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svlad2 [7]
3 years ago
8

is a constant related to the size and composition of the ball as well as the viscosity of the syrup. Find the rate at which grav

itational energy is converted to thermal energy once the ball reaches terminal velocity.
Physics
1 answer:
Neporo4naja [7]3 years ago
4 0

Answer:

Thermal = 0.48 W

Explanation:

Given:

- The complete question is:

" Starting from rest, a 0.0367-kilogram steel ball sinks into a vat of corn syrup. The thick syrup exerts a viscous drag force that is proportional to the ball's velocity: where C = 0.270 N- s/m is a constant related to the size and composition of the ball as well as the viscosity of the syrup. Find the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity. After the ball reaches terminal velocity its kinetic energy is constant Any further work done by gravity is therefore directly transferred to the syrup and ball as thermal energy The rate at which the energy is converted is the power supplied by gravity: where F is the force, v is the speed of the ball and B is the relative angle between the force and velocity vectors. The terminal velocity of the ball can be determined from the given information using Newton's Laws. "

Solution:

- Once the terminal velocity Vt of the body is reached, it begins to move downward  at constant speed. The drag force Fd then attains a constant value:

                                         Fd = - C • Vt

- Instant the rate of conversion of gravitational potential energy to thermal  energy becomes equal to the power dissipated by the drag force at the terminal  velocity is given by:

                                        Fd • Vt = - C • Vt ²

- To get the general velocity V , solve the equation of motion:

                                        mv' = -m*g - C*V

                                        v' = -g - [C/ m] * V

Where,

                                        v' = dv / dt

                                        dv / dt =  -g - [C/ m] * V

- We will denote [C/ m]  as constant B:

                                        dv / dt =  -B*[ g/B +  V ]

- Separate variables:

                                        dv / [ g/B +  V ]  =  -B*dt

- Integrate both sides:

                                        Ln | g/B +  V | = -B*t + C

                                        g/B +  V = C*e^( -B*t )

                                        V = C*e^( -B*t ) - g/B

- The above expression for velocity gives the velocity of the steel ball at any time t. We know that terminal velocity Vt is achieved when t - > ∞. Then we have:

                                       Vt = 0 - g/B = -m*g/C

- Then from power equation we have:

                                        P = - C*Vt^2

                                        P = -C ( -m*g/C )^2

                                        P = -(mg)^2 / C

                                        P = -(0.0367*9.81)^2 / 0.27

                                        P = - 0.48 W

- The rate at which gravitational energy is lost to surrounding gives rise to thermal energy of surrounding fluid Thermal = 0.48 W .

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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity v=0 \ m/s

initial velocity u= 71 m/s  distance  

Distance covered in the slowing down phase = 250 m

v^2-u^2=2as

a= \frac {(v^2-u^2)}{2s}

a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2

v=u+at

t= \frac {(v-u)}{a}

= \frac {(0-71)}{(-10.082)}=7.042 s

t_1=7.042 s

The car is in the pit stop for 5s t_2=5 s

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2

v=u+at

t= \frac{(v-u)}{a}

t= \frac{(71-0)}{6.81}= 10.425 s

t_3=10.425 s

total time= t_1 +t_2+t_3=7.042+5+10.425=22.467 s

Distance covered by the Mercedes Benz during this time is given by s=vt=71 \times 22.467= 1595.157 m

Distance covered by the Thunderbird during this time=250+350=600 m

Difference between distance covered by the Mercedes  and Thunderbird

= 1595.157-600=995.157 m

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0
3 years ago
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