Answer:
There is a lot of instruments used for measurement, may I ask which one are you referring to?
Señalar la importancia de las capacidades fisico-motiz que se desarrollan en el futbol de salon y dar un ejemplo para cada uno
La respuesta correcta para esta pregunta abierta es la siguiente.
A pesar de que no anexas opciones o incisos para responder, podemos comentar lo siguiente.
La importancia de las capacidades físico-motriz que se desarrollan en el futbol de salón son determinantes para desarrollar o maximizar las actividades propias de este deporte con objeto de rendir al máximo y aspirar al mejor de los resultados.
Estas capacidades físico-motrices son las que le permiten a un jugador realizar su máximo esfuerzo, mejorar su desempeño físico y conseguir resultados positivos.
Estamos hablando de la fuerza, la velocidad y la resistencia.
La velocidad es la aceleración que el jugador de futbol necesita para aumentar su velocidad de un punto A, a un punto B, en el menor tiempo posible.
La resistencia es la capacidad del jugador de futbol para mantener ese nivel de aceleración y desempeño, sin bajar su rendimiento. Su capacidad física debe ser resistente para ser constante en su rendimiento físico.
La fuerza es la potencia con la que desempeña los movimiento físicos dentro de la cancha.
1. Compared to most other metals, what properties do the alkali metals have? А low melting points and high densities B low melting points and low densities C high melting point and high densities D high melting points and low densities
Answer:
B. low melting points and low densities
Explanation:
Alkali metals are any of the monovalent elements found in Group IA of the periodic table. They readily lose their one valence electron to form ionic compounds with nonmetals. Examples of alkali metal are Potassium (K), Lithium (L), and Sodium (Na).
Compared to most other metals, the chemical properties that alkali metals have are low melting points (28.5°C) and low densities that is typically less than 1 grams per cubic centimeters.
what is rotation and revolution
A 75.0 kg man pushes backward on a 300.0 kg boat with a force of 150.0N causing the boat to accelerate backward at 0.5m/s^2. What is the acceleration of the man?
A) 150.0 m/s^2
B) 8.00 m/s^2
C) 2.00 m/s^2
D) 4.00 m/s^2
Answer:
C) 2.00 m/s^2
Explanation:
F = m*a
150N = 75kg(a)
a = 150N/75kg
a = 2.0m/s²
Magnets are formed from iron, nickel, or cobalt when the ______________________ line up in the same direction.
Answer:
electrons
In substances such as iron, cobalt, and nickel, most of the electrons spin in the same direction. When you rub a piece of iron along a magnet, the north-seeking poles of the atoms in the iron line up in the same direction. The force generated by the aligned atoms creates a magnetic field, hence it acts as a magnet.
the materials that can be magnetized this way are called ferromagnetic materials, but it is not permanent and after some times it loses all the magnetic properties.
please if you find this answer helpful mark it as brainliest
Magnets are formed from iron, nickel, or cobalt when the electrons in their atoms' outer energy levels line up in the same direction that is present in Option A, as the alignment creates a magnetic field that can attract or repel other magnets or magnetic materials.
What are magnets?At the atomic level, electrons orbit the nucleus in shells or energy levels, and electrons have a property called spin, which creates a tiny magnetic field around them. Normally, electrons in an atom have random spin directions, so their magnetic fields cancel each other out and the atom has no overall magnetic field. In some materials, such as iron, nickel, and cobalt, the electrons in the outer energy level can line up in the same direction, creating a net magnetic field. This is known as ferromagnetism.
Hence, magnets are formed from iron, nickel, or cobalt when the electrons in their atoms' outer energy levels line up in the same direction that is present in Option A,
Learn more about the magnets here.
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question is incomplete, complete question is below
Magnets are formed from iron, nickel, or cobalt when the ______________________ line up in the same direction.
A)electrons in their atoms' outer energy levels
B)proton explodes
Una muestra de agua (líquida) de 1220 kg se encuentra a 0° C y baja se temperatura hasta -29°C mientras se congela en el proceso. ¿Cuánta energía es liberada al ambiente (Mega Joules) ?
Answer:
-148,6 MJ
Explanation:
Dado que la liberación de calor al medio ambiente es;
H = mcθ
Dónde;
m = masa de agua
c = capacidad calorífica específica del agua
θ = aumento de temperatura
H = 1220 kg × 4200 × [-29-0]
H = -148,6 MJ
State Pascal’s law. (2) b) The area of one end of a U-tube is 0.01 m2 and that of the other
end the force 1 m2, when a force was applied on the liquid at the first end, the force experienced at
the other end was
Answer:
Pascal Law's says that:
If the area of one end of a U-tube is A, and the area of the other end is A'. then if we apply a force F in the first end (the one of area A), the force experienced at the other end must be:
F' = F*(A'/A).
b) Now we can apply this to our particular case:
if the area of one end is 0.01m^2, and the area of the other end is 1m^2
Then we have:
A = 0.01m^2
A' = 1m^2
So, if now we apply a force F in the first end, the force experienced at the other end will be:
F' = F*(1m^2/0.01m^2) = F*100
This means that the force in the other end must be 100 times the force in the first end.
