Answer:
1. [tex] f = 2 Hz [/tex]
2. [tex] f = 0.011 Hz [/tex]
3. [tex] f = 0.067 Hz [/tex]
4. [tex] t = 4 s [/tex]
Explanation:
1. The frequency of rotation is given by:
[tex] f = \frac{\omega}{2\pi} [/tex]
Where:
ω: is the angular speed = 50 rotations (revolutions) in 25 s.
We need to convert the units of ω.
[tex] \omega = \frac{50 rev}{25 s}*\frac{2\pi rad}{1 rev} = 4\pi rad/s [/tex]
Now, the frequency is:
[tex] f = \frac{4\pi rad/s}{2\pi} = 2 Hz [/tex]
2. The frequency is:
We know:
5 laps = 5 revolutions
t: time = 450 s
[tex] f = \frac{\omega}{2\pi} = \frac{\frac{5 rev}{450 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.011 Hz [/tex]
3. The frequency of the pendulum is:
[tex]f = \frac{\omega}{2\pi} = \frac{\frac{1 rev}{15 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.067 Hz[/tex]
4. We have:
θ: number of revolutions = 48 rev
f = 12 Hz
t =?
The time can be calculated as follows:
[tex] f = \frac{\omega}{2\pi} = \frac{\theta}{2\pi t} [/tex]
[tex] t = \frac{\theta}{2\pi f} = \frac{48 rev*\frac{2\pi rad}{1 rev}}{2\pi*12 Hz} = 4 s [/tex]
I hope it helps you!
El peso(w) de una persona en la Tierra es 550N. Determinar la masa (m) de la persona y el peso (w) de la misma en la Luna
The question is: The weight (w) of a person on Earth is 550N. Determine the mass (m) of the person and the weight (w) of the person on the Moon.
Answer: The mass (m) of the person is 550 N and the weight (w) of the person on the Moon is 891 kg.
Explanation:
Given: Mass = 550 N
When a person moves towards moon then its mass will remain the same but there will occur change in its weight because the acceleration due to gravity on moon is different than that on Earth.
Therefore, formula used to calculate the weight of person on moon is as follows.
[tex]W = m \times g_{m}[/tex]
where,
m = mass
[tex]g_{m}[/tex] = acceleration due to gravity on moon = 1.62 [tex]m/s^{2}[/tex]
Substitute the values into above formula as follows.
[tex]W = m \times g_{m}\\= 550 N \times 1.62 m/s^{2}\\ (1 N = 1 kg \times m/s^{2})\\= 891 kg[/tex]
Thus, we can conclude that the mass (m) of the person is 550 N and the weight (w) of the person on the Moon is 891 kg.
5. A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 6400 m away?
Answer:
8.89 secs
Explanation:
720m = 1 sec
6400m = ?
6400/720 = 8.89 secs
If an object of mass 70kg falls from a height of 500 m, what is the maximum velocity of the object?
Answer:
H = 1/2 * g * t^2 since initial velocity is zero
v = g * t where v is the final velocity
t = v / g
H = 1/2 g * v^2 / g^2 = 1/2 v^2 / g
v = (2 * H * g)^1.2
v = (2 * 500 * 9.8)^1/2 = 99 m/s
Check: t = v / g = 99 / 9.8 = 10.1 sec
H = 1/2 * 9.8 * 10.1^2 = 500 m
A constant force (3,0,4)N moves a body from point (-5,3,1)m to the point (4,3,3)m. If the initial velocity of the body is (6,1,1)m/s, determine the work done by the force and the angle between the force and the initial velocity of the body
The net displacement of the body is
∆x = (4, 3, 3) m - (-5, 3, 1) m = (9, 0, 2) m
so the work done by F = (3, 0, 4) N in the direction of ∆x is
F • ∆x = (3, 0, 4) • (9, 0, 2) N•m = (27 + 0 + 8) N•m = 35 J
The angle between the force and initial velocity v₀ is θ, such that
F • v₀ = ||F|| ||v₀|| cos(θ)
(3, 0, 4) • (6, 1, 1) N•m/s = √(3² + 0² + 4²) √(6² + 1² + 1²) cos(θ) N•m/s
==> cos(θ) = (18 + 0 + 4) / (√25 × √38) = 22/(5√38)
==> θ = arccos(22/(5√38)) ≈ 44.5°
is driving with a velocity of 5 m/s and speeds up to a velocity of 10 m/s over 5 seconds. What is Mr. DaCosta's acceleration?
