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maksim [4K]
3 years ago
10

What affect does a tripling of the net force have upon the acceleration of the object ? Be quantitative

Physics
1 answer:
vovikov84 [41]3 years ago
5 0
The formula of net Force is:
F = ma
where m is the mass of the object
a is the acceleration of the object

so if we triple the net force applied to the object:
3F = ma
a = 3F / m

so the acceleration will also be tripled. because from the equation, the force is directly proportional to the acceleration
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Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

Here, Tension= m*g = m*9.81

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Linear mass density= \frac{8.25*10^-3}{65*10^-2}

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3 0
3 years ago
An initially motionless test car is accelerated uniformly to 120 km/h in 8.28 s before striking a simulated deer. The car is in
Anestetic [448]

Answer:

The value is   a =  4.0 \ m/s^2

Explanation:

From the question we are told that  

  The velocity  is  a =  120 \  km/h =  [tex]\frac{120 *  1000}{3600} =  33.3 \  m/s[/tex]

  The time taken is t  =  8.28 \  s

    The  time taken for contact is  t_c  =  0.815 \  s

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From the equation of kinematics we have that  

       v =  u  + at

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3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

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Let. convert to revolution /sec²

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b. Time Taken to complete 30revolution

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Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

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