Explanation:
Visible light communication (VLC) is a wireless method that uses light emitted by LEDs to deliver networked, mobile, high-speed communication similar to Wi-Fi, leading to the term Li-Fi. It can be used as standalone solution or in a supplementary role to radio-frequency (RF) or cellular network communication.
Inside a car that was at 273 K, a bottle with a pressure at 100,000 pascals warms up to 350 K. If the volume of the bottle remains constant, what is the pressure, in pascals, inside the hot water bottle!
Answer:
128.205.1
Explanation:
What is the speed in kilometers per hour of a 27.78m/s baseball pitch?
Answer:
I actually don't know
Explanation:
ok chillll
Two teams are competing in a tug-of-war contest. Team A is pulling at 4000N and Team B is pulling at 4900N is the opposite direction.
Answer:
There is 8900N of force on the rope and team B wins
Explanation:
A car drives 0.696 km North, and then turns and drives 4.326 km East. Calculate the magnitude of the car's resultant displacement. Include equation, substitution with units, and answer with units.
Answer:
4.3186km
Explanation:
If a car drives 0.696 km North, and then turns and drives 4.326 km East, the magnitude of the car's resultant displacement will be gotten using the Pythagoras theorem as shown;
d² = x²+y²
Given;
x = 4.326km
y = 0.696km
Substitute into the formula and get d;
d² = (4.326)²+(0.696)²
d² = 18.714276+0.484416
d² =19.198692
d = √19.198692
d = 4.3816km
Hence the magnitude of the car's resultant displacement is 4.3186km
When the car was stopped by the tree, its change in velocity during the collision was -6 meters/second. This change in velocity occurred in 2 seconds. What was the acceleration of the car?
how is an object's motion related to speed, velocity, and acceleration?
Answer:
Speed describes only how fast an object is moving, whereas velocity gives both the speed and direction of the object’s motion being a vector quantity. The standard unit to measure velocity is meter per second (m/s or ms-1). For example, “10 meters per second” is scalar and “10 meters per second West” is vector. Acceleration
In a container there are 2 liters of liquid, of which 75% is alcohol and the remaining 25% is water.
Determine how many cm3 of alcohol must be added to bring the alcohol percentage to 80%.
Answer:
There must be added 500 cm3 of alcohol to bring the concentration to 80%.
Explanation:
Concentration
The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solution.
In the container, there are 2 liters (2000 cubic centimeters) of a solution, 75% of which is alcohol (solute) and 25% of water (solvent).
The original amount of alcohol is:
H = 2000*75% = 1500 cc
The original amount of water is:
W = 2000*25% = 500 cc
When we add x cc of pure alcohol, there are 1500+x cc of alcohol out of 2000+x cc of solution.
The new concentration is calculated as:
[tex]\displaystyle \frac{1500+x}{2000+x}[/tex]
And it's known this concentration is 80% (0.8), thus:
[tex]\displaystyle \frac{1500+x}{2000+x}=0.8[/tex]
Multiplying by 2000+x:
[tex]1500+x = 0.8(2000+x)[/tex]
Operating:
[tex]1500+x = 1600+0.8x[/tex]
Rearranging:
[tex]x-0.8x=1600-1500[/tex]
[tex]0.2x=100[/tex]
[tex]x=100/0.2[/tex]
x= 500
There must be added 500 cm3 of alcohol to bring the concentration to 80%
Density Questions
How many blocks of ice-cream each 10cm x 10 cm x 4cm can be stored in the compartenent of a deep-freeze measuring 40 cm x 40cm x 20cm?
Explanation:
let's take the volume of each first:
ice cream: 10×10×4=400cm^3
freezer: 40×40×20=32,000cm^3
now we just divide: 32000/400=80
Which of the following equations is balanced?
a. 2Al2O3 -----> 3Al + 3O2
b. 3Al2O3 -----> 3Al + 2O2
c. 2Al2O3 -----> 4Al + 3O2
d. 2Al2O3 -----> Al + 3O2
Answer: c
Explanation:
The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.
a. 2×2 atoms of Al ≠ 3×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!
b. 3×2 atoms of Al ≠ 3×1 atoms of Al
3×3 atoms of O ≠ 2×2 atoms of O
c. 2×2 atoms of Al = 4×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH ARE EQUAL, CORRECT ANSWER!!!
d. 2×2 atoms of Al ≠ 1×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
How do I graph distance ,speed ,velocity, and acceleration?