Look at the image of a sports car. It has less resistance than a normal car because it is. What word fills the gap?
Answer:
It is streamlined
Explanation:
The speed of a car as well as its air resistance and fuel consumption has a lot to do with its shape.
A car that is streamlined glides through air with less resistance, moves with a higher speed and consumes less fuel.
The essence of streamlining is to reduce the aerodynamic drag on the car thereby increasing its speed, decreasing its air resistance and fuel consumption.
Answer:
it is streamlined
Explanation:
What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What is the direction of this force?
Answer:
88.34 N directed towards the center of the circle
Explanation:
Applying,
F = mv²/r................... Equation 1
F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.
But,
v = 2πr/t................... Equation 2
Where t = time, π = pie
Substitute equation 2 into equation 1
F = m(2πr/t)²/r
F = 4π²r²m/t²r
F = 4π²rm/t²............. Equation 3
From the question,
Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s
Constant: π = 3.14
Substitute these values into equation 3
F = 4(3.14²)(0.7)(0.8)/0.5²
F = 88.34 N directed towards the center of the circle
Which of the following are correct statements about the way an atom is put
together?
Check all that apply.
A. The valence shell is the innermost shell of electrons in an atom.
B. Electrons are found in the nucleus of an atom.
O C. There are two electrons in all electron shells.
O D. Electrons are in layers around the nucleus.
***** multiple choice ******
Answer:
valenc e shell
Explanation:
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament is a blackbody radiator? The filament of a particular electric lamp can be considered as a 90%blackbody radiator. calculate the energy per second radiated when its temperature is 2000k if its surface area is 10∧-6 m²
Answer:
(a) [tex]\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}[/tex]
(b) P = 0.816 Watt
Explanation:
(a)
The power radiated from a black body is given by Stefan Boltzman Law:
[tex]P = \sigma AT^4[/tex]
where,
P = Energy Radiated per Second = ?
σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴
T = Absolute Temperature
So the ratio of power at 250 K to the power at 2000 K is given as:
[tex]\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}[/tex]
(b)
Now, for 90% radiator blackbody at 2000 K:
[tex]P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4[/tex]
P = 0.816 Watt
Question 1 of 35
Which statement applies only to magnetic force instead of both electric and
magnetic forces?
A. It can push objects apart.
B. It acts between a north pole and a south pole.
C. It acts between objects that do not touch.
D. It can pull objects together.
A 1kΩ resistor is placed across potential difference of 100 V. Find the voltage drop across resistor.
Answer: 100 V
Explanation:
Given
Resistance [tex]R=1\ k\Omega[/tex]
Potential difference [tex]V=100\ V[/tex]
If the resistor is placed across the potential difference, then it receives a voltage of 100 V as it parallel to the source.
Voltage drop across resistor is 100 V.
plz help me yesterdays answer was wrong....
Name the Thermal expansion of fluids such as liquid and gas
Answer:
water
mercury in thermometer
volume gas thermometer
jar lids
power lines
Calculate the power of a train that uses 2MJ of chemical energy in 4
seconds.
Answer:
500 kW
Explanation:
Given that,
Energy of the train, E = 2 MJ = 2×10⁶ J
Time, t = 4 seconds
We need to find the power of the train. We know that,
Power = energy/time
So,
[tex]P=\dfrac{2\times 10^6}{4}\\\\P=500000\ W\\\\or\\\\P=500\ kW[/tex]
So, the power of the train is equal to 500 kW.
Which property of an object determines the strength of the buoyant force acting on the object in a fluid? A) weight B)surface area C) volume
what is the formula to calculate liquid pressure?/
1. set up the equation
2. multiply the variables
3. analyze your results
The momentum of an object is 35 kg•m/s and it is travelling at a speed of 10 m/s.
a) What is the mass of the object?
Answer:
[tex]{ \bf{momentum = mass \times velocity}} \\ \\ { \tt{35 = m \times 10}} \\ { \tt{mass = 3.5 \: kg}}[/tex]
A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.0 m/s without any friction. When a car is going 31.8 m/s on this curve, what minimum coefficient of static friction is needed if the car is to navigate the curve without slipping?
Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of
a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²
There are two forces acting on the car in this situation:
• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n
• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)
By Newton's second law, the net forces in the perpendicular and parallel directions are
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) = ma
==> sin(θ) = a/g ==> θ = arcsin(a/g) ≈ 17.8°
(Notice that in the paralell case, the positive direction points toward the center of the curve.)