What is the answer to this question?
Answer:
Explanation:
In a velocity/time (aka acceleration) graph, the slope of a line indicates the value of the acceleration in m/s/s. Acceleration is the change in velocity over the change in time. From 0 - 2 seconds, there is no change in velocity, so the acceleration during this interval is 0 (which is the same as the slope of the line). From 2 - 4 seconds, the slope of the line is -2, so the acceleration during the time interval from 2 to 4 seconds is -2 (negative because David is slowing down but is still going the same direction: to the right).
The electric motor of an elevator(lift) uses 630kJ of electric energy when raising the elevator and passengers, of total mass 12,500N, through a vertical height of 29m.Calculate the efficiency of the elevator. pppppplease need helpp willl markk the brrainliest
Answer:
Efficiency = 0.575 = 57.5%
Explanation:
First, we will calculate the output energy of the elevator, which is equal to the potential energy acquired by the load.
[tex]O = Potential\ Energy\\O = mgh[/tex]
where,
O = Output = ?
mg = W = Weight = 12500 N
h = height = 29 m
Therefore,
[tex]O = (12500\ N)(29\ m) \\O = 362500\ J = 362.5\ KJ[/tex]
The input is given as:
I = 630 KJ
Thus the efficiency will be:
[tex]Efficiency = \frac{O}{I} = \frac{362.5\ KJ}{630\ KJ}\\\\[/tex]
Efficiency = 0.575 = 57.5%
A battery provides a potential difference of 20 V. What size current will flow through a 5 ohm resistor in
this circuit
In which of the following is there more than one path for electrons to flow?
A.An open Circuit
B.A parallel circuit
C.A series circuit
D.A closed circuit
B
because in parallel combination there are more than one paths for the current to flow .
What is meant by electric current?
Explanation:
the electric current is ratio of electric poer to the voltage
Which resource would be the best choice to learn more information about studying martial arts?
Dance studio
Local bait shop
Phone book
Public Golf Course
Answer:
Dance studio
Explanation:
Martial art use to defend ourself from any dangerous. Dance is a way to learn it
Answer:
The answer is A.) Dance studio.
Explanation:
See screenshot below
Hope this helps! :D
(Can you mark me brainliest?)
- ❤ 7272033Alt ❤
what happens when a candle burns?
a) oxygen is used up
b) nitrogen is used up
c) hydrogen is produced
d) carbon dioxide is consumed
Answer:
oxygen is used up is the answer
Explanation:
These vaporized molecules are drawn up into the flame, where they react with oxygen from the air to create heat, light, water vapor (H2O) and carbon dioxide (CO2).
if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use w=mgh since work done and mass of a person is same everywhere
Answer:
[tex]h_{moon} = 11.8\ m[/tex]
Explanation:
Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:
[tex]W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} = 11.8\ m[/tex]
Un punto se ubica en un sistema de coordenadas polar por las coordenadas r = 7.6 m y = 30°. Encuentre las coordenadas x y y de este punto (en m), suponiendo que los dos sistemas de coordenadas tienen el mismo origen.
Answer:
Las coordenadas del punto son [tex](x,y) = (6.582\,m, 3.8\,m)[/tex].
Explanation:
La relación entre un vector en formato polar y uno con formato rectangular con respecto al mismo origen queda sintetizado por la siguiente expresión:
[tex](x, y) = r\cdot (\cos \theta, \sin \theta)[/tex] (1)
Donde:
[tex]r[/tex] - Magnitud de la distancia del punto con respecto al origen, medido en metros.
[tex]\theta[/tex] - Dirección del punto con respecto del semieje +x, medido en grados sexagesimales.
[tex]x,y[/tex] - Coordenadas rectangulares del punto con respecto al origen, medidas en metros.
Si sabemos que [tex]r = 7.6\,m[/tex] y [tex]\theta = 30^{\circ}[/tex], entonces las coordenadas rectangulares del punto con respecto al origen son:
[tex](x,y) = (7.6\,m)\cdot (\cos 30^{\circ}, \sin 30^{\circ})[/tex]
[tex](x,y) = (6.582\,m, 3.8\,m)[/tex]
Las coordenadas del punto son [tex](x,y) = (6.582\,m, 3.8\,m)[/tex].
