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AleksAgata [21]

2 years ago

11

A race car driver must average 200km/hr for four laps to qualify for a race. Because of engine trouble, the car averages only 17

0km/hr over the first two laps. What is the average speed must be maintained for the last two laps

1 answer:

vampirchik [111]2 years ago

5 0

The average speed would have to be 260 km/hr due to the driver originally going 30 km/hr too slow the first two laps

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A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th

Tju [1.3M]

Answer:

$a = \frac{2}{3}g$

$T = \frac{mg}{3}$

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

$mg - T = ma$

also by torque equation we can say

$TR = I\alpha$

$TR = \frac{1}{2}mR^2(\frac{a}{R})$

$T = \frac{1}{2}ma$

Now we have

$mg - \frac{1}{2}ma = ma$

$mg = \frac{3}{2}ma$

$a = \frac{2}{3}g$

Also from above equation the tension force in the string is

$T = \frac{1}{2}ma$

$T = \frac{mg}{3}$

7 0

2 years ago

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

5 0

2 years ago

You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm

maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0

2 years ago

An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init

elena-14-01-66 [18.8K]

Answer:

Approximately $71\%$ .

Explanation:

The formula for the kinetic energy $\rm KE$ of an object is:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ ,

where

$m$ is the mass of that object, and
$v$ is the speed of that object.

Important: Joule ( $\rm J$ ) is the standard unit for energy. The formula for $\rm KE$ requires two inputs: mass and speed. The standard unit of mass is $\rm kg$ while the standard unit for speed is $\rm m \cdot s^{-1}$ . If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

$m = 63\; \rm g = 0.063\; \rm kg$ .

Initial $\rm KE$ of this arrow:

$\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}$ .

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

$\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%$ .

8 0

2 years ago

A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point

denis-greek [22]

Answer:

V0=27.4 m/s; t=0.8 s

Explanation:

Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:

$y_f=v_0t-0.5*g*t^2$ We solve for initial speed

$v_0=\frac{y_f+0.5gt^2}{t}=\frac{37+4.9*2.3^2}{2.3}=27.4m/s$

Now, using the same expression we estimated time to first reach 18.5 m :

$18.5=27.4t-4.9t^2$ Second order equation with solutions

t1=0.8 s and t2=4.8 s

The first time corresponds to the first reach.

7 0

2 years ago