Answer:
1) Force is the action of pull or push that changes or tends to change the position of a body. Its SI unit is newton.
2) The types of force are:
• Pulling and Pushing force
• Frictional force
• Gravitational force
•Muscular force
• Centripetal and Centrifugal force
Calculate the efficiency of the following appliances:
1. A radiator that converts 1000) of electrical energy into 900J
of heat energy and 100J of light energy
2. A torch that converts 100J of chemical energy into 35) of
light energy and 65J of heat energy
3. A car that converts 10,000J of chemical energy into 6000) of
kinetic energy and 4000J of heat energy.
4. An energy saver light converts 1,000J of electrical energy
into 7003 of light energy and 300J of heat energy.
5. A speaker converts 100J of electrical energy into 50J of
sound energy and 50J of heat and kinetic energy.
we
Answer:
1. The efficiency of the radiator is 90 %
2. The efficiency of the torch is 65 %
3. The efficiency of the car is 40 %
4. The efficiency of the energy saver is 70 %
5. The efficiency of the speaker is 50 %
Explanation:
Efficiency = (Useful energy out ÷ Total energy in) × 100 J
1. Useful energy = 900 J
The total energy in = 1000 J
The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %
2. Useful energy = 65 J
The total energy in = 100 J
The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %
3. Useful energy = 4,000 J
The total energy in = 10,000 J
The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %
4. Useful energy = 700 J
The total energy in = 1,000 J
The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %
5. Useful energy = 50 J
The total energy in = 100 J
The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %
in a cathode-ray tube, particles are fired at the screen. What are these particles
A toy car with a mass of 5.5 kg is moving horizontally over flat ground at a speed of 2.1 m/s. An unknown force then directly pushes the car for a distance of 3 meters, after which the car has a speed of 7.3 m/s. You may assume that air resistance and friction are both negligible. What was the magnitude of the unknown force
Answer:
The magnitude of the unknown force is 44.8 N.
Explanation:
The force can be found with Newton's second law:
[tex] F = ma [/tex]
Where:
m: is the mass of the toy car = 5.5 kg
a: is the acceleration
F: is the force =?
We can calculate the acceleration with the following kinematic equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex] v_{f} [/tex]: is the final speed = 7.3 m/s
[tex] v_{0} [/tex]: is the initial speed = 2.1 m/s
d: is the distance traveled = 3 m
Hence, the acceleration is:
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{(7.3 m/s)^{2} - (2.1 m/s)^{2}}{2*3 m} = 8.15 m/s^{2} [/tex]
Finally, the magnitude of the force is:
[tex]F = ma = 5.5 kg*8.15 m/s^{2} = 44.8 N[/tex]
Therefore, the magnitude of the unknown force is 44.8 N.
I hope it helps you!
Please help me with these I might need more than only 1 person to answer
Explanation:
a) copper
b) olive oil
Hope it helps✌✌
7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.
[tex] \\ \tt \longmapsto \: h = - 4.8 {t}^{2} + 8.2t + 1.8 \\ \\ \tt \longmapsto \: h = - 4.8(1.7) {}^{2} + 8.2 \times 1.7 + 1.8 \\ \\ \tt \longmapsto \: - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 17.06[/tex]
importance of measurement in our dairy life
Answer:
in order to know the length of something
Una partícula efectúa un MAS cuya ecuación es: x=0,3 cos (2t + /6) m. Determinar: Amplitud, frecuencia angular, fase inicial, periodo, frecuencia de oscilación y posición en t=0,25 s
Answer:
The answer is below
Explanation:
Una partícula efectúa un MAS cuya ecuación es: x=0.3cos (2t + π/6) m. Determinar: Amplitud, frecuencia angular, fase inicial, periodo, frecuencia de oscilación y posición en t=0.25 s
Solution:
La ecuación de la onda es:
x = A cos (ω t + Ф), donde:
A = amplitud, ω = frecuencia angular = 2 π / T = 2 π f, Ф = fase inicial, f = frecuencia, T = período
Por lo tanto, comparando la ecuación de la onda con x = 0.3cos (2t + π / 6), obtenemos:
a) A = 0.3
b) ω = 2 rad / s
c) Ф = π / 6 rad
d) ω = 2π / T
2 = 2π / T
T = 3.14 s
e) ω = 2πf
2 = 2πf
f = 0.32 Hz
f) en t = 0.25 s:
x (t) = 0.3cos (2 * 0.25 + π / 6) = 0.22 m
a teacher and two students are mesuring the speed of sound. the teacher makes loud sound by hitting two cymbals together. each students starts a stopwatch then they see the teacher hit the cymbals, they each stop their stop watch when they hear the sound. describe how a sound wave moves through the air.
