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forsale [732]
3 years ago
5

Due Sun 06/09/2019 11:59 pm Skip Navigation Questions correct Q 1 [1/1] correct Q 2 [1/1] correct Q 3 [1/1] untried Q 4 (0/1) un

tried Q 5 (0/1) untried Q 6 (0/1) untried Q 7 (0/1) untried Q 8 (0/1) untried Q 9 (0/1) untried Q 10 (0/1) Grade: 3/10 Print Version Start of Questions Two cyclists, 42 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?
Physics
1 answer:
astraxan [27]3 years ago
5 0

Due Sun 06/09/2019 11:59 pm <u><em>(you're already more than a week late)</em></u> Skip Navigation Questions, correct Q 1 [1/1], correct Q 2 [1/1], correct Q 3 [1/1], untried Q 4 (0/1), untried Q 5 (0/1), untried Q 6 (0/1), untried Q 7 (0/1), untried Q 8 (0/1), untried Q 9 (0/1), untried Q 10 (0/1), Grade: 3/10, Print Version, <u><em>Start of Questions</em></u>:

Questions Two cyclists, 42 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

<u><em>Start of Answer:</em></u>

-- When they started, they were 42 miles apart.  When they met, 2 hours later, they were no miles apart.  The distance between them shrank at the rate of 21 miles per hour.  Since they rode directly toward each other, the sum of their individual speeds must have been 21 miles per hour.

-- One biker was two times the speed of the other.  Slower biker: 1 time.  Faster biker: 2 times.  Sum of their speeds:  3 times = 21 miles per hour.  Each 'time' = 7 miles per hour.

-- Slower cycler = 1 time = 7 mi/hr

<em>Faster cycler = 2 times = 14 mi/hr</em>

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Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.
fredd [130]

Energy to lift something =

               (mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm  =  5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

            =  2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

7 0
2 years ago
A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if
diamong [38]

Answer:

I believe that the answer is d.

Explanation:

Because there is nothing to make the aircraft accelerate or decelerate, it is going to stay in constant motion with no acceleration.

3 0
3 years ago
A light ray incident from medium 1 to medium 2, where n1&gt;n2. When the incident angle exceed the critical angle ac, the refrac
vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence

{\theta_r} is the angle of refraction = 90°

{n_2} is the refractive index of the refraction medium

{n_1} is the refractive index of the incidence medium

Thus,

n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}

The formula for the calculation of critical angle is:

{sin\theta_{critical}}=\frac {n_2}{n_1}

Where,  

{\theta_{critical}} is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0
3 years ago
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

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For isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

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The density of the flow at the exit is 2.2721 kg/m³

4 0
2 years ago
Student pushes a 50 N block across the floor for a distance of 15 m how much work was done to move the block
Talja [164]

Answer:

750 J

Explanation:

We have a student that pushes a 50N block  across the floor for a distance of 15m. The question is asking how much work was done to move the block.

To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS

F being the force and S being the distance.

W = FS

W = (50)(15)

W = 750

Therefore, 750 joules is our answer.

7 0
3 years ago
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