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Sati [7]
3 years ago
9

8) Calculate the kinetic energy of a truck that has a mass of 6100 kg and is moving at 55 m/s.

Physics
1 answer:
Inessa [10]3 years ago
7 0

Answer:

<h2>9,226,250 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

v is the velocity

m is the mass

From the question we have

k =  \frac{1}{2}  \times 6100 \times  {55}^{2}  \\  = 3050 \times 3025 \\  = 9226250

We have the final answer as

<h3>9,226,250 J</h3>

Hope this helps you

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The diameter of a baseball is 7.4 cm and its mass is 0.15 kg. a) If a pitcher throws the baseball at a velocity of 44.3 m/s (100
Tema [17]

Answer:

drag force F_D = 1.5 \  N

Velocity (V) = 40.169 m/s

Explanation:

The drag force F_D is given  by the formula:

F_D = C_D * \frac{1}{2}* \rho * V^2*A

where:

C_D = drag coefficient depending on the Reynolds number

Reynolds number Re = \frac{\rho *V*D}{ \mu}

Let's Assume that the air is in room temperature at 25 °C ; Then

density of the air \rho = 1.1845  kg/m³

viscosity of fluid or air \mu = 1.844 × 10⁻⁵ kg/ms

diameter of the baseball D = 7.4 cm

Velocity V = 44.3 m/s

Replacing them into the equation of Reynolds number ; we have :

Re = \frac{1.1845 \ kg/m^3*44.3 m/s*0.074 m}{1.844*10^{-5}kg/ms}\\\\Re = 2.1*10^5

A = Projected Area

From the diagram attached below which is gotten from NASA for baseball;

the drag coefficient which depends on Reynolds number is read as:

C_D = 0.3

Projected Area A = \frac{\pi D^2}{4}

A = \frac{\pi 0.074^2}{4}

A = 0.0043 m²

Finally, drag force is then calculated as ;

F_D = C_D * \frac{1}{2}* \rho*V^2*A\\\\F_D = 0.3* \frac{1}{2}*1.1845 \ kg/m^3*(44.3 \ m/s) ^2*0.0043 m^2\\\\F_D = 1.5 \ N

b)

- F_D = ma

since acceleration a = \frac{dV}{dt}

Then;

-F_D = m \frac{dV}{dt}

Also;

velocity (V) = \frac{dx}{dt}

Then;

- F_D = \frac{md_2x}{dt^2}

\frac{d_2x}{dt^2}  = \frac {- F_D}{m}

F_D = 1.5 \ N\\m = 0.15 \ kg

Then;

\frac{d_2x}{dt^2}  = \frac {- 1.5 }{0.15}

\frac{d_2x}{dt^2}  =- 10

Integrating the above equation ; we have :

\frac{dx}{dt}= - 10 t + C\\

when time (t) = 0 ; then \frac{dx}{dt}= V = 44.3

44.3 = - 10 × 0 + C

C = 44.3

\frac{dx}{dt}= V = -10 t + 44.3

Time (t) =

\frac{distance }{velocity} \\\\= \frac{18.3 m}{44.3 m/s}\\\\= 0.413 s

∴ Velocity ; \frac{dx}{dt}= V = - 10t +44.3

\frac{dx}{dt}= V = - 10(0.413 s) +44.3

Velocity (V) = 40.169 m/s

6 0
3 years ago
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