Answer:
distance is how much space is between, speed is how fast they are going, velocity is money and acceleration is going
Explanation:
5. Explain how forming an ionic bond makes metal and nonmetal atoms more stable.
6. Describe how metallic bonds contribute to the physical characteristics of a metal.
7. Describe the type of ions created through metallic bonding.
8. List the three types of atomic bonds and describe the types of atoms-metal
or nonmetal-that form each type of bond.
A baseball is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stop watch measures the stone's time in the air until it hits the ground to be 4.3 s. How far does the baseball travel horizontally?
Complete Question
A baseball is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stop watch measures the ball's time in the air until it hits the ground to be 4.3 s. How far does the baseball travel horizontally?
Answer:
The distance covered horizontally is [tex]d = 43 \ m[/tex]
Explanation:
From the question we are told that
The initial horizontal velocity is [tex]u = 10 \ m/s[/tex]
The time taken by the ball in the air is [tex]t = 4.3 \ s[/tex]
Generally the distance traveled by the ball is
[tex]d = u * t[/tex]
=> [tex]d = 10 * 4.3[/tex]
=> [tex]d = 43 \ m[/tex]
A box sitting still on the ground by itself has a Normal Force of 700N, what is the mass? (gravity’s acceleration is 9.80 m/s²)
a baby carriage is sitting on the top of a hill that is 21 m high. the carriage with a baby 75kg. the carriage has what energy? calculate it
Answer:
It has potential energy
PE = 15450.75 J
Explanation:
we know its potential energy because the carriage is not moving, although it can
PE = mgh
in this case, m is 75, h is 21, and g is always 9.81
plug these in: PE = (75) (21) (9.81)
your answer: PE = 15450.75 J
Under what conditions can an object have forces acting on it, but its motion does not change ?
Answer: An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion.
Explanation: I hoped that helped!!
The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this impressive position is traveling with an initial velocity of 8.0 m/s before accelerating. Then the rider travels 40.0 m at a constant acceleration of 2.00 m/s2. What is the rider's velocity after the acceleration?
Answer:
14.97m/sExplanation:
Given
Initial velocity of the car u = 8m/s
Distance travelled by the rider S = 40m
Acceleration a = 2m/s²
Required
rider's velocity after the acceleration v
Using the equation of motion
v² = u²+2as
v² = 8²+2(2)(40)
v² = 64+160
v² = 224
v = √224
v = 14.97m/s
Hence the rider's velocity after the acceleration is 14.97m/s
A 200 N net force acts on a 50 kg box. What is the acceleration of the box?
Answer:
a= 4m/s^2
Explanation:
Fnet =200N
m=50kg
a=?
Fnet = ma
200=50a then divide both sides by 50 to get acceleration
Taking into account the Newton's second law, the acceleration of the box is 4 [tex]\frac{m}{s^{2} }[/tex].
In first place, you have to know that Newton's second law, called the fundamental law or fundamental principle of dynamics, states that a body accelerates if a force is applied to it.
Then, the amount of force necessary is proportional to the acceleration that the body acquires, and the constant of proportionality is the mass of the body.
Mathematically, Newton's second law is:
F= m×a
where:
F = Force [N] m = Mass [kg] a = Acceleration [m/s²]In this case, you know:
F= 200 N m= 50 kg a= ?Replacing in the Newton's second law:
200 N= 50 kg×a
Solving:
a=200 N÷ 50 kg
a= 4 [tex]\frac{m}{s^{2} }[/tex]
In summary, the acceleration of the box is 4 [tex]\frac{m}{s^{2} }[/tex].
