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Nitella [24]
3 years ago
7

Calculate the acceleration of a 90-kg skydiver with a net force of -883 N

Physics
1 answer:
Helen [10]3 years ago
6 0
F=ma, in this case your force=-883 N and mass = 90 kg. F/m=a therefore acceleration=-883/90=-9.81 m/s/s
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A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
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Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

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Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

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