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Marianna [84]
2 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]2 years ago
3 0

Answer:

An attachment has been added

Explanation:

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storchak [24]
<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s
 
vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

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The peregrine falcon is the world's fastest known bird and has been clocked diving downward toward its prey at constant vertical
Sergio [31]
100m / 97.2m/s = 1.0288 seconds
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1 year ago
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Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​
Alinara [238K]

The volume of the room is the product of its dimensions:

10\times 12 \times 4 = 480\text{ m}^3

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7 0
2 years ago
Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌
NemiM [27]

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

(m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]

7 0
1 year ago
How much heat must be absorbed by 375 grams of water to raise its temperature by 25°c
Nesterboy [21]
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2 years ago
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