Match each description to one of the graphs below.
A) A child runs at a constant speed for 40 s before stopping.
B)A panda moves slowly at a constant speed for 40 s before stopping.
C)A cheetah stalks his prey, moving slowly, stopping occasionally.
D)A rabbit runs away from a farmer, not stopping for anything!
Answer:
I) the correct one is A
II) the alternatives C
III) corresponds to option D
IV) alternative B
Explanation:
Various information can be obtained from the speed versus time graphs
* If the graph is an inclined line, the speed is constant
* if the graph is a horizontal line the velocity is zero
In this case, the first graph (I) has two parts: an inclined line that corresponds to a uniform movement with speed and a second one, so that it is a horizontal line that indicates that the body is stopped.
If we see the different alternatives, the correct one is A
The second graph (II) has a movement with constant speed second period of being stopped
This graph corresponds to the alternatives C
The third graph (III) the body moves at constant speed without stopping
This corresponds to option D
The fourth graph (IV) the body moves with a slow speed and after 40 s it stops
This graph is associated with alternative B
2. 2. A drawing that shows the outline of an object is called
What is the process of combining the sperm and egg cell called and what does it produce? Anyone know the answer ?
Answer:
fertilization and it produces a Zygote
A sled drops 20 mts in height on a hill. The rider is going 20 m/s at the bottom of the hill. Where can you find kinetic and potential energy? In which moments?
Explanation:
Potential energy is the energy occupied by an object or substance due to its position.
For example, a sled drops 20 meters in height on a hill shows that a decrease in height is taking place.
Hence, potential energy is involved there.
Kinetic energy is the energy acquired by an object due to its motion.
For example, a rider is going 20 m/s at the bottom of the hill shows that rider is in motion due to which it has kinetic energy.
converts light energy to electrical energy
this electric current produced can be stored in another rechargeable cell.
If the collector of the cell is covered the voltage is higher
The current shown is real current
All of the above
None of the above
Answer:
If the collector of the cell is covered, the voltage is higher
Explanation:
The collector of the a passive, active, or photovoltaic (PV) solar cell is the device that takes the Sun's energy by the absorption of the Sun's radiation which is then transformed into electricity and heat. The ability to absorb the Sun's energy is given by the material type and the coating of solar collector
Therefore, the device requires access to energy from Sunlight to get the energy which can be stored in a rechargeable cell and well the collector's cell is covered, access to Sunlight is denied and therefore energy is not supplied to the cells and current is not generated and there is no flow of electricity and the voltage in the circuit is lower or there is no voltage in the circuit
The statement which is wrong about the solar cell is therefore, that if the collector of the cell is covered, the voltage is higher
A car starts from the state of xestIf its velocity becomes 70 km/hr in 6 minutes, i) what is the accordine acceleration of F the car? (ii) what is the the distance cover ded by the car?
Answer: [tex]3.5\ km[/tex]
Explanation:
Given
Car starts from the state of rest and acquires a velocity of [tex]70\ km/hr[/tex] in 6 minutes
Final velocity in m/s is [tex]v=70\approx 19.44\ m/s[/tex]
Using equation of motion
[tex]v=u+at\\\Rightarrow 19.44=0+a(6\times 60)\\\Rightarrow a=0.054\ m/s^2[/tex]
Distance covered in 360 s
[tex]\Rightarrow v^2-u^2=2as\\\Rightarrow 19.44^2-0=2\times 0.054\times s\\\Rightarrow s=3500.64\ m\approx 3.5\ km[/tex]
Which of the following exercises would be BEST for improving muscular strength and endurance?
A.
walking
B.
running
C.
push-ups
D.
bicycling
Answer:
Push-ups
explanation:
Answer:
the anser for your qestion is
c. push- ups
The diagram shows a stone tied to a string in circular
motion. If the string breaks, what will be the direction of
motion of the stone?
Answer:
Newton's First Law of motion states that an object will continue to move in a straight line or remain at rest unless a force acts on it
The stone is able to move in circular motion because a (centripetal) force acts through the string on it, which maintains the circular motion, by turning the stone to the circular path as the stone moves with momentary velocity along a straight line which is (therefore) tangential to the circle
Therefore, at the point the spring breaks, the centripetal force is no longer acting on the stone to change its path and the motion of the continues on the tangent to the initial circular motion path at that point
Explanation:
Question 2 of 10
A wave has a wavelength of 0.05 m and a wave speed of 10 m/s. What is the
frequency of the wave?