Answer:
Slowly and smoothly lol
Answer:
S.I. on Rs. 1600 = T.D. on Rs. 1680. Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%. Time =100 x 80year=1year = 4 months.1600 x 153
Explanation:
Is this you are?
Group elements number 11 to 20 as either metallic, non metallic or metalloid.
Answer:
The elements are grouped into the different substances by color. As you can see, Lithium, Beryllium, Sodium, Magnesium, Aluminum, Potassium, and Calcium are metals out of the first 20 elements.
Hydrogen, Helium, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Phosphorus, Sulfur, Chlorine, and Argon, are non-metals within the first 20 elements.
Boron and Silicon count as Metalloids in the Periodic Table (properties of both metals and non-metals)
reference- socatric q and a
Explanation:
Answer:
(11-13) - Metals
(14) - Metalloid
(15-18) - Non- metals
(19-20) - Metals
Explanation:
The elements from 11 to 13 are metals. They are Sodium(Na), Magnesium(Mg) and Aluminum(Al) respectively.
The element 14 is a metalloid. It’s Silicon(Si).
The elements from 15 to 18 are non metals. The 18th element is a noble gas known as Argon(Ar).
The elements from 15 to 17 are Phosphorus(P), Sulphur(S) and Chlorine(Cl).
The elements from 19 to 20 are metals. They are Potassium(K) and Calcium(Ca).
Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iron
Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up
Write down the conservation of momentum?
A 250g object hangs from a spring that has a spring constant of 48.0 N/m and oscillates with an amplitude of 5.42cm
1)The magnitude of the objects acceleration when the displacement is 4.27 cm (down) is __ m/s^2
2)Given that the object has an amplitude of 5.42 cm the maximum speed that the object is __m/s
Explanation:
Given that,
Mass of an object, m = 250 g = 0.25 kg
Spring constant, k = 48 N/m
The amplitude of the oscillation, A = 5.42 cm = 0.0542 m
1. At equilibrium,
ma = kx
Where
a is the acceleration of the object
So,
[tex]a=\dfrac{kx}{m}\\\\a=\dfrac{48\times 0.0542}{0.25}\\\\a=10.4\ m/s^2[/tex]
2. The maximum speed of the object is :
[tex]v=A\omega\\\\v=A\sqrt{\dfrac{k}{m}}\\\\v=0.0542\times \sqrt{\dfrac{48}{0.25}}\\\\v=0.75\ m/s[/tex]
Hence, this is the required solution.
Assuming Faraday constant to be 96500c/mol and relative atomic mass of copper 63,calculate the mass of copper liberated by 2A current in 5min.ans 0.196gm
Answer: The mass of copper liberated is 0.196 g.
Explanation:
The oxidation half-reaction of copper follows:
[tex]Cu\rightarrow Cu^{2+}+2e^-[/tex]
Calculating the theoretical mass deposited by using Faraday's law, which is:
[tex]m=\frac{M\times I\times t(s)}{n\times F}[/tex] ......(1)
where,
m = actual mass deposited = ? g
M = molar mass of metal = 63 g/mol
I = average current = 2 A
t = time period in seconds = 5 min = 300 s (Conversion factor: 1 min = 60 sec)
n = number of electrons exchanged = 2
F = Faraday's constant = 96500 C/mol
Putting values in equation 1, we get:
[tex]m=\frac{63 g/mol\times 2A\times 300s}{2\times 96500 C/mol}\\\\m=0.196g[/tex]
Hence, the mass of copper liberated is 0.196 g.
