Explanation:
A four-dimensional space is a mathematical extension of the concept of three-dimensional or 3D space. Three-dimensional space is the simplest possible abstraction of the observation that one only needs three numbers, called dimensions, to describe the sizes or locations of objects in the everyday world.
According to the special theory of relativity, the fourth dimension is time. The special theory of relativity, proposed by Albert Einstein in 1905, revolutionized our understanding of space and time.
It states that space and time are interconnected and form a four-dimensional framework known as spacetime. In this framework, three dimensions correspond to space (length, width, and height), and the fourth dimension is time. Einstein's theory introduced the concept of time dilation, which means that time can appear to pass differently for observers in relative motion. This idea challenges the notion of time as an absolute quantity and demonstrates that it is influenced by an observer's relative motion and gravitational fields.
To visualize the fourth dimension, consider a space-time diagram where an object's position in space is plotted against time. Each point on the diagram represents an event, combining its spatial coordinates with the time at which it occurred. This four-dimensional representation allows us to understand the relationship between space and time in a unified framework.
Learn more about Special Theory of Relativity here:
https://brainly.com/question/14140033
#SPJ2
True or False: A higher vapor pressure (evaporates easily) corresponds to strong intermolecular forces. (dont answer)
anyways free p o i n ts...
Answer:
truuuuuuuuuuuuuuuuuuuuuuuuujuuuuuuuur
Answer:
true
Explanation:
I took a test this year and I know the answer
which feature is used to classify galaxies
age
color
shape
size
Answer:Shape
I took the quiz
Answer:shape
Explanation:
I took the quiz
The value of gravitational pull on the moon is 1.6 m/s/s. What is the weight of a 75 kg
astronaut on the moon?
Answer:
120N
Explanation:
Weight=mass x acceleration due to gravity (on the planet)
Mass of an astronaut is 75kg and value of acceleration due to gravity on moon is 1.6m/s2
Weight of astronaut on moon=mass of astronaut x value of acceleration due to gravity on moon
75kg x 1.6m/s2
=120N
what instrument used measure
to volume of water
Answer: A Graduated Cylinder or A Beaker ?
Explanation:
A juggler is throwing balls up in the air and catching them. An observer in the crowd makes the statement that the potential and kinetic energy of the ball is the same when the ball is halfway down from its highest point. In one paragraph, using your own words, explain using the relationship of mechanical energy, potential energy, and kinetic energy why the observer is correct.
Answer:
do you have the answer?
Explanation:
if so whats the answer?
An object with a mass of 1298 kg is accelerating at a rate of 17 m/s² when a net force is applied to it. What is the magnitude of the net force that is acting on the object?
Answer:
Explanation:
This is a simple F = ma equation. We have the mass and we have the acceleration, so
F = 1298(17) and
F = 22066 which rounds correctly to
F = 22000N
The magnitude of the net force of object with a mass of 1298 kg is accelerating at a rate of 17 m/s² is 22066N
HOW TO CALCULATE NET FORCE:
The net force of a body can be calculated by multiplying the mass of the body by its acceleration. That is;Force (N) = mass (kg) × acceleration (m/s²)According to this question, an object with a mass of 1298 kg is accelerating at a rate of 17 m/s². The net force can be calculated as follows:F = 1298Kg × 17m/s²F = 22066NTherefore, the magnitude of the net force of object with a mass of 1298 kg is accelerating at a rate of 17 m/s² is 22066N.Learn more at: https://brainly.com/question/23245710?referrer=searchResults
What helps night vision? Answer fast please.
Answer:
Halos, blurriness, and night blindness causes night vision (symptoms) while vitamin-A,corrective lenses, such as eyeglasses or contacts, can improve or help night vision.
Explanation:
I bet you will get full Marks if you write it!!
(please note that you will have to mark me as brainliest if it helped you)£
NEED HELP The main difference between kinetic energy and potential energy is thatSingle choice.
(2 Points)
kinetic energy involves position, and potential energy involves motion.
kinetic energy involves motion, and potential energy involves position.
although both energies involve motion, only kinetic energy involves position.
although both energies involve position, only potential energy involves motion.
Answer:
kinetic energy involves motion, and potential energy involves position.although both energies involve position, only potential energy involves motion.hope it is helpful to you ☺️☺️✌️
Elements in Group
are common alkaline earth metals.
Answer here
A worker leaves home at 9:00 AM, travels 40 km to the office, and then returns home at 5:00 PM. What is the magnitude of the worker's displacement during this 8-hour span of time?
Answer:
The answer is 0km
Explanation:
An electric pencil sharpener is powered by 6 volts. The current through the pencil sharpener is 0.5 amps. What's is the resistance?