Learn more about the Newton's second law:
brainly.com/question/23845187?referrer=searchResults brainly.com/question/13959891?referrer=searchResultsGOOD MFRNING
PAGE NO.:
DATE:
cap a b
body have larger mess be holter than another of smaller
maws if they are same temperature why,
at the
Answer:
umm im confused
Explanation:
Answer:
larger mass more the molecule more heat absorbed
and vice versa .Thats the reason ok?
in a free body diagram the weight is
Answer:
Explanation:
a. WEIGHT: As we have seen, weight is the gravitational force exerted on an object by the Earth (or any other celestial body). If an object is near the Earth's surface and has mass, then the object has a weight. The magnitude of its weight is w = mg and its direction is toward the center of the Earth.
what would be the velocity
when a dog of 10kg and it's kinetic energy is 20J
Answer:
2 m/s
Explanation:
An object is projected straight upward with an initial speed of 25.0 m/s. What's the maximum height obtained by the object ?
Answer:
The maximum height obtained by the object is approximately 31.855 m
Explanation:
The vertical velocity of the object = 25.0 m/s
The height reached by the object, is given by the following formula;
v² = u² - 2 × g × h
Where;
u = The initial velocity of the object = 25.0 m/s
v = The final speed of the object = 0 m/s at maximum height
h = The maximum height obtained by the object
g = The acceleration due to gravity = 9.81 m/s²
Substituting the values, gives;
0² = (25.0 m/s)² - 2 × 9.81 m/s² × h
2 × 9.81 m/s² × h = (25.0 m/s)²
h = (25.0 m/s)²/(2 × 9.81 m/s²) ≈ 31.855 m
The maximum height obtained by the object, h ≈ 31.855 m.
Due to the friction between a tire and the road surface, the temperature of the air inside a tire increases as one drives. If the tire pressure is 37 psi when the car is in the driveway at 25° C, What is the pressure of the tire when the temperature increases to 50° C?
Answer:
40.10 atm .
Explanation:
Here volume of tire is assumed to be nearly constant .
Modified gas law ,
P₁ / T₁ = P₂ / T₂
37 / ( 273 + 25 ) = P₂ / ( 273 + 50)
37 / 298 = P₂ / 323
P₂ = 37 x 323 / 298
= 40.10 atm .
Three vectors , , and , each have a magnitude of 52.0 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 29.0 ˚, 191 ˚, and 311 ˚, respectively. What are (a) the magnitude and (b) the angle of the vector (relative to the +x direction in the range of (-180°, 180°)), and (c) the magnitude and (d) the angle of in the range of (-180°, 180°)? What are (e) the magnitude and (f) the angle (in the range of (-180°, 180°)) of a fourth vector such that ?
Answer:
a) A´= A
b) θ₁´ = 29º, θ₂´ = - 169º , θ₃´ = -49º
Explanation:
In this exercise you are asked to give the magnitudes and angles of the vectors from another system of
reference
a) The magnitudes
The magnitude of a vector, the size of which is a scalar, this does not depend on the reference system, since it is obtained by subtracting the coordinates of the end point minus the coordinate of the origin of the vector
A = [tex]x_{f}[/tex]- x₀
if the vectors are measured in another reference frame
x_{f}´ = xx_{f}- U
x₀´ = x₀ -U
where U is the distance between the two reference frames
A´ = x_{f}´ - x₀´
we substitute
A´ = (x_{f} - U) - (x₀-U) = x_{f} - x₀
A´ = A
it does not change
b) Angles
The given angles are measured from the positive part of the x axis in a counterclockwise direction, it is asked to give these angles from the x axis
θ₁ = 29º
does not change
θ₁´ = 29º
θ₂ = 191º
we measure clockwise
θ₂´ = θ₂ - 360
θ₂´ = 191 - 360
θ₂´ = - 169º
θ₃ =311º
we measure clockwise
θ₃´ = 311 -360
θ₃´ = -49º
A 2,100kg car drives toward a 55kg cart that has a velocity of 0.50m/s west. The two objects collide giving the car a final velocity of 6.31m/s east and the shopping cart a velocity of 13.5m/s east what the initial velocity of the car?
