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liq [111]
3 years ago
11

Achilles and the tortoise are having a race. The tortoise can run 1 mile (or whatever the Hellenic equivalent of this would be)

per hour. Achilles runs ten times as fast as the tortoise so the tortoise gets a head start of 1 mile. The race begins! By the time Achilles reaches the 1 mile mark, the tortoise is .1 miles ahead. By the time Achilles runs this extra tenth of a mile, the tortoise is still .01 miles ahead. This process continues; each time Achilles reaches the point where the tortoise was, the tortoise has moved ahead 1/10 as far. Can Achilles ever catch the tortoise? If so, when? If not, who would you bet on?
Physics
1 answer:
fenix001 [56]3 years ago
7 0

Answer:

Surely Achilles will catch the Tortoise, in 400 seconds

Explanation:

The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).

To calculate the time, we use kinematic expression for constant speed:

x_{final}=x_{initial}+t_{tor}v_{tor}=1+t_{tor}\\x_{final}=x_{initial}+t_{ach}v_{ach}=10t_{ach}

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):

1+t=10t\\t=1/9 hour=0.11 hours

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geniusboy [140]
1) 20 min = 1/3 h
2) distance = speed * time = 12 * 1/3 = 4 km

3 0
3 years ago
A resistor, inductor, and capacitor are connected in series, each with effective (rms) voltage of 65 V, 140 V, and 80 V respecti
Morgarella [4.7K]

Answer:

The value of the effective (rms) voltage of the applied source in the circuit is 132 V

Explanation:

Given;

effective (rms) voltage of the resistor, V_R = 65 V

effective (rms) voltage of the inductor, V_L = 140 V

effective (rms) voltage of the capacitor, V_C = 80 V

Determine the value of the effective (rms) voltage of the applied source in the circuit;

V= \sqrt{V_R^2 + (V_L^2-V_C^2} )\\\\V= \sqrt{65^2 + (140^2-80^2} )\\\\V = \sqrt{4225+ 13200} \\\\V = \sqrt{17425} \\\\V = 132 \ V

Therefore, the value of the effective (rms) voltage of the applied source in the circuit is 132 V.

6 0
4 years ago
how many days does it take a team of drivers to drive a truck california and back to lansing if the drivers average 15m/s includ
Scilla [17]

Answer:

t = 2.2 [days] and is there is a round trip, it will be double time t = 4.4 [days]

Explanation:

First, we need to arrange the problem to work in the same unit system (SI).

We need to convert the 1800 [miles] to meters, therefore:

1800[miles] * \frac{1609.34[m]}{1[mile]} }=2896812[m] = 2896.8[km]

Now using the following equation of kinematics, for the avarage velocity  we have:

v=\frac{x}{t} \\where \\v=velocity [m/s]\\t = time [s]\\x=distance traveled [m]\\

therefore:

t=\frac{x}{v} \\t=\frac{2896812}{15}\\ t=193120.8[s]

Now we can convert from seconds into days.

193120.8[s]*\frac{1[hr]}{3600[s]}*\frac{1[day]}{24[hr]}\\  t = 2.2[days]

Now if the truck has the need to come back, the team will spend double time.

t= 4.4 [days]

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4 0
3 years ago
Read 2 more answers
assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th
finlep [7]

Answer:

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

y=y_o+v_osin\theta-\frac{1}{2}gt^2      (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2 (2)

You use the quadratic formula to obtain the value of t:

t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

x_{max}=v_ocos\theta t      

x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}     (3)

By replacing in the equation (3) the values of all parameters you obtain:

y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m

The maximum height reached by the cannon ball is 220.33m

3 0
3 years ago
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