Answer:
Resistance in pencil sharpener = 12 ohms
Explanation:
Given:
Voltage in pencil sharpener = 6 volts
Current in pencil sharpener = 0.5 amps
Find:
Resistance in pencil sharpener
Computation:
Resistance = Voltage / Current
Resistance in pencil sharpener = Voltage in pencil sharpener / Current in pencil sharpener
Resistance in pencil sharpener = 6 / 0.5
Resistance in pencil sharpener = 12 ohms
Please please help me :)
Answer:
Explanation:
2) From F=ma
Force =15×40=600N or kgm/s2
3)From the same equation making acceleration the subject of the formula will give
a=f÷m
=24÷4=6m/s2
4)m=f÷a
=45÷15=3kg
How are wave properties and energy related?
give me examples
Answer:
They don’t ‘represent’ anything, they are properties of the wave.
Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.
If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.
For many waves, we don’t perceive them at all (e.g. radio waves).
For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.
Define Acceleration.
Ty!
Answer:
Acceleration is the name we give to any process where the velocity changes. Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both.
Susan gently pushes the tip of her finger against the eraser on her pencil and the pencil does not move. Which of the
following figures best illustrates the interaction of forces between Susan's finger and her pencil?
Answer: D. ➡️⬅️
Explanation: I just knew the answer ;)
Please select the word from the list that best fits the definition
Does not easily transfer electricity
conductor
insulator
circuit
current
Answer:
conductor
Does not easily transfer electricity
Two vehicles A and B accelerate uniformly from rest.
Vehicle A attains a maximum velocity of 30ms - in los
while B attains a maximum velocil) ol 40ms in the same
time. Both vehicles maintain these velocities for 6s belore
they are decelerated to rest in 6s and 4s respectively
Sketch on the same axes, velocity time graphs
for the motion of the vehicles
Calculate the velocity of each vehicle 18s aftur
the start. (VA
= 20ms -land vs
and vs = 20ms-')
How far will the two vehicles be from one
another during the moment in (ii) above?
(SA = 380m and SB
= 500m: SAB
120m). plz help
Answer:
(i) Please find attached the required velocity time graphs plotted with MS Excel
(ii) The velocity of vehicle A at the 18th second = 20 m/s
The velocity of vehicle B at the 18th second = 0 m/s
(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters
Explanation:
The given parameters of the motion of vehicles A and B are;
The acceleration of vehicles A and B = Uniform acceleration starting from rest
The maximum velocity attained by vehicle A = 30 m/s
The time it takes vehicle A to attain maximum velocity = 10 s
The maximum velocity attained by vehicle B = 30 m/s
The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s
The time duration vehicle A maintains its maximum velocity = 6 s
The time duration vehicle B maintains its maximum velocity = 4 s
(i) From the question, we get the following table;
[tex]\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}[/tex]
From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here
(ii) The velocity of vehicle A at the start = 0 m/s
After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s
The maximum velocity is maintained for 6 seconds which gives;
At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s
The time it takes vehicle A to decelerate to rest = 6 s
The deceleration of vehicle A, [tex]a_A[/tex] = (30 m/s - 0 m/s)/(6 s) = 5 m/s²
Therefore, we get;
v = u - [tex]a_A[/tex]·t
At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s
u = 30 m/s
∴ v₁₈ = 30 - 5 × 2 = 20
The velocity of vehicle A at the 18th second, [tex]V_{18A}[/tex] = 20 m/s
For vehicle B, we have;
At the 14th second, the velocity of vehicle B = 40 m/s
Vehicle B decelerates to rest in, t = 4 s
The deceleration of vehicle B, [tex]a_B[/tex] = (40 m/s - 0 m/s)/(4 s) = 10 m/s²
For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s
∴ [tex]v_{18B}[/tex] = 40 m/s - 10 m/s² × 4 s = 0 m/s
The velocity of the vehicle B at 18th second, [tex]v_{18B}[/tex] = 0 m/s
(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;
The area triangle A₁ = (1/2) × 10 × 30 = 150
Area of rectangle, A₂ = 6 × 30 = 180
Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50
The distance covered in the 18th second by vehicle [tex]S_A[/tex] = A₁ + A₂ + A₃
∴ [tex]S_A[/tex] = 150 + 180 + 50 = 380
The distance covered in the 18th second by vehicle [tex]S_A[/tex] = 380 m
The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;
Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440
The distance covered by the trapezoid, [tex]S_B[/tex] = 440 m
The distance of the two vehicles apart at the 18t second, [tex]S_{AB}[/tex] = [tex]S_B[/tex] - [tex]S_A[/tex]
∴ [tex]S_{AB}[/tex] = 440 m - 380 m = 60 m
The distance of the two vehicles from one another at the 18th second, [tex]S_{AB}[/tex] = 60 m.
