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3 years ago

According to the work-energy theorem, the amount of work done can be determined using which formula?

Physics

2 answers:

Tema [17]3 years ago

8 0

W = KE = 1/2m(v2f-v2i) also known as D just took the test

Marina CMI [18]3 years ago

8 0

Answer:

$W=K_f -K_i = \frac{1}{2}mv_f^2 -\frac{1}{2}mv_i^2$

Explanation:

The work-energy theorem states that the amount of work done is equal to the variation of kinetic energy of the object, therefore:

$W=K_f -K_i = \frac{1}{2}mv_f^2 -\frac{1}{2}mv_i^2$

Where:

m is the mass of the object

$v_i$ is the initial velocity of the object

$v_f$ is the final velocity of the object

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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

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 $54.66\ \text{revolutions}$ .

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3 years ago

A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially a

olasank [31]

Answer:

M1 V1 = M1 V2 + M2 V3 conservation of momentum

V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign

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2 years ago

Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work

LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

$\frac{\Delta x}{\Delta t}$

Analysing the units, meters (displacement) and seconds (time) are basic units:

$\frac{m}{s}$

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

$F = ma$

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

$a=\frac{\Delta v}{\Delta t}$

Analysing the units:

$\frac{\frac{m}{s}}{s}=\frac{m}{s^2}$

Therefore, the unit of force is:

$kg\frac{m}{s^2}$

Pressure:

Pressure is given by the equation:

$P=\frac{F}{S}$ where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

$[l^2]=m^2$ , the unit of area

Dividing the aforementioned unit of force by the unit of area:

$\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}$ , the unit of pressure

Work:

Work is given by the equation:

$W=\vec{F}\cdot \vec{x}$ , (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

$[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}$ , the unit of work.

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C. an electrical resistor

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