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iogann1982 [59]
3 years ago
10

20-kilogram canoe is floating downriver at 2 m/s. what’s the kinetic energy?

Physics
2 answers:
ziro4ka [17]3 years ago
6 0

Answer:

40J

Explanation:

Kinetic energy = 1/2 mv^2

Where m = mass and

v = velocity

Given mass = 20kg

v = 2m/s

K.E = 1/2 x 20 x2^2

= 1/2 x 20 x 2 x 2

= 80/2

= 40J

ryzh [129]3 years ago
3 0

To a physicist floating next to it in another canoe, its KE is zero.

To a physicist standing on the riverbank watching it go by, its KE is 40J.

They're both correct for their own reference frame.

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Which of the following objects exerts a gravitational force?
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the earth exerts a gravitational force

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2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this wee
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Answer:

  • 62 seconds
  • no

Explanation:

The total travel time Suzette experiences is the sum of the times in each hallway. Using

  time = distance/speed

we can add the times.

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It takes Suzette 62 seconds to get to class. She does not beat the bell.

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Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

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\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

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