Starting circuit One battery. 2 light bulbs in parallel; switch What is the voltage across the battery? What is the voltage across light bulb 1? What is the voltage across light bulb 2?
Answer:
The voltage across light bulb 1 and light bulb 2 is the the same i.e V
Explanation:
In a parallel circuit, the Voltage is same across all the components of the circuit and the current flowing through each component is added to get the total current across the circuit.
Let us say, the voltage across the circuit is V. The voltage across light bulb 1 and light bulb 2 is the the same i.e V
Which of the following settings describes a displacement vs. time graph that
looks like this?
Displacement
Time
Check all that apply
O A. Constant velocity
OB. Increasing velocity
I C. Zero acceleration
O D. Decreasing acceleration
I NEED HELP ASAP
Answer:
c and a
Explanation:
ape x
A man is driving his 1100 kg car at 36 km/h on a straight freeway. After accelerating for 30 seconds, the car
has a speed of 108 km/h. How much work did the engine do during the 30 seconds?
Answer:
W = 439998 J = 439.99 KJ
Explanation:
First, we will calculate the acceleration of the car by using the first equation of motion:
[tex]v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}[/tex]
where,
a = acceleration = ?
vf = final speed = [tex]108(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})[/tex] = 30 m/s
vi = initial speed = [tex]36(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})[/tex] = 10 m/s
t = time = 30 s
Therefore,
[tex]a = \frac{30\ m/s - 10\ m/s}{30\ s}[/tex]
a = 0.67 m/s²
Now, we will calculate the force applied by the engine:
F = ma
where,
F = force = ?
m = mass = 1100 kg
Therefore,
F = (1100 kg)(0.67 m/s²)
F = 733.3 N
Now, we will calculate the distance covered by the car by using the second equation of motion:
[tex]s = v_it+\frac{1}{2}at^2\\\\s = (10\ m/s)(30\ s)+\frac{1}{2} (0.67\ m/s^2)(30\ s)^2[/tex]
s = 600 m
Now, the work done (W) by engine can be calculated as follows:
W = Fs
W = (733.3 N)(600 m)
W = 439998 J = 439.99 KJ
differences between weightlessness in space and weightlessness in earth
Answer:
it depends on a person's own weight
Answer:
The gravity is roughly the same. The inverse square law applies to gravity and that means that being 60 miles above Earth is about the same as being on Earth. The thing is, they are going really fast in an orbit and thus falling around the Earth. The weightlessness comes from a sort of cancellation of gravity pulling them down and their going perpendicular to that force. The forces are cancelling and making them relatively “weightless”. It’s not really weightlessness but just an equilibrium reached between the two accelerations of gravity and their motion against the pull of the Earth.
Explanation:
True or False?
________________The fact that ocean crust is younger the farther away it is from an ocean ridge supports the theory of plate tectonics
Answer:
true is the answer of the question
A very long cylindrical shell made of copper has an inner radius 2 cm and an outer radius of 3 cm. This shell concentrically surrounds a very long charged aluminum rod of radius 1 cm with a charge density of 4 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod:
Question:
Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod:
(a) 0.5 cm, (b) 1.5 cm, (c) 2.5 cm, (d) 3.5 cm, (e) 7.0 cm
Answer:
(a) 0
(b) 4.795 N/C
(c) 2.877 N/C
(d) 2.055 N/C
(e) 1.027 N/C
Explanation:
The given parameters are;
The inner radius of the copper = 2 cm
The outer radius of the copper - 3 cm
The radius of the aluminum rod, r = 1 cm
The radius of the aluminum rod = 4 pC/m
a) The charge at the 0.5 cm from the center of the aluminum rod, E = 0
There is no charge inside the rod given that the aluminum road is a conductor of electricity, having all the charge on its surface
b) At 1.5 cm, we have;
[tex]E = \dfrac{2 \cdot \lambda}{4 \cdot \pi \cdot r \cdot \epsilon_0}[/tex]
Where;
[tex]\dfrac{ \lambda}{4 \cdot \pi \cdot \epsilon_0} = 8.99 \times 10^9 \times 4 \times 10^{-12}[/tex]
r = 1.5 cm = 0.015 m
∴ E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.015 ≈ 4.795
The magnitude of the electric field, E = 4.795 N/C
The direction of the charge depends on the charge sign on the rod
c) At 2.5 cm, we have;
r = 2.5 cm = 0.025 m
E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.025 ≈ 2.877
The magnitude of the electric field, E ≈ 2.877 N/C
d) At r = 3.5 cm, we have;
r = 3.5 cm = 0.035 m
E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.035 ≈ 2.055
The magnitude of the electric field, E ≈ 2.055 N/C
e) At
r = 7.0 cm = 0.070 m
E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.07 ≈ 1.027
The magnitude of the electric field, E ≈ 1.027 N/C
Si un conductor tiene una longitud de 4 metros y una sección de 2 mm^2 calcule su resistencia, si su coeficiente de resistividad es de 0.10(Ωmm^2/m)
Answer:
R = 0.2 Ω
Explanation:
Para aplicar el concepto de resistencia de un conductor electrico, consideramos la siguiente formula:
R = ρ . l / S
Donde l es la longitud del conductor
S la superficie
y ρ el coeficiente de resistividad
Primero necestiamos hacer ciertas conversiones:
ρ = 0.10 ×10⁻⁶ (Ω . m²/m)
S = 2 mm² → 2 ×10⁻⁶ m²
R = 0.10 ×10⁻⁶ (Ω . m²/m) . 4m / 2 ×10⁻⁶ m²
Dentro de la formula simplificamos las unidades, los metros de ρ con la longitud y ambos m². De esa forma la unidad de R queda en Ω (ohmios).
