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lara31 [8.8K]
3 years ago
13

What is the avarage velocity of a car that travels 30 kilometers due west in 0.5 hours

Physics
2 answers:
Fed [463]3 years ago
6 0
Velocity is distance over time v= \frac{d}{t}. If it wants it in km/h you're good to go otherwise if its m/s you need to convert km to m by multiplying 30 by 1000 and t in hours to seconds by multiplying 0.5 by 3600.
Nikolay [14]3 years ago
5 0
Average velocity=distance/time
average v=s/t
averagev=0.3/30 min
 
so use your calculator 30min into seconds(30/60)
and divide your kilometre with your time(v=distance/time
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The initial velocity of a micro van is 15 m/s. It gains a velocity of 40 ms in 10 seconds. Calculate the average velocity and ac
PSYCHO15rus [73]

Answer:

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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}

Substitute the value in the above equation.

\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}

Thus, the mass of the box is 12.9 kg.

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Answer:

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