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Marianna [84]
3 years ago
7

Question must be solved using only symbolic algebra.During the experiment if you could triple the breakaway magnetic force with

all other quantities left unchanged, what is the new value for the critical velocity if it was v0 (initial velocity), initially? (b) Now if you halved the radius with all other quantities left unchanged, what is the new critical velocity if it was v0 (initial velocity), initially? (c) If during the experiment, critical velocity quadrupled with all other quantities left unchanged, what is the new breakaway force if its magnitude was initially F0,?

Physics
1 answer:
Papessa [141]3 years ago
3 0

Answer:

An attachment has been added

Explanation:

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Ok so I am taking a test I need help please
Aloiza [94]

Answer:

i know its definetly either clockwise or counter clockwise

Explanation:

6 0
3 years ago
An object completes one and half revolution of a circle of radius R calculate the displacement and distance
Papessa [141]

Answer:

Distance is path length covered by particle. When particle moves along half circle, it covers half the circumference therefore distance covered is (2×pi×r)/2 = pi× r. ... Hence displacement is equal to diameter or 2 times the radius of circle.

8 0
3 years ago
Look at the image to answer the question correctly.
r-ruslan [8.4K]

Answer:

1-  b: 2- a : 3- c : 4- d

Explanation:

it starts 2 move away from strting point, then no motion, then moves toward the start, the slows up.

3 0
3 years ago
How do you do this question? Please include free body diagrams and clear explanation, so I can understand.
vagabundo [1.1K]

Explanation:

Draw a free body diagram for each disc.

Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.

∑F = ma

86.5 N − T₁ − Wa = 0

Wa = 86.5 N − T₁

ma × 9.8 m/s² = 86.5 N − 55.6 N

ma = 3.2 kg

Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.

∑F = ma

T₁ − T₂ − Wb = 0

Wb = T₁ − T₂

mb × 9.8 m/s² = 55.6 N − 36.5 N

mb = 1.9 kg

Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.

∑F = ma

T₂ − T₃ − Wc = 0

Wc = T₂ − T₃

mc × 9.8 m/s² = 36.5 N − 9.6 N

mc = 2.7 kg

Disc D has two forces acting on it: T₃ up and Wd down.

∑F = ma

T₃ − Wd = 0

Wd = T₃

md × 9.8 m/s² = 9.6 N

md = 0.98 kg

3 0
3 years ago
A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and
Airida [17]

Answer:

Explanation:

Given

N_1=1 rev/s

angular velocity \omega =2\pi N_1=6.284 rad/s

Combined moment of inertia of stool,student and bricks =6\ kg.m^2

Now student pull off his hands so as to increase its speed to suppose N_2 rev/s

\omega _2=2\pi N_2  

After Pulling off hands so final moment of inertia is

I_2=2\ kg-m^2

Conserving angular momentum  as no external torque is applied

I_1\omega _1=I_2\omega _2

6\times 6.284=2\times \omega _2

\omega _2=18.85\ rad/s

N_2=3 rev/s

7 0
3 years ago
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