Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
[tex]9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}[/tex]
W = 3.266 N
The mass of the meters stick is :
[tex]m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg[/tex]
So, the mass of the meter stick is 0.333 kg.
The mass of the meterstick is calculated using the principle of torque balance. The meterstick must have a mass that creates an equal opposite torque to the one created by the 1-kg rock. The calculation shows that the meterstick has a mass of 1/3 kg, which is not listed in the options, so the answer is (E) none of the above.
Explanation:The mass of the meterstick can be found using the principle of moments, also known as the principle of leverage, which states that for an object to be in rotational equilibrium, the sum of the clockwise moments about the pivot (fulcrum) must equal the sum of the counterclockwise moments. In this case, the 1-kg rock is hanging from the 0-cm mark and the meterstick balances when the fulcrum is at the 12.5-cm mark. The torque created by the 1-kg mass is given by its mass multiplied by its distance from the fulcrum, or 1 kg × 12.5 cm.
For the meterstick to balance, its center of mass must be located directly above the fulcrum. Since the meterstick is uniform, its center of mass is in the middle, at the 50-cm mark. This means that the mass of the meterstick acts 50 cm - 12.5 cm = 37.5 cm away from the fulcrum. Let's denote the mass of the meterstick as 'M'. The counterbalance torque provided by the meterstick is given by M kg × 37.5 cm.
Setting the torques equal for a balanced system:
M kg × 37.5 cm = 1 kg × 12.5 cm, which simplifies to M = 1/3 kg. Since 1/3 kg is equivalent to approximately 0.33 kg, the correct answer is (E) none of the above, as none of the provided options matches this result.
https://brainly.com/question/17014679
#SPJ3
A form of charge designated as negative positive or neutral is called
Answer:
Electrical charge
Explanation:
Electrical Charge is a form of charge, designated negative, positive, or neutral (without charge) that is found on the subatomic particles that make up all atoms
How many significant figures are present in the number 50,670?
El vector a tiene componented ax= 2.70 cm, ay=2.25 cm; y el vector b tiene componentes bx= 0.30 cm, by= 1.75 cm. Calcular las componentes de la resultante a+b
Answer:
Explanation:
Given that,
A vector A has x component to be 2.7cm and y component to be 2.25cm
Then,
A = 2.7•i + 2.25•j
A vector B has x component of 0.30cm and y component of 1.75cm
B = 0.3•i + 1.75•j
So, we want to find A+B
Addition of vectors
Generally
(a•i + b•j) + (c•i + d•j) = (a+c)•i +(b+d)•j
Vectors are added component wise.
So,
A + B = (2.7•i + 2.25•j) + (0.3•i + 1.75•j)
A + B = (2.7 + 0.3)•i + (2.25 + 1.75)•j
A + B = 3•i + 4•j
We can also find it magnitude and direction
Generally,
A = a•i + b•j
|A| = √(a²+b²)
<A = arctan(b/a)
So,
|A+B| = √(3²+4²) = √9+16 = √25
|A+B| = 5
And it's direction
< = arctan(y/x)
< = arctan(4/3)
< = 53.13°
The top and bottom surfaces of a metal block each have an area of A = 0.030 m 2, and the height of the block is d = 0.11 m. At the top surface of the block, a force F1 is applied to the right, while at the bottom surface of the block, a force F2 is applied to the left, causing a shear in the metal block. If F1 = F2 = 30 ⨯ 106 N and the displacement between the two edges due to the shear is 1.12 10-3 m, what is the shear modulus of the metal
Answer:
Shear modulus is equal to [tex]9.82\times 10^{10}N/m^2[/tex]
Explanation:
We have given area [tex]A=0.030m^2[/tex]
Force is given [tex]F_1=F_2=30\times 10^6N[/tex]
Height of the block d = 0.11 m
Change in height of the block [tex]\Delta d=1.12\times 10^{-3}m[/tex]
Stress is given by
[tex]stress=\frac{force}{area}[/tex]
[tex]stress=\frac{30\times 10^6}{0.030}=10^9N/m^2[/tex]
Strain is equal to
[tex]strain=\frac{\Delta d}{d}[/tex]
[tex]strain=\frac{1.12\times 10^{-3}}{0.11}=10.18\times 10^{-3}[/tex]
Shear modulus is equal to
Shear modulus [tex]=\frac{stress}{strain}[/tex]
[tex]=\frac{10^9}{10.18\times 10^{-3}}=9.82\times 10^{10}N/m^2[/tex]
In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following pressure p, in atm, as a function of volume V, in liters: p = p0 V-6/5. Here p0 is a constant of units atm⋅L6/5. show answer Incorrect Answer 50% Part (a) Write an expression for the work W done on the gas when the gas is compressed from a volume Vi to a volume Vf.
The work done on the gas during an adiabatic compression process is calculated by integrating the pressure as a function of volume from the initial volume to the final volume. In this case, the integral ∫from V(i) to V(f) p₀ V^(-6/5) dV provides the value for the work done.
Explanation:In an adiabatic process, the work done W on the gas when it's compressed from a volume V(i) to a volume V(f) can be calculated by integrating the pressure-volume equation over the given volume range.
