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mamaluj [8]
3 years ago
8

A car cruises at a constant rate of 50 miles per hour. How much time will it take to go 600 miles?

Physics
1 answer:
Len [333]3 years ago
7 0
Speed is constant. 50 miles = 1 hour. 600/50 = 12. 1hr(12) = 12 hours.
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Many particles, such as dust, are inhaled when people breathe. Which part of the respiratory system filters out the majority of
sergey [27]

Answer:

A: Nose

Explanation:

Because your nose has mucus covered hairs that catch some of the dust particles that you inhale.

8 0
3 years ago
In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m a
balu736 [363]

Answer:

The value is  \lambda  = 214.3  \ nm

Explanation:

From the question we are told that

   The  slit separation is  d =  3.00 * 10^{-5} m

    The  distance of the screen is  D =   2.00\ m

    The  order of fringe is  n  =  7

    The path difference is  y =  10.0 \ cm  =  0.1 \  m

    Generally the path difference is mathematically represented as

      y =  \frac{n *  \lambda  *  D}{ d}

=>   0.1 =  \frac{7 *  \lambda  * 2.00 }{ 3.00 * 10^{-5}}

=>  \lambda  =  \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }

=>   \lambda  =  \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }

=>   \lambda  = 2.143 *10^{-7} \  m    

=>    \lambda  = 214.3  \ nm

6 0
3 years ago
Do the math for 30%÷75
Dafna1 [17]

" 30% " means " 0.30 "

30% ÷ 75  means  (0.30 / 75) = <em>0.004</em>

3 0
3 years ago
How do u convert 7.68 cal/sec to kcal/min
stich3 [128]
<span>First you have to convert 7.68 cal/sec to cal/min to do that, just multiply by 60 So it will be 460.8 cal/min now convert that to kcal/min kcal = kilocalories = 1000 calories. so just divide by 1000 and youll have your answer in kcal/min </span>
5 0
4 years ago
Read 2 more answers
After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.
GarryVolchara [31]

Answer:

Imp = 11.666\,\frac{kg\cdot m}{s}

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

(0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m) = \frac{1}{2}\cdot (0.320\,kg)\cdot v_{A}^{2}

v_{A} \approx 19.304\,\frac{m}{s}

After collision:

\frac{1}{2}\cdot (0.320\,kg)\cdot v_{B}^{2} = (0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (15\,m)

v_{B} \approx 17.153\,\frac{m}{s}

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

Imp = (0.32\,kg)\cdot [(17.153\,\frac{m}{s} )-(-19.304\,\frac{m}{s} )]

Imp = 11.666\,\frac{kg\cdot m}{s}

5 0
3 years ago
Read 2 more answers
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