La respuesta correcta para esta pregunta abierta es la siguiente.
A pesar de que se te olvidó especificar el país al que te refieres y a la fecha específica para saber de qué época estás hablando, te podemos ayudar comentando lo siguiente.
¿Qué juegos se practicaban en esa Época?
Anterior al surgimiento de la era digital y la era de los "gammers," los niños y la juventud en general salían a la calle o al parque a realizar diferentes actividades físicas y recreativas. Había juegos de ronda, juegos deportivos, juegos de destreza, y juegos de mesa.
¿Qué nombre tenían los juegos?
Había juegos que se llamaba "Doña Blanca," "La Roña," "Lobo estás ahí," "Las escondidas," "canicas," el yoyo," "el trompo," "el burro castigado," "las cebollitas," "el látigo," además de las "cascaritas" que eran juegos de soccer o futbol americano en las calles y con equipos formados por tus amigos.
¿En qué consistía cada uno de los juegos?
Si jugabas canicas, tenías que golpear una canica con otra para desplazarla y llevarla a cierto lugar. O tenías que meterla en agujeros.
Si jugabas, "Doña Blanca," formabas un círculo entrelazado de las manos. Mientras cantabas la canción de Doña Blanca, alguien que estaba afuera del círculo trataba de romperlo, tratando de soltar algunas de las manos entrelazadas.
La verdad era sumamente divertido, creativo, te reías, mucho, y lo mejor de todo era que hacías ejercicio y no te quedabas sentado todo el día frente a una pantalla de computadora.
Parallel incident rays appear to bounce like they have all originated from the same point. What is this point called?
A. cross point
B. midpoint
C. bounce point
D. focal point
How do you use the periodic table to recall the ionic charge of an alkali metal, an alkaline earth metal, or aluminum?
The positive charge is the group number.
The negative charge is the group number.
The positive charge is the period number.
The negative charge is the period number.
Answer:
the positive charge is the period number
Explanation:
I might be wrong
Answer:
The positive charge is the group number.
Explanation:
A 15 cm length of wire is moving perpendicularly
through a magnetic field of strength 1.4 T at the rate
of 0.12 m/s. What is the EMF induced in the wire?
A. OV
C. 0.025 v
B. 0.018 V
D. 2.5 V
Answer: C or B
Explanation:
The EMF induced in the wire moving perpendicularly through a magnetic field is 0.025V. The correct option is C.
What is EMF?The EMF is the electro motive force which causes the current to induce in the object moving in the magnetic field.
Given is the length of wire L =15cm =0.15m, magnetic field strength B = 1.4T and velocity of wire V =0.12 m/s
EMF is related to the length of wire, magnetic field strength and velocity of wire proportionally.
ε = B x L x V
Plug the values, we get
ε = 1.4 x 0.15 x 0.12
ε = 0.025 Volts
Thus, the correct option is C.
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In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?
The force of the wall on the car and the car on the wall are equal
The force of the wall on the car is greatest
The force of the car on the wall is greatest
There is not enough information to tell
Answer:
A...................................
The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.
According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.
The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.
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The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?
a. The force of the wall on the car and the car on the wall are equal
b. The force of the wall on the car is greatest
c. The force of the car on the wall is greatest
d. There is not enough information to tell"
in a series circuit, how does the voltage supplied by the battery compare to the voltages on each load?
Answer:
In a series circuit, how does the voltage supplied by the battery compare to the voltage on each load? The voltage of the battery is equal to the voltage of each load added together. ... The voltage across the two resistors must both have the same voltage of the battery.
Explanation:
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Answer:
The voltage of the battery is equal to the voltage of each load added together. The voltage across the two resistors must both have the same voltage of the battery.
>3
At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?