O A. 5.0 Hz
B. 200 Hz
C. 2.0 Hz
O D. 0.50 Hz
Explanation:
everything can be found in the picture
Why don't gorilla evolve to human if scientist was about to said that human evolve from gorilla to human?
Answer:
"Man's mind is far too advanced for his survival needs."
See also "Wallace", Darwin's junior partner who couldn't explain the gap in intelligence.
Whay is the other name of negative acceleration.
Answer:
deceleration? I think because the prefix de means to like not or to slow down
How does the "lollipop moment", exemplify for us the opportunity we all have to exercise leadership on a normal everyday basis?
Answer:
WHat do you mean by that
Explanation:
I dont get what you re asking
ASAP
A light wave travels through glass (n=1.50) at an angle of 35°. What angle
does it have when it passes from the glass into the air (n = 1.0)?
A. 59.4
B. 0.860
C. 22.5
D. 0.382
Answer:
A. 59.4
Explanation:
The refractive index of the glass, n₁ = 1.50
The angle of incidence of the light, θ₁ = 35°
The refractive index of air, n₂ = 1.0
Snell's law states that n₁·sin(θ₁) = n₂·sin(θ₂)
Where;
θ₂ = The angle of refraction of the light, which is the angle the light will have when it passes from the glass into the air
Therefore;
θ₂ = arcsin(n₁·sin(θ₁)/n₂)
Plugging in the values of n₁, n₂ and θ₁ gives;
θ₂ = arcsin(1.50 × sin(35°)/1.0) ≈ 59.357551° ≈ 59.4°
The angle the light will have when it passes from the glass into the air, θ₂ ≈ 59.4°.
Answer:
A. 59.4
Explanation:
got it right, trust
You attend a concert and have a seat that is 70 m away from the stage where the sound is produced. How long does it take for the sound to reach your ears? You take a picture with your phone and send it to your friend who could not come to the concert. She lives in a town 45 miles away. How long does it take until she gets the picture on her phone? Please, include the assumptions that you make.
Given:
distance of observer from the sound source, [tex]s=70~m[/tex]
distance between the source of image information sender and the receiver, [tex]d=45~miles=72420.48~m[/tex]
Assumptions:
speed of sound, [tex]v=343~m/s[/tex]
speed of the image information being sent, [tex]c=3\times 10^8~m/s[/tex] (we assume it to be nearly the speed of light in vacuum)
Solution:
Time taken by the sound to reach the observer:
[tex]time=\frac{distance}{speed}[/tex]
[tex]time=70\div 343[/tex]
[tex]time \approx 0.2~s[/tex]
Time taken by the image to reach the destination:
[tex]time=\frac{72420.48}{3\times 10^8}[/tex]
[tex]time=2.4\times 10^{-4}~s[/tex]
Enunciado del ejercicio n° 1
Se lanza un cuerpo verticalmente hacia abajo con una velocidad inicial de 7 m/s.
a) ¿Cuál será su velocidad luego de haber descendido 3 s?
b) ¿Qué distancia habrá descendido en esos 3 s?
c) ¿Cuál será su velocidad después de haber descendido 14 m?
d) Si el cuerpo se lanzó desde una altura de 200 m, ¿en cuánto tiempo alcanzará el suelo?
e) ¿Con qué velocidad lo hará?
Usar g = 10 m/s²
Desarrollo
Datos:
v0 = 7 m/s
t = 3 s
y = 200 m
h = 14 m
Fórmulas:
(1) vf = v0 + g·t
(2) y = v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 2
Se lanza un cuerpo verticalmente hacia arriba con una velocidad inicial de 100 m/s, luego de 4 s de efectuado el lanzamiento su velocidad es de 60 m/s.
a) ¿Cuál es la altura máxima alcanzada?
b) ¿En qué tiempo recorre el móvil esa distancia?
c) ¿Cuánto tarda en volver al punto de partida desde que se lo lanzo?
d) ¿Cuánto tarda en alcanzar alturas de 300 m y 600 m?