Please help (I didn’t know what subject to put this in but it’s earth science)
Un auto se desplaza por una carretera recta a una velocidad de 85 km/h. Al cabo de 2 horas, ¿qué distancia habrá recorrido, en metros?
Answer:
Distancia, D = 170,000 metros
Explanation:
Dados los siguientes datos;
Velocidad = 85 km/h Tiempo = 2 horasPara encontrar la distancia recorrida, en metros;
Matemáticamente, la distancia recorrida por un objeto se calcula mediante la fórmula;
Distancia = velocidad * tiempo
Sustituyendo los valores en la fórmula, tenemos;
Distancia = 85 * 2
Distancia = 170 kilómetros
A continuación, convertiríamos el valor en kilómetros a metros;
Conversión:
1 kilómetros = 1000 metros
170 kilómetros = D metros
Multiplicación cruzada, tenemos;
D = 170 * 1000
D = 170,000 metros
Por lo tanto, habría viajado 170,000 metros después de 2 horas.
An ant moves towards the plane mirror with speed of 2 m/s & the mirror is moved towards the ant with the same speed. What is the relative velocity between the ant and its image?
We know
[tex]\boxed{\sf Relative\:velocity(V_{AB})=V_A-V_B}[/tex]
[tex]\\ \sf\longmapsto V_{AB}=2-2[/tex]
[tex]\\ \sf\longmapsto V_{AB}=0m/s[/tex]
The vector difference between the velocities of two bodies : the velocity of a body with respect to another regarded as being at rest compare relative motion
[tex]Relative velocity $\left(\mathrm{V}_{\mathrm{AB}}\right)=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$$$\begin{aligned}&\longmapsto \mathrm{V}_{\mathrm{AB}}=2-2 \\&\longmapsto \mathrm{V}_{\mathrm{AB}}=0 \mathrm{~m} / \mathrm{s}\end{aligned}$$[/tex]
What is relative velocity and its unit?The relative velocity of an object with respect to another is the velocity with which one object moves with respect to another object. The unit of velocity can be referred to as the ratio of unit of distance and that of time. The SI unit of Relative velocity is meter per second.
What is absolute velocity?The concept of absolute velocity is mainly used in turbomachinery design and defines the velocity of a fluid particle in relation to the surrounding, stationary environment. Together with the relative velocity (w) and the circumferential speed (u), it forms the velocity triangle.
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Một con lắc lò xo gồm 1 quả nặng có m= 0,2kg treo vào lò xo có độ cứng k= 100N/m, cho vật dao động điều hoà theo phương thẳng đứng với biên độ A= 1,5 cm. Lực đàn hồi cực đại có giá trị
1. Imagine a bowling ball with a mass of 5,0 kg that is dropped from a height of 100m,
a. At the moment the bowling ball is dropped it is at a height of 100m, What is its
potential energy?
b. At the moment the bowling ball is dropped, it is not yet moving. Therefore, its
kinetic energy is what?
c. What is the total of both potential and kinetic energy?
a.
Given,
height (h) = 100m
mass (m) = 5kg
acceleration due to gravitation (g) = 9.8ms^-2
Potential energy
= mgh
= 5kg × 9.8ms^-1 × 100m
= 4900 kgm²s^-2
= 4900 J
b.
Since the ball is not moving yet, its kinetic energy is 0.
c.
The total of potential and kinetic energy at every point of the journey is same, i.e., 4900 J.
What is Plancks Constant? Need it ASAP
Answer:
The Planck constant, or Planck's constant, is a fundamental physical constant denoted h, and is of fundamental importance in quantum mechanics. A photon's energy is equal to its frequency multiplied by the Planck constant. Due to mass–energy equivalence, the Planck constant also relates mass to frequency.
Theory:Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.
E = hf
[tex] \sf \: h = 6.626 \times {10}^{ - 34} \: kg \: {m}^{2} {s}^{ - 1} [/tex]
solve two questions under inclined plane using Newton 2nd law
Explanation:
If the mass of the inclined plane is large enough, could N ever be equal to mgcosθ. Reasons?