Answer:
The mass of the dog food added is 9.03 kg
Explanation:
Given;
mass of the shopping cart, m₁ = 14.5 kg
let the mass of the bag added = m₂
the force applied, F = 12 N
initial velocity of the cart-bag system, u = 0
distance traveled by the system, d = 2.29 m
time of motion of the system, t = 3.0 s
The acceleration of the system is calculated as;
[tex]d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2[/tex]
The total mass of the system (M) is calculated as follows;
F = Ma
M = F/a
M = (12)/(0.51)
M = 23.53 kg
The mass of the dog food added is calculated as;
m₂ = M - m₁
m₂ = 23.53 kg - 14.5 kg
m₂ = 9.03 kg
which forces are capable of affecting particles or objects from large distance
Answer:
only long-range force that affects all particles is the gravitational force.
Explanation:
In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.
The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.
Consequently the only long-range force that affects all particles is the gravitational force.
Which energy store is increased when an object is heated?
Answer:
Kinetic Energy
Explanation:
help me please
only if you really know
Assume R is measured in meters (m) and M in kilograms (kg). Then
R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²
so t ² is indeed proportional to R ²/(GM).
The rhinestones in costume jewelry are glass with index of refraction 1.50. To make them more reflective, they are often coated with a layer of silicon monoxide of index of refraction 2.00. What is the minimum coating thickness needed to ensure that light of wavelength 576 nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference
Answer:
[tex]T=62.9*10^{-9}[/tex]
Explanation:
From the question we are told that:
Index of refraction of Rinestones [tex]\gamma_1 =1.5[/tex]
Index of refraction of silicon [tex]\gamma_2 =2.0[/tex]
Wavelength [tex]\lambda=576nm=576*10^{-9}[/tex]
Let each layer have thickness T
Therefore
Total Thickness =2T
Generally the equation for Constructive interference is mathematically given by
[tex]2T=(m+0.5)\frac{l\lambda}{\gamma_2}[/tex]
Where
[tex]M=0[/tex]
[tex]2T=(0+0.5)\frac{576*10^{-9}}{2*2.0}[/tex]
[tex]T=62.9*10^{-9}[/tex]
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How much force (in Newtons) does a baseball pitcher have to exert on a 250g baseball to make it accelerate to 50 m/s the
Instant that it leaves his hand?
Answer:
Force = 12.5 Newton
Explanation:
Given the following data;
Mass = 250 g to kilograms = 250/1000 = 0.25 kg
Acceleration = 50 m/s²
To find the force;
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
[tex] Acceleration = \frac {Net \; force}{mass} [/tex]
Making force the subject of formula, we have;
[tex] Force = mass * acceleration [/tex]
Substituting into the formula, we have
[tex] Force = 0.25 * 50 [/tex]
Force = 12.5 Newton
why are circuit breakers used in parts of national grid ?
Answer:
To protect control circuits or small devices with insufficient cutting power
Explanation:
Use the universal law of gravitation to solve the following problem.
Hint: mass of the Earth is = 5.97 x 1024 kg
A scientific satellite of mass 1300 kg orbits Earth 200 km above its surface. If Earth has a radius of 6378 km, what is the force of gravity acting on the scientific satellite?
a. Write out the formula for this problem.
b. Plug in the values from this problem into the formula.
c. Solve the problem, writing out each step.
d. Correct answer
Answer:
a.
[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]
b.
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]
c.
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]
d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N
Explanation:
a. The formula for finding the force of gravity, F, acting object on an object is given as follows;
[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]
Where;
F = The force acting between the Earth and the object
G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
M = The mass of the Earth = 5.97 × 10²⁴ kg
m = The mass of the object
r = The distance between the center of the Earth and the object
b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;
The given mass of the satellite, m = 1,300 kg
The distance between the center of the Earth and the center of the satellite, r = The length of the radius of the Earth + The height of orbit of the satellite
The given height of orbit of the satellite, h = 200 km
∴ r = R + h = 6,378 km + 200 km = 6,578 m
Therefore, by plugging in the values, we get;
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]
c. Solving the above equation gives;
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]
d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton
HELP, SCIENCE QUESTION I AM STUCK
6. Which of the following is NOT part of a circuit?
A. rim B. load C. power source D. conductor
please help me..im begging you
Answer: The equations in column A is matched with gas laws in column B as follows:
21. PV = nRT : (g) Ideal gas law
22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law
23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law
24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law
25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law
26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law
27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion
Explanation:
(A) Ideal gas law: It states that the product of pressure and volume is directly proportional to the product of number of moles and temperature.
So, PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant
T = temperature
Boyle's law: At constant temperature, the pressure of a gas is inversely proportional to volume.So, [tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Charles' law: At constant pressure, the volume of a gas is directly proportional to temperature. So,[tex]V \propto T\\\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\V_{1}T_{2} = V_{2}T_{1}[/tex]
Gay-Lussac's law: At constant volume, the pressure of a gas is directly proportional to temperature.So, [tex]P_{1}T_{2} = P_{2}T_{1}[/tex]
Avogadro's law: At same temperature and pressure, the volume of gas is directly proportional to moles of gas.So, [tex]V_{1}n_{2} = V_{2}n_{1}[/tex]
Combined gas law: When Boyle's law, Charles' law, and Gay-lussac's law are combined together then it is called combined gas law. So,[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\or, P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex]
Graham's law of effusion: It states that the rate of effusion of a gas is inversely proportional to the square root of mass of its particles.[tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex]
Thus, we can conclude that equation in column A is matched with gas laws in column B as follows:
21. PV = nRT : (g) Ideal gas law
22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law
23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law
24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law
25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law
26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law
27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion
Rewrite the false statements correctly
1.If an object sinks in one liquid and floats on another liquid,it implies that the density of second liquid is less than the first liquid.
2.The immersed volume of body in a liquid depends on density of the liquid.
3.Relative density of a body is usually expressed in kgm^-3
Explanation:
1. if an object sjnks in one liquid and floats on another liquid it implies that the density of second liquid is greater than the density of first liquid
The force of friction acting on a sliding crate is 223 N.
How much force must be applied to main- tain a constant velocity?
Answer:
Friction Opposes Motion of an Object.
Now
To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force
Net force = Pushing Force - Frictional Force
Recall
Net Force; F=Ma
Ma = P - Fr
Now the question asked for How Much force Must be applied to Maintain a Constant velocity.
In a Constant Velocity Motion... Acceleration do not change... Its Zero
So Putting this into the formula above
M(0) = P - Fr
0=P - Fr
Fr = P.
This means
That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force
Since Frictional Force; Fr =223N
The Applied Force(Pushing Force) Must be equal to 223N too.
Which type of wave causes particles of matter to vibrate in a direction
perpendicular to the direction of its motion?
O A. Sound
B. Transverse
C. Longitudinal
D. Compression
Answer:
C.) Longitudinal
Lamp is placed in the lamp holder. The switch is closed. The lamp glows brightly for a short time and then the lamp does not work. Explain these observations
Solution :
It is given that the lamp glows brightly for a shorter period of time when the switch is closed on it the switch is put on. But after the some time the lamp goes off and it stops working.
This is because as soon as we on the switch, the current start flowing to the lamp which makes the filament of the lamp to glow, but due to some issue, the current stop flowing even when the switch is on and this stops the lamp from glowing and hence the lamp does not work.
In this graph, calculate the speed of
segment A in m/s?
Answer:
The answer is Speed=2m/s
Explanation:
S=D/T
S=10m/5s
S=2m/s
A educação física, enquanto componente curricular da educação básica. Qual a tarefa que educação física?? Alguem me ajuda por fvr ??
Answer:
como assim qual a tarefa que educação física? se você me explicar melhor eu consigo te responder !!
Explanation:
id really appreciate it if you helped (I'll give you brainlist)
Explanation:
answer is in the picture above
A swimmer is capable of swimming at 1.4m/s in still water. a. How far downstream will he land if he swims directly across a 180m wide river? b. How long will it take him to reach the other side.
