Answer:
A particle that has gained: Anion
A particle that has lost: Cation
Explanation:
![{}](https://tex.z-dn.net/?f=%7B%7D)
Answer:
class
Explanation:
An examination of a tool mark is essential when it comes to "Criminalistics" <em>(the study of physical evidence in a crime). </em>It allows the forensic investigator to know what kind of tool was used in the crime scene. There are only two kinds of characteristics of tool marks and these are: individual and class. <em>Striated and molded are types of tool marks. </em>They are not characteristics.
The "class" characteristic of a tool mark allows the investigator to know the <u>shape of tool, its dimensions and its type</u>. However, <em>it doesn't point to the exact tool that was used, </em>but it gives him an idea how the mark was actually created.
Answer:
A) 17.7 m/s
B) 15.98 m
C) Zero
E) 9.8 m/s²
Explanation:
given information
distance, h = - 34 m
time, t = 5 s
A) What is the initial speed of the egg?
h - h₀ = v₀t -
t², h₀ = 0
- 34 = v₀ 5 - \frac{1}{2} 9.8 5²
- 34 = 5 v₀ - 122.5
v₀ = 122.5 - 34/5
= 17.7 m/s
B) How high does it rise above its starting point?
v² = v₀² - 2gh
v = 0 (highest point)
2gh = v₀²
h = v₀²/2g
= 17.7²/2 (9.8)
= 15.98 m
C) What is the magnitude of its velocity at the highest point?
v = 0 (at highest point)
E) What are the magnitude and direction of its acceleration at the highest point?
g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.
Answer:
0.136 C
Explanation:
This question uses the simple idea of conservation of energy. We work with the assumption that energy loss is zero
At the top of the waterfall the water has maximum potential energy
![P_{e} = mgh](https://tex.z-dn.net/?f=P_%7Be%7D%20%20%3D%20mgh)
Where
= Mass of water
= Acceleration due to gravity
= Height
Now at the bottom just before impact all of the potential energy is converted to kinetic energy and right after impact all this kinetic energy is converted to thermal energy (
)
![T_{e} = mc_{w} (dT)](https://tex.z-dn.net/?f=T_%7Be%7D%20%3D%20mc_%7Bw%7D%20%28dT%29)
Where
= Mass of water
= Specific Heat Capacity of Water
= Change in Temperature
Equating the two equations
![P_{e} = T_{e} \\ mgh = mc_{w}(dT)\\ gh = c_{w}(dT)\\ dT = \frac{gh}{c_{w}}](https://tex.z-dn.net/?f=P_%7Be%7D%20%20%3D%20T_%7Be%7D%20%5C%5C%20mgh%20%3D%20%20mc_%7Bw%7D%28dT%29%5C%5C%20gh%20%3D%20c_%7Bw%7D%28dT%29%5C%5C%20dT%20%3D%20%5Cfrac%7Bgh%7D%7Bc_%7Bw%7D%7D)
Since the value of
is a constant, it is equal to 4186 J / (kg * C). We input all the given values into the equation and find the answer
![dT = \frac{(9.81)(58)}{4186}}\\ dT = 0.136 C](https://tex.z-dn.net/?f=dT%20%3D%20%5Cfrac%7B%289.81%29%2858%29%7D%7B4186%7D%7D%5C%5C%20dT%20%3D%200.136%20C)
Answer:
7650 m.
Explanation:
Ocean floor depth, d is:
d = v * t,
where,
d = the distance from the vessel to the ocean floor (or the depth)
v = 1530 m/s = velocity of the ultrasonic sound
t = t_echo/2 = time that the ultrasonic sound needs to reach the ocean floor
t_echo = 10 s = time that the ultrasonic sound needs to reach the ocean floor and return back to the vessel.
d = v * t
= v * t_echo/2
= 1530 * 10/2
= 7650 m.