Resolvemos:
R = 0.10 ×10⁻⁶ Ω . 4 / 2 ×10⁻⁶
R = 0.2 Ω
Un cuerpo se mueve por una carretera que tiene rozamiento.¿Podríamos aplicar el principio de conservación de la energía mecánica? ¿Qué requisitos se han de cumplir para poder aplicarla?
Answer:
Sí se puede aplicar el Principio de Conservación de la Energía Mécanica siempre y cuando se considere los efectos de pérdida de energía a causa del rozamiento, de lo contrario, no se cumpliría el requisito de conservación derivado del citado principio de conservación.
Explanation:
Consideremos que el cuerpo representa nuestro sistema de estudio, el cual se desplaza por una carretera y experimenta rozamiento, significando cambios en la energía cinética y potencial gravitacional. La presencia de rozamiento indica una disipación de la energía del cuerpo, cuya energía mecánica total disminuye, significando un sistema no conservativo.
Sí se puede aplicar el Principio de Conservación de la Energía Mécanica siempre y cuando se considere los efectos de pérdida de energía a causa del rozamiento, de lo contrario, no se cumpliría el requisito de conservación derivado del citado principio de conservación.
During which radioactive decay process does the mass number of a nucleus change?
A. Beta Decay (Electron)
B. Beta Decay (Positron)
C. Alpha Decay
D. Gamma Decay
Answer:
Alpha decay
Explanation:
Answer:
C
Explanation:
Alpha Decay
1 point
The Sun is a massive star at the center of our Solar System. The planets
closest to the Sun are terrestrial planets, and those farther away are gas
giants. Which force maintains this arrangement of the Solar System? *
Electromagnetism
Friction
Gravity
Torsion
Answer:
Gravity
Explanation:
It is gravity because the sun's gravity pulls all the planets towards itself and keeps the planets in the sun's orbit
Below is a single strand of DNA, what is the complementary base pair for the strand?
AAG, GCA, TCA, GAT
Describe the general relationship between the weight of adult men and the average number of Calories needed daily to maintain a steady weight.
Answer:
The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking.
The amount of calories contained in the food we eat every day must represent the amount of calories eliminated by the body in that time to have a steady weight.
Explanation:
The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking. If total quantity of calories in the food we consume every day is higher that the average number of calories needed daily, then weight increases by fat accumulation.
Dos masas con valores de 8kg y 12kg cada una se encuentran separadas por una distancia de 1m. Determinar la fuerza de atraccion gravitacional entre ellas si se compara con el valor de sus pesos Como es este valor?
The question is: Two masses with values of 8kg and 12kg each are separated by a distance of 1m. Determine the force of gravitational attraction between them if compared with the value of their weights. What is this value?
Answer: The force of gravitational attraction between them if compared with the value of their weights is [tex]640.704 \times 10^{-11} N[/tex].
Explanation:
Given: [tex]m_{1}[/tex] = 8 kg, [tex]m_{2}[/tex] = 12 kg
Distance = 1 m
Formula used is as follows.
[tex]F = \frac{G \times m_{1} \times m_{2}}{r^{2}}[/tex]
where,
F = force of gravitational attraction
G = gravitational constant = [tex]6.674 \times 10^{-11} N m^{2}/kg^{2}[/tex]
[tex]m_{1}[/tex] = mass of object 1
[tex]m_{2}[/tex] = mass of object 2
r = distance between the centers of these two objects
Substitute the values into above formula as follows.
[tex]F = \frac{G \times m_{1} \times m_{2}}{r^{2}}\\= \frac{6.674 \times 10^{-11} Nm^{2}/kg^{2} \times 8 kg \times 12 kg}{(1 m)^{2}}\\= 640.704 \times 10^{-11} N[/tex]
Thus, we can conclude that the force of gravitational attraction between them if compared with the value of their weights is [tex]640.704 \times 10^{-11} N[/tex].