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3 years ago

When may it be a good idea to use scientific notation or Si prefixes?

1 answer:

6 0

When you're dealing with quantities that produce very large or very small numbers in the units that you're using; like meters to Neptune, or kilograms in one Hydrogen atom.

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The learning about Universe is called ________ ...

ASHA 777 [7]

Answer:

The learning about Universe is called

Astronomy.

Explanation:

<h2>Hope it Helps you!! </h2>

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3 years ago

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How to test the pH level of a soap?

olchik [2.2K]

Just test the soap book

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3 years ago

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If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
η = 1 - Tc / Th = (Th - Tc) / Th 
where 
Tc is the absolute temperature of the cold reservoir, and 
Th is the absolute temperature of the hot reservoir. 

In this case, given is η=22% and Th - Tc = 75K 
Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). 

Th = (Th - Tc) / η 
Th = 75 / 0.22 = 341 K (rounded to closest number) 
Tc = Th - 75 = 266 K 

Lower temperature is Tc = 266 K 
Higher temperature is Th = 341 K

6 0

3 years ago

Help me do this question

Zepler [3.9K]

Answer:

c20800

Explanation:

go to bear khana bro your book looks cheap go and study in durbar kanda school and take somee cash and do ash ah boy

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3 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

5 0

3 years ago