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lana [24]
3 years ago
5

Are you ready for a brain twister? murphy can row at 5 kmph in still water. if the velocity of current is 1 kmph and it takes hi

m 1 hour to row to a place and come back, how far is the place?
Physics
1 answer:
ipn [44]3 years ago
7 0
Answer: 2.4km

In this case, Murphy is <span>rowing to a place and come back so he was going 2x of the place distance. That means when he goes to the place the current velocity is helping him(5+1 kmph) and when he comes back the current velocity is slowing him(5-1kmph).
To calculate the place distance, the calculation would be:
X/6kmph + X/4kmph= 1 hour
</span>X/6kmph + 1.5X/6kmph= 1 hour
<span>2.5X/6kmph= 1hour
2.5X= 1 hour x 6km/ hour
X= 6km/2.5
X= 2.4km
</span>
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What is the net force?(please don't forget the direction and magnitude)​
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Answer: In mechanics, the net force is the vector sum of forces acting on a particle or object. The net force is a single force that replaces the effect of the original forces on the particle's motion.

Explanation: Hope dis helps u

8 0
3 years ago
Suppose an endothermic reaction has a positive change in entropy greater than the heat absorbed. What can be said about the spon
pentagon [3]
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4 years ago
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A 1200 kg car is put into neutral and pushed by 2 students. The first student pushed the car forward with 200 N of force. The se
sukhopar [10]

Explanation:

∑F = ma

200 N + 250 N = (1200 kg) a

a = 0.375 m/s²

3 0
3 years ago
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Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
MariettaO [177]

Answer:

The separation between the two lowest levels =  1.24 * 10^{-39}J

The values of n where the energy of molecule reaches 1/2 kT at 300K = 2.2 * 10^{9}

The separation at this level = 1.8 * 10^{-30}J

Explanation:

Knowing the formula

En = \frac{n^{2} h^{2}  }{8 mL^{2} }

Mass of oxygen molecule

m (O2) = 32 amu * \frac{1.6605 * 10^{-27 kg} }{1 amu}

So the energy diference between the two lowest levels:

E2 - E1 = \frac{3h^{2} }{8mL^{2} }

E2 - E1 =  \frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2}   } = 1.24 * 10^{-39}J

Now we should find n where the energy of molecule reaches 1/2 kT

En = \frac{n^{2} h^{2}  }{8 mL^{2} } = \frac{1}{2}kT

\frac{h^{2}  }{8 mL^{2} } = 4.13 * 10^{-14}J

n^{2} *  (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K

n = 2.2 * 10^{9}

by the end is necessary to calculate the separation of the level

En - En-1 = (n^{2} - (n - 1)^{2}) * \frac{h^{2}  }{8 mL^{2} }

              = 1.8 * 10^{-30}J

4 0
3 years ago
If a beam of white light in air strikes a sheet of this glass at 66.0 ∘∘ with the normal in air, what will be the angle of dispe
Hunter-Best [27]

Answer:

The angle between the extreme red and extreme violet light in the glass is 1.18°

Explanation:

According the Snell's law:

n_{a} sin\theta _{a} =n_{b} sin\theta _{b}\\\theta _{b} =sin^{-1}  (\frac{n_{a}sin\theta  a}{n_{b} } )

Where

na = refractive index of air = 1

θa = angle of incidence in air = 66°

For red light nb = 1.61

Replacing:

\theta _{b} =sin^{-1} (\frac{1*sin66}{1.61} )=34.57

For violet light nb = 1.66

\theta _{b} =sin^{-1} (\frac{1*sin66}{1.66} )=33.39

So, the angle between the extreme red and extreme violet light in the glass is 34.57-33.39 = 1.18°

4 0
3 years ago
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