The cycle is a process that returns to its beginning, but it does not repeat

Determine the maximum theoretical speed that may be achieved over a distance of 66 m by a car starting from rest, knowing that t

My name is Ann [436]

Answer:

v=32.49 m/s

Explanation:

Given that

Distance ,d= 66 m

Initial speed of the car ,u = 0 m/s

Coefficient of friction ,μ = 0.8

Lets take the total mass of the car = m

The acceleration of the car is given as

a = μ g ( g= 10 m/s² )

Now by putting the values in the above equation we get

a= 0.8 x 10 m/s²

a= 8 m/s²

We know that ,final speed is given as

v²= u ²+ 2 a d

Now putting the value

v²=0² + 2 x 8 x 66

v²= 1056

v=32.49 m/s

3 0

2 years ago

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

vivado [14]

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 $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

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 $54.66\ \text{revolutions}$ .

3 0

2 years ago

Problem:

gayaneshka [121]

This is confusing yes

3 0

2 years ago

Please Help!! This is very tough for me. Try you best to answer these problems!!Pleaseee

Andrews [41]

Answer:

everyone else does this to me so lol

Explanation:

7 0

2 years ago

Read 2 more answers

Describe a procedure that would increase the potential energy of two magnets if like poles are used. Explain why the energy of t

zalisa [80]

Answer:

If you apply a force to separate 2 opposite poles, the potential energy of the system increases.

5 0

1 year ago