A. 6.68m/s east
B. 7.19m/s west
C. 6.68m/s west
D. 7.19m/s east
Answer: A. 6.68m/s east
Explanation: just did it on ap3x
A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?
Answer:
Workdone = 465766038 Joules.
Explanation:
Given the following data;
Mass = 1167
Initial velocity = 10m/s
Final velocity =28m/s
To find the workdone;
We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.
Mathematically, it is given by the equation;
W = Kf - Ki
W = ½MVf² - ½MVi²
Substituting into the equation
W = ½(1167)*28² - ½(1167)*10²
W = ½ * 1361889* 784 - ½ * 1361889 * 100
W = 533860488 - 68094450
Workdone = 465766038 Joules.
An object with a mass of 43 kg measured on Earth is taken to the Moon. What is the mass of the object on the Moon's surface if the acceleration due to gravity on the Moon is one-sixth of that on Earth? O a. 16 kg O b. 43 kg O c. 7.9 kg O d. 26 kg
Answer:
b 43 kg
Explanation:
mass doesn't differ from place to place.
but the weight may
What would be the y-component of a velocity vector that had a
magnitude of 37m/s in a direction 30° from the x-axis?
Answer:
18.5 m/s
Explanation:
Just did it on EdPuzzle
The y-component of a velocity vector that had a magnitude of 37 m / s in a direction 30° from the x - axis would be 18.5 m / s .
What is a vector quantity?The quantities that contain the magnitude of the quantities along with the direction are known as the vector quantities.
Examples of vector quantities are displacement, velocity acceleration, force, etc.
As given in the problem we have to find out the y-component of a velocity vector that had a magnitude of 37 m / s in a direction 30° from the x - axis.
The y-component of a velocity vector = 37 × sin 30
= 18.5 m / s
Thus , y-component of a velocity vector that had a magnitude of 37 m / s in a direction 30° from the x - axis would be 18.5 m / s.
To learn more about the vector quantity here , refer to the link given below ;
https://brainly.com/question/15516363
#SPJ2
Someone help me plz. I have no idea.
Answer:
fault im pretty sure
Explanation:
Answer:
fault............
...........
En un año se otorgará uno de tres premios (a la investigación, la enseñanza y el servicio) a algunos de los estudiantes, de un grupo de 25, de posgrado del departamento de estadística. Si cada estudiante puede recibir un premio como máximo, ¿cuántas selecciones posibles habría?
Answer:
Habría 13,800 selecciones posibles.
Explanation:
Permutar es colocar elementos en distintas posiciones.
Se llama permutación de m elementos a cada una de las diferentes ordenaciones que se pueden hacer con esos elementos. En otras palabras, se llama permutaciones de m elementos en n posiciones a las distintas formas en que se pueden ordenar los m elementos ocupando únicamente las n posiciones.
En el caso de la permutación no se repiten los elementos, ya que de repetirse o ser iguales entre si, al intercambiarlos no se genera una nueva permutación.
El número de permutaciones de n elementos elegidos k a la vez es calculado como:
[tex]P=\frac{n!}{(n-k)!}[/tex]
Como los premios son distinguibles, la permutación es usada en este caso, donde n=25 y k=3
Entonces:
[tex]P=\frac{25!}{(25-3)!}[/tex]
Resolviendo:
[tex]P=\frac{25!}{(22)!}[/tex]
Recordando que el factorial es una operación matemática que indica que se deben multiplicar todos los números enteros y positivos que hay entre el número al cual deseas calcular el factorial y el número 1, entonces se obtiene:
P= 13,800
Habría 13,800 selecciones posibles.
La selección de los alumnos se puede realizar de 2300 formas.
Teniendo en cuenta eso, tenemos 25 estudiantes graduados para ser seleccionados para tres tipos diferentes de premios, como pabellones de investigación, docencia y servicio.
Ahora;
25C3 = 25! / 3! (25 - 3)!
25C3 = 25! / 3! 22!
25C3 = 2300
La selección de los alumnos se puede realizar de 2300 formas.
Más información: https://brainly.com/question/13387529
if you had to be an animal, which would you be and why?