When rounding the curve at 31.8 m/s, the car's radial acceleration changes to
a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²
and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) + f = ma
The first equation gives
n = mg cos(θ)
and substituting into the second equation, we get
mg sin(θ) + µmg cos(θ) = ma
==> µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12
Answer:
Explanation:
You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is
[tex]F_c=\frac{mv^2}{r}[/tex] where
[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,
m is the mass of the car,
v is the velocity with which the car is moving, and
r is the radius of the circle that the car is moving around.
For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become
[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by
f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).
Going on and getting buried even deeper,
[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:
μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:
μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:
μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is
μ = 1.4
The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,
On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.
Accelaration is a derived quantity .Give reason
Answer:
The accerlation is a derived from the other quantities like change in velocity/time take etc.
How are ions formed ?
Answer:
Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the ..
Answer:
Ions are formed when atoms lose or gain electrons, simply when electrons from the metal transfers electrons from its outermost shell( forming a positively charged metallic ion) to the outermost shell of the non-metallic atom( forming a negatively charged ion)
PLEASEE HELPP!!!
Describe melting.
Describe evaporation.
Describe boiling.
Describe condensation.
Describe freezing.
Melting: the substance changes back from the solid to the liquid
Evaporation: the process by which water changed from a liquid to a gas.
Boiling: the process by which a liquid turns into a vapor when it is heated to it's boiling point.
Condensation: the substance changed from a gas to a liquid
Freezing: the substance changed from a liquid to solid.
Answer:
Melting is a process that causes a substance to change from a solid to a liquid.
Evaporation is the process of turning from liquid into vapour.
Boiling is the rapid vaporization of a liquid, which occurs when a liquid is heated to its boiling point, the
Condensation is the process of water vapor turning back into liquid water
Freezing is a phase transition where a liquid turns into a solid when its temperature is lowered below its freezing point
which principle is used in mass spectrograph to estimate the mass of a charged particle
An object dropped from a cliff falls with a constant acceleration of 10 m/s2.Find its speed 5 s after it was dropped.
Answer:
50m/s2
Explanation:
u=0m/s
a=10m/s
v=?
t=5s
v=u+at
=0+(10×5)
=0+50
=50m/s
A car is travelling at 60m/s. and brakes to a speed of 14m/s, in 2.7 seconds. What is the deceleration?
Answer:
by using v = u + at equation we can find "a"
14 = 60 - 2.7a
2.7a = 60 - 14
2.7a = 46
decceleration = 17.03
A point charge of +3.0 X 10-7 coulomb is placed 2.0 X 10-2 meter from a second point charge of +4.0 X 10-7 coulomb. What is the magnitude of the electrostatic force on the charges?
A. 6.0 X 10-12 N
B. 3.0 X 10-10 N
C. 5.4 X 10-2 N
D. 2.7 N
Answer:
D. 2.7 N
Explanation:
Applying
F = kq'q/r²................ Equation 1
Where F = force, k = coulomb's constant, q' = first charge, q = second charge, r = distance between the charge
From the question,
Given: q' = +3.0×10⁻⁷ C, q = +4.0×10⁻⁷C, r = 2.0×10⁻² m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 1
F = ( +3.0×10⁻⁷)(+4.0×10⁻⁷)(8.98×10⁹)/(2.0×10⁻²)²
F = 26.94×10⁻¹ N
F = 2.694 N
F ≈ 2.7 N
The amount of work required to increase the distance between -6μC and 4μC from 6 cm to 18 cm will be
Answer:
W = 1.8 J
Explanation:
The amount of work required to move the given charges can be found by using the following formula:
[tex]W = \frac{kq_1q_2}{\Delta r} \\\\[/tex]
where,
W = Work done = ?
k = Colomb's constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 6 μC = 6 x 10⁻⁶ C
q₂ = magnitude of second charge = 4 μC = 4 x 10⁻⁶ C
Δr = change in distance = 18 cm - 6 cm = 12 cm = 0.12 m
Therefore,
[tex]W = \frac{(9\ x\ 10^9\ Nm^2/C^2)(6\ x\ 10^{-6}\ C)(4\ x\ 10^{-6}\ C)}{0.12\ m}[/tex]
W = 1.8 J
which force is responsible for the moon revolving around the sun?
Brains or Brawl? List the reason why you choose what you choose.
What is the work input to lift a box of 810 with a 50% mechanical efficiency?
Answer:
Work input = 1620
Explanation:
Given the following data;
Work output = 810
Efficiency = 50%
To find the work input;
Mathematically, efficiency is given by the formula;
[tex] Efficiency = \frac {Out-put \; work}{In-put \; work} * 100 [/tex]
Substituting into the formula, we have;
[tex] 50 = \frac {810}{WI} * 100 [/tex]
Cross-multiplying, we have;
[tex] 50I = 810 * 100 [/tex]
[tex] 50I = 81000 [/tex]
[tex] WI = \frac {81000}{50} [/tex]
Work input = 1620