So far I've come up with: mgcosθ−N=ma meaning if N were to equal mgcosθ, ma=0. Since the surfaces are frictionless and since N is acting on the inclined plane too(it'll have a horizontal component), this isn't possible. Am I right? Where am I going wrong?
what is the relation between centre of gravity and stability
Explanation:
tilting it will raise the height of its center of gravity.
. A radio station transmitting at a frequency of 200KH, emits waves of wavelength 1.5 km.the velocity of the radio waves is
Answer:
Explanation: as,
v=f∧
v=200×10³×1.5×10³
v=2.7×10^7ms⁻¹
The velocity of the radio waves is 3 × 10⁸ ms⁻¹.
A radio station frequency = 200 KH
wavelength = 1.5 km
How the velocity of the radio waves are calculated?Radio waves in the electromagnetic spectrum has the longest wavelength and it will always be below 300GHz. The radio waves can be generated with acceleration through some charged particles. Only through the transmitter via antenna the radio waves can gets transmitted. Radio waves can be used in all the electronic devices as mobile phones, radio communication, radars and navigations.
V = f λ
Velocity, v = ( 200 × 10³ ) × ( 1.5 × 10³ )
= ( 300 × 10⁶ )
v = 3 × 10⁸ ms⁻¹.
Hence, the velocity of the radio waves is 3 × 10⁸ ms⁻¹.
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In a Rutherford scattering experiment, alpha parti- cles having kinetic energy of 7.70 MeV are fired toward a gold nucleus that remains at rest during the collision. The alpha particles come as close as 29.5 fm to the gold nucleus before turning around. (a) Calculate the de Broglie wave- length for the 7.70-MeV alpha particle and compare it with the distance of closest approach, 29.5 fm. (b) Based on this comparison, why is it proper to treat the alpha particle as a particle and not as a wave in the Rutherford scattering experiment
(a) The de Broglie wavelength is approximately 5.175 × 10⁻¹⁵ meters. The wavelength is lesser than the distance of closest approach
(b) It is proper to treat the alpha particle as a particle and not as wave because the distance of closest approach is much larger than and not comparable to its wavelength for the alpha particle for the alpha particle to be treated as a wave
The given parameters are;
The kinetic energy of the alpha particles = 7.70 MeV = 1.23368 × 10⁻¹² J
The distance from the gold nucleus the alpha particles reach = 29.5 fm
(a) The de Broglie wavelength of a particle is given as follows;
[tex]\mathbf{\lambda = \dfrac{h}{p}}[/tex]
Where;
λ = The wavelength
h = Planck's constant = 6.62607004 × 10⁻³⁴ m²·kg/s
p = The momentum of the particle = Mass of an electron, m × Velocity, v
The mass of an alpha particle, m ≈ 6.645 × 10⁻²⁷ kg
Therefore;
[tex]\lambda = \dfrac{h}{m \times v}[/tex]
The kinetic energy of the alpha particle, K.E. = (1/2)·m·v²
∴ v = √(2 × K.E./m)
Therefore;
[tex]\lambda = \dfrac{h}{m \times \sqrt{2 \times \dfrac{K.E.}{m} } } = \dfrac{h}{ \sqrt{2 \times m \times K.E.} }[/tex]
Plugging in the values of the variables gives;
[tex]\lambda = \dfrac{6.62607004 \times 10 ^{-34} }{ \sqrt{2 \times 6.645 \times 10 ^{-27} \times 1.23368 \times 10^{-12} } } \approx 5.175 \times 10^{-15}[/tex]
The de Broglie wavelength of the alpha particle, λ ≈ 5.175 × 10⁻¹⁵ m
The distance of closest approach = 29.5 fm = 29.5 × 10⁻¹⁵ m
Compared to the distance of closest approach, the wavelength of the alpha particle is lesser than the distance of closest approach
(b) Given that the distance of closest approach is six times larger than the wavelength of the alpha particle, and alpha particle behaving as waves are expected to approach closer to the gold nucleus in the region of their wavelength before deflection, therefore, the larger distance of closest approach is indicative of a charged particle to charged particle interaction, and therefore, particle behavior of alpha particles.
