A weather pattern is defined as a period of time when the weather remains consistent. In the lab, a lot of observation about weather is obtained
A weather pattern is defined as a period of time when the weatherremains consistent. Weather changes are crucial to humanexistence.
because they influence our everyday activities and provide moisture for crops.
The rain does not always end within the day, and gloomy days might last just as long as sunny days. Tornadoes and hurricanes, for example, may inflict tremendous damage.
In the lab the following observation about weather is obtained;
1. We will find the graphs and statistics that indicate signs of climate change and engage with an interactivegraphic.
2. You'll also look at and debate maps of global temperature and precipitation patterns that are changing.
3. This lab will teach you about Earth's biomes and the close relationship that exists between them and the climates that serve to define them.
To learn more about the weatherpattern refer to the link;
The question pertains to meteorology, climatology, and atmospheric science. These are disciplines that study weather and climate, respectively, and their effects on the planet. Atmospheric Science is a broad field that includes both and employs physics principles.
The question refers to the subjects of meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric impacts on the Earth's weather. It involves the prediction of weather in the short term based on thousands of measurements of variables such as air pressure and temperature.
Climatology, on the other hand, is the study of climate, which involves analyzing averaged weather conditions over longer time periods using atmospheric data. Unlike meteorologists, climatologists focus on patterns and effects that occur over longer timescales of decades, centuries, and millennia.
Atmospheric Science is a broad field that encompasses both meteorology and climatology, as well as other disciplines that study the atmosphere. This discipline is typically based heavily on physics and involves the study of weather and climate patterns, predictions of developments in weather and climatic events, and the analysis of the effects of these events on the planet and its inhabitants.
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b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
Answer:
μ = 0.423
Explanation:
To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body
Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis
x = v₀ t + ½ a t²
The body starts from rest so its initial speed is zero
a = 2 x / t²
a = 2 0.5 /0.5²
a = 4 m / s²
Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive
X axis
fr - Wₓ = m a (1)
Y Axis
N- = 0
N = W_{y}
We use trigonometry to find the components of the weight
sin 45 = Wₓ / W
cos 45 = W_{y} / W
Wₓ = W sin 45
W_{y} = W cos 45
The out of touch has the expression
fr = μ N
fr = μ W_{y}
We substitute in 1
μ mg cos 45 - mg sin 45 = m a
W_{y} = (a + g sin 45) / g cos 45
μ = a / g cos 45 + 1
We calculate
Acceleration goes down the plane, so it is negative
a = -4 m / s²
μ = 1- 4 / (9.8 cos 45)
μ = 0.423
Answer:
The μ = 0.422
Explanation:
The distance travelled by the mass is equal to:
The sum of forces in y-direction equals zero:
∑Fy = 0
N - (m * g * cosθ) = 0
N - (1 * 9.8 * cos45) = 0
N = 6.93 N
The sum of forces in x-direction is equal to:
∑Fx = ma
(m * g * sinθ) - fk = m * a
(1 * 9.8 * sin45) - fk = 1 * 4
fk = 2.93 N
fk = μ * N
2.93 = μ * 6.93
μ = 0.422
Explanation:
The relation between resistance and resistivity is given by :
is resistivity of material
l is length of wire
A is area of cross section of wire
Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is
To solve this problem it is necessary to apply to the concepts related to energy conservation. For this purpose we will consider potential energy and kinetic energy as the energies linked to the body. The final kinetic energy is null since everything is converted into potential energy, therefore
Potential Energy can be defined as,
Kinetic Energy can be defined as,
Now for Conservation of Energy,
Therefore the highets position the car reaches above the bottom of the hill is 40.02m
Answer:
Explanation:
A charge within an electric field E experiences a force proportional to the field whose module is F = qE, whose direction is the same, if the charge is negative, it experiences a force in the opposite direction to the field and if the charge is positive, experience a force in the same direction of the field.
In our case we are interested in the magnitude of the force, therefore the sign of the charge has no relevance
You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final² = (V)initial²+2a(dx)
Given:
a=0.500m/s²
dx=6.32 m
(V)intial=0m
(V)final= UNKNOWN
(V)final= 2.51396m/s