Given the pressure equation p = p₀ V^(-6/5), the work done on the gas (W) can be computed by the integral formula for work: W = ∫p dV = ∫from Vi to Vf p₀ V^(-6/5) dV. Evaluating this integral gives the work done on the gas during the compression process in the adiabatic process.
Learn more about the Adiabatic Process here:https://brainly.com/question/14930930
#SPJ3
In physics how much energy is required to raise the temp of 9.78kg of water from 40.82c to 52.07c? The specific heat of water in this state is 4186j/kg-c.
Answer:
469.6KJ
Explanation:
Heat energy required can be calculated using the formula
H = mc∆t where
m is the mass of the water
c is the specific heat capacity of the water
∆t is the change in temperature of the water
Given m = 9.78kg
c = 4186j/kg-c.
∆t = 52.07°C - 40.82°C
∆t = 11.25°C
H = 9.78 × 4186 × 11.25
H = 460,564.65Joules
= 460.6KJ
Answer:
the amount of energy required to raise the temperature of the water is 460564.65 J
Explanation:
The energy required to raise the temperature of water can be calculated as follows;
Q = mcΔθ
where;
Q is the quantity of heat or energy required to raise the temperature of water
m is mass of water
c is specific heat capacity of water
Δθ is change in temperature = T₂ - T₁
Given;
m = 9.78kg
c = 4186j/kg-c
Δθ = T₂ - T₁ = 52.07°C - 40.82°C = 11.25°C
Q = mcΔθ
Q = (9.78)(4186)(11.25)
Q = 460564.65 J
Therefore, the amount of energy required to raise the temperature of the water is 460564.65 J
Sound waves are converted into mechanical movements by the
Answer:
a.auditory ossicles
b.oval window
c.Round window
d.tympanic membrane
Answer is tympanic membrane
Explanation:
The tympanic membrane otherwise called the ear drum is a membrane shaped like a cone,it connects the outside to the inner ear,it serves to convert vibration from air into fluid membrane vibration a good example of mechanical waves for onwatd transmission into the cochlea of the inner ear through the oval window
Sound waves are converted into mechanical movements by the ear, specifically through structures in the middle and inner ear. The eardrum, ossicles, and cochlea play significant roles in this process, with the final conversion to electrical signals occurring in the cochlea.
Explanation:Sound waves are converted into mechanical movements by the ear, specifically the structures within the middle and inner ear.
When sound waves enter the ear, they strike the eardrum or tympanic membrane in the middle ear, causing it to vibrate. These vibrations are then transferred to three tiny bones in the middle ear called the ossicles, consisting of the malleus, incus, and stapes. The vibrations move these bones, with the stapes pushing into the oval window of the cochlea in the inner ear.
The cochlea is filled with fluid and lined with tiny hair-like structures called stereocilia. The mechanical movement from the ossicles creates waves in this fluid, which cause the stereocilia to move. This movement is converted into electrical signals that the brain interprets as sound.
Learn more about Sound conversion in ear here:https://brainly.com/question/32551108
#SPJ6
A 0.100-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone
Answer:
Change in momentum of the stone is 3.673 kg.m/s.
Explanation:
Given:
Mass of the ball on the horizontal the surface, m = 0.10 kg
Velocity of the ball with which it hits the stone, v = 20 m/s
According to the question it rebounds with 70% of the initial kinetic energy.
We have to find the change in momentum i.e Δp
Before that:
We have to calculate the rebound velocity with which the object rebounds.
Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".
⇒ [tex]KE_1=0.7\times \frac{mv^2}{2}[/tex]
⇒ [tex]KE_1=0.7\times \frac{0.10\times (20)^2}{2}[/tex]
⇒ [tex]KE_1=0.7\times \frac{0.10\times 400}{2}[/tex]
⇒ [tex]KE_1=14[/tex] Joules (J).
Rebound velocity "v1".
⇒ [tex]KE_1=\frac{m(v_1)^2}{2}[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2KE_1}{m} }[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2\times 14}{0.10} }[/tex]
⇒ [tex]v_1=16.73[/tex]
⇒ [tex]v_1=-16.73[/tex] m/s ...as it rebounds.
Change in momentum Δp.
⇒ [tex]\triangle p= m\triangle v[/tex]
⇒ [tex]\triangle p= 0.10\times (20-(-16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (20+16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (36.73)[/tex]
⇒ [tex]\triangle p = 3.673[/tex] Kg.m/s
The magnitude of the change in momentum of the stone is 3.673 kg.m/s.
Luis is trying to push a box of new soccer balls across the floor. In the illustration, the arrow on the box is a vector representing the force that Luis exerts . If the box is not moving, which of the filling must be true
Answer:
Complete question
Luis is trying to push a box of new soccer balls across the floor.
If the box is not moving, which of the following must be true?
A. The box is exerting a larger force on Luis than he is exerting on the box
B. There is another force acting on the box that balances Luis's force.
C. Luis is applying a force that acts at a distance.
D. There is no force of friction acting on the box.
Explanation:
Using newton second law of motion
ΣF = ma
Now, for a body not to move when a force is acting on it means that the body is in equilibrium and acceleration a = 0
Therefore