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Một mạch kín hình vuông cạnh 20cm đặt vuông góc với 1 từ trường đều có độ lớn thay đổi theo thời gian. Trong khoảng thời gian 0,01s , cho độ lớn cảm ứng từ tăng đều từ 0 lên đến 0,05T. Biết điện trở của mạch là 0,5. Tính cường độ dòng điện cảm ứng trong mạch
Answer:
please do write it in english
i. The lift raises a car to a height of 1.8 m using a force of 5500 N. How much work does the lift
perform? (1 point)
Work = force x distance
Work = 5500 x 1.8
Work = 9900 N
The work measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.
Work = force x distance
Work = 5500 x 1.8
Work = 9900 N
therefore, work does 9900 N
What is work?Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter (N · m).
What is work and energy?Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.
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express in standard form
(I)0.000038
Answer:
3.8 x 10^-5
Explanation:
all you doing is putting it in the standard form I hope this help
A 111 kg linebacker running at 1.9 m/s and a 82 kg quarterback running at 3 m/s have a head-on collision in midair. The linebacker grabs and holds onto the quarterback. (a) What is their combined speed immediately after the collision
Answer:
Explanation:
mass of linebacker, m = 111 kg
initial velocity, u = 1.9 m/s
mass of quarterback, m' = 82 kg
initial velocity, u' = 3 m/s
(a) Let they are moving in the same direction, the velocity is v after collision.
Use conservation of momentum
m u + m' u' = (m + m') v
111 x 1.9 + 82 x 3 = (111 + 82) v
v = 2.4 m/s
Let they are moving in opposite direction, the velocity is v after collision.
Use conservation of momentum
m u - m' u' = (m + m') v
111 x 1.9 - 82 x 3 = (111 + 82) v
v = - 0.18 m/s
A bullet with a mass mb=13.5 g is fired into a block of wood at velocity vb=245 m/s. The block is attached to a spring that has a spring constant k of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass mw of the wooden block.
Momentum is conserved, so the sum of the momenta of the bullet and block before collision is equal to the momentum of the combined bullet-block system,
[tex]m_bv_b+m_wv_w = (m_b+m_w)v[/tex]
where v is the speed of the bullet-block system. The block starts at rest so it has no initial momentum, and solving for v gives
[tex]v = \dfrac{m_b}{m_b+m_w} v_b[/tex]
The total work W performed by the spring on the bullet-block system as it is compressed a distance x is
[tex]W = -\dfrac12kx^2[/tex]
where k is the spring constant, and the work done is negative because the restoring force of the spring opposes the bullet-block as it compresses the spring.
By the work-energy theorem, the total work done is equal to the change in the bullet-block's kinetic energy ∆K, so we have
[tex]W_{\rm total} = W = \Delta K[/tex]
The bullet-block starts moving with velocity v found earlier and comes to a stop as the spring slows it down, so we have
[tex]-\dfrac12kx^2 = -\dfrac12(m_b+m_w)v^2 \implies kx^2 = \dfrac{{m_b}^2}{m_b+m_w}{v_b}^2[/tex]
Solve for [tex]m_w[/tex]:
[tex]m_w=\dfrac1k\left(\dfrac{m_bv_b}x\right)^2-m_b[/tex]
[tex]m_w=\dfrac1{205\frac{\rm N}{\rm m}}\left(\dfrac{(0.0135\,\mathrm{kg})\left(245\frac{\rm m}{\rm s}\right)}{0.350\,\rm m}\right)^2-0.0135\,\mathrm{kg}\approx \boxed{0.422\,\mathrm{kg}}[/tex]
what is efficiency of a machine?
Answer:
Efficiency of a machine is defined as the ratio of output work to input work in a machine . It is expressed in percentage and denoted by
η ( eta).
define mass defect and binding energy
Answer:
About Transcript. Nuclear binding energy is the energy required to split an atom's nucleus into protons and neutrons. Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. The binding energy of a system can appear as extra mass, which accounts for this difference.
Explanation:
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