Usar g = 10 m/s²
Desarrollo
Datos:
v0 = 100 m/s
vf = 60 m/s
t = 4 s
y1 = 300 m
y2 = 600 m
Fórmulas:
(1) vf = v0 + g·t
(2) y = v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 3
Un observador situado a 40 m de altura ve pasar un cuerpo hacia arriba con una cierta velocidad y al cabo de 10 s lo ve pasar hacia abajo, con una velocidad igual en módulo pero de distinto sentido.
a) ¿Cuál fue la velocidad inicial del móvil?
b) ¿Cuál fue la altura máxima alcanzada?
Usar g = 10 m/s²
Desarrollo
Datos:
t = 10 s
y = 40 m
Fórmulas:
(1) vf = v0 + g·t
(2) y = y0 + v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 4
Desde un 5° piso de un edificio se arroja una piedra verticalmente hacia arriba con una velocidad de 90 km/h, ¿cuánto tardará en llegar a la altura máxima?
Usar g = 10 m/s²
Desarrollo
Datos:
v0 = 90 km/h
v0 = 25 m/s
Fórmulas:
(1) vf = v0 + g·t
(2) y = v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 5
Un auto choca a 60 km/h contra una pared sólida, ¿desde qué altura habría que dejarlo caer para producir el mismo efecto?
Usar g = 10 m/s²
Desarrollo
Datos:
vf = 60 km/h
vf = 16,67 m/s
v0 = 0 m/s
Fórmulas:
(1) vf = v0 + g·t
(2) y = v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 6
Se lanza una pelota hacia arriba y se recoge a los 2 s, calcular:
a) ¿Con qué velocidad fue lanzada?
b) ¿Qué altura alcanzó?
Usar g = 10 m/s²
Desarrollo
Datos:
t = 2 s
Fórmulas:
(1) vf = v0 + g·t
(2) y = v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 7
Se lanza una pelota de tenis hacia abajo desde una torre con una velocidad de 5 m/s.
a) ¿Qué velocidad tendrá la pelota al cabo de 7 s?
b) ¿Qué espacio habrá recorrido en ese tiempo?
Usar g = 10 m/s²
Desarrollo
Datos:
v0 = 5 m/s
t = 7 s
Fórmulas:
(1) vf = v0 + g·t
(2) y = v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 8
Se lanza un cuerpo verticalmente hacia arriba con una velocidad de 60 km/h, se desea saber la altura máxima alcanzada, la velocidad que posee al cabo de 4 s y 30 s, la altura alcanzada a los 8 s, el tiempo total que se encuentra en el aire.
Desarrollo
Datos:
v0 = 60 km/h = (60 km/h)·(1.000 m/km)·(1 h/3.600 s) = 16,67 m/s
t1 = 4 s
t2 = 30 s
t3 = 8 s
Usar g = 10 m/s²
Fórmulas:
(1) vf = v0 + g·t
(2) y = v0·t + ½·g·t²
(3) vf² - v0² = 2·g·h
Enunciado del ejercicio n° 9
Se dispara verticalmente hacia arriba un objeto desde una altura de 60 m y se observa que emplea 10 s en llegar al suelo. ¿Con que velocidad se lanzo el objeto?
Desarrollo
Datos:
h0 = 60 m
t = 10 s
g = 9,81 m/s²
Fórmulas:
Δy = v0·t + ½·g·t²
Enunciado del ejercicio n° 10
Se lanza verticalmente hacia abajo una piedra de la parte alta de un edificio de 14 pisos, llega al suelo en 1,5 s, tomando en cuenta que cada piso mide 2,6 m de altura. Calcular la velocidad inicial de la piedra y la velocidad al llegar al piso.
Desarrollo
Datos:
Número de pisos = 14
Altura de cada piso = 2,6 m
t = 1,5 s
g = 9,81 m/s²
Fórmulas:
1) Δh = v0·t + ½·g·t²
2) vf = v0 + g·t
*xfv se que es mucho pero e visto videos pero no me sale muy bien los resultados con mis compañeros. xfv alguien que me ayude
Answer:
34
Explanation:
Which is a characteristic of atoms?
A) They come in ten major types.
B) They are the smallest units of matter.
C) They can be found in nonliving things only.
D) They cannot combine together.
Answer:
c
Explanation:
c