Answer:
t = 180 / 1.4 = 129 sec (time to swim horizontally across river)
S = 129 sec * V where V is speed of current and S is the distance he will be carried downstream
The problem does not specify V the speed of the river
Answer:
Explanation:
From the given information:
a) the distance(D) showing how dar downstream he will land can be computed as follows:
Assuming the current of the river = 0.2 m/s
[tex]D = \dfrac{180 \ m \times 0.2 \ m/s}{1.4 \ m/s}[/tex]
D = 36 m ÷ 1.4
D = 25.71 m
The required time (t) to reach the other side is:
time (t) = 180 m/ 1.4 m/s
time (t) = 128.57 seconds
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The current in a resistor is 5 A and the voltage between its terminals is 40 V. Calculate the resistance.
An iron wire has a resistance of 24 Ω. If the voltage across its ends is 12 V, calculate the current in the wire.
Answer:
1=8 ohms 2=0.5 Amps
Explanation:
A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10 m/s^2 find the maximum height reach by the stone and what is the net displacement and distance covered by the stone.
Explanation:
u=40
v=?
h=?
v²-u²=2gs
0²-40²=2×10×s
160=20s
s=160/20
=80m/s
total distance= upward distance ×downward distance
=80+80
=160m
total displacement=0 because u and v is the same.
Answer:
The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.
Explanation:
Final velocity v = 0
Initial velocity u = 40m/s
We know that,
Using equation of motion
[tex]v^{2} =u^{2} +2gh[/tex]
[tex]0-40^{2} =2[/tex] × [tex]10[/tex] × [tex]h[/tex]
The maximum height is:
[tex]h=80[/tex] [tex]m[/tex]
The stone will reach at the top and will come down
Therefore, the total distance will be:
[tex]s=h_{1} +h_{2}[/tex]
[tex]s=80m-80m=160m[/tex]
The net displacement is:
[tex]D=h_{1} -h_{2}[/tex]
[tex]D=80m-80m=0[/tex]
Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.
hope this helps.....
What type of electromagnetic waves do heat lamps give off?
A. infrared
B. ultraviolet
C. microwaves
D. radio waves
How to calculate displacement, velocity, acceleration.
how do you do question 2?
Answer:
Look at explanation
Explanation:
a) Weight is another term for how much gravity is on an object and can be calculated by using the local gravity*mass so in order to find weight divide 1200 by g
m= 1200/9.8= 122.45kg
b) We know Fnet=ma and Fnet=Fapp+Fresistivity so Fapp+Fresistivity=ma
Plug in values
500+Fresistivity=122.45*2
solve for Fresistivty= 244.9-500=-255.1N (the reason it is negative is because it is in the opposite direction
c)Power= ΔE/Δt and we also know ΔE=ΔWork so Power= ΔWork/Δt. If a person pulls harder, they have a greater force and since mass is constant, acceleration is greater and since the amount of time needed to cover A to B is reduced since x-x0=v0t+1/2at²(v0=0, when you solve for t it will be lower because acceleration increases). If t decreases than Power increases by inverse proportionality. Work =Fd if the amount of Force increases by distance travelled remains constant than work also increases so power will also increase.
A truck travelling at 30m/s decelerates at 1.5m/s². How far does it travel during the 10th second after the brakes are applied?
Answer
225 meters.
Explanation:
x=x0+30t-(1/2)(1.5)t^2
x=0+30(10)-(1/2)(1.5)(10)^2
x=300-75
x=225
A ship anchored at sea is rocked by waves that have crests Lim apart the waves travel at 70m/S, at what frequency do the waves reach the ship?
Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?
Answer:
0.7 Hz
Explanation:
Applying,
v = λf............... Equation 1
Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave
make f the subject of the equation
f = v/λ................. Equation 2
From the question,
Given: v = 70 m/s, λ = 100 m ( distance between successive crest)
Substitute these values into equation 2
f = 70/100
f = 0.7 Hz
Hence the frequency at which the wave reach the ship is 0.7 Hz
