 = -1

W2 = - fr X

W2 = - 81.9 4.4

W2 = -360.4 J

4- The normal one has a direction perpendicular to the displacement, so its angle is 90º and the cos 90 = 0, which implies that the work is zero

W3 =0

5- To calculate total work we just have to add the work of each force

W all = W1 + W2 + WN + Ww

W all = 360.4 + (- 360.4) + 0+ 0

W all = 0

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3 years ago

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Brut [27]

I don't understand what you mean but whatever

It is often used for baking bread rolls<span>, </span>pastries<span> and flat products such as </span>cookies<span>, </span>sheet cakes<span>, </span>swiss rolls<span> and </span>pizzas<span>.</span>

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2 years ago

Two manned satellites approach one another at a relative speed of 0.230m/s, intending to dock. The first has a mass of 4.50×103k

aliya0001 [1]

Answer:

$v_{f1}=-0.345m/s$ , $v_{f2}=0.23m/s$

Explanation:

The collision is elastically that's means that both mass restore their same value after the collision and don't have the same speed however the collision both satellites approach at the same velocity but in different direction to try to dock so:

$p_i=p_f$

$m_1*v_1+m_2*v_2=m_1*v_{f1}+m_{2}*v_{f2}$

$k_i=k_f$

$1/2*m_1*v_1^2+1/2*m_2*v_2^2=1/2*m_1*v_{f1}^2+1/2*m_1*v_{f2}^2$

Two equations and two variables to find solve to both finals velocities:

$v_{f1}=\frac{m_1-m_2}{m_1+m_2}*v_1+\frac{2*m_2}{m_1+m_2}*v_2$

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$v_{f2}=\frac{2*m_1}{m_1+m_2}*v_1+\frac{m_2-m_1}{m_1+m_2}*v_2$

$v_{f2}=0.23m/s$

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2 years ago