<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
What’s the answer choices to pick from?
The acceleration of the box up the ramp is 9.65 m/s².
<h3>
What is the magnitude of acceleration of the box?</h3>
The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;
F(net) = ma
where;
- m is the mass of the box
- a is the acceleration of the box
The net force on the box is calculated as follows;
F(net) = F - Ff
F(net) = F - μmgcosθ
where;
- θ is the inclination of the plane
- μ is coefficient of friction
F(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.7
The acceleration of the box is calculated as;
a = F(net) / m
a = (144.7) / (15)
a = 9.65 m/s²
Thus, the acceleration of the box up the ramp is 9.65 m/s².
Learn more about acceleration here: brainly.com/question/14344386
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Answer: 0.006in/s
Explanation:
Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s
Also let the rate at which the radius of the balloon is increasing be dr/dt
Given r = 4.7in and Π = 3.14
Applying the chain rule method
dV/dt = dV/dr × dr/dt
If the volume of the sphere is 4/3Πr³
V = 4/3Πr³
dV/dr = 4Πr²
If r = 4.7in
dV/dr = 4Π(4.7)²
dV/dr = 277.45in²
Therefore;
1.68 = 277.45 × dr/dt
dr/dt = 1.68/277.45
dr/dt = 0.006in/s
<span>7.21 ft/s^2
Since you're looking for average acceleration, you can simply divide the change in velocity by the time. To make the calculation more reasonable, first convert the speed of 173 mi/h into ft/sec by multiplying by 5280 to convert from mi/h to ft/h and then dividing by 3600 to convert from ft/h to ft/s.
173 * 5280 / 3600 = 253.7333 ft/s
Now divide the change in velocity by the time in seconds.
253.7333 ft/s / 35.2 s = 7.208333 ft/s^2
Rounding result to 3 significant figures gives 7.21 ft/s^2</span>
