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3 years ago

11

Three taxi cabs make a complete trip from downtown to the airport and back in 15 , 33 and 55 minutes, respectively. If all three

cabs leave at the same time, what is the shortest time that must pass before they are all together again?

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:

165 minutes

Explanation:

We can break the numbers into prime numbers

15 = 3 * 5

33 = 3 * 11

55 = 5 * 11

The shortest time that must pass before they are all together again would be the least multiple that is divisible by all 3 number, which is

3 * 5 * 11 = 165

Hence after 165 minutes the 3 cars should all meet together again.

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Minchanka [31]

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3 years ago

You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes. That is half of your blood

Zina [86]

Answer:

8 times

Explanation:

Given that You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes.

That means the heart will pump 10 quarts in 2 minutes.

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3 years ago

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu

Gemiola [76]

Answer:

$\displaystyle w=3.478\ rad/sec$

$M=0.0182\ J$

$v=0.398\ m/s$

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

$\displaystyle w=\sqrt{\frac{g}{L}}$

We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

And

$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

The potential energy is

$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

The mechanical energy is, then

$M=K+U=0+U=U$

$M=0.0182\ J$

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

$\displaystyle K=\frac{mv^2}{2}$

Equating to the mechanical energy of the system (M)

$\displaystyle \frac{mv^2}{2}=0.0182$

Solving for v

$\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}$

$v=0.398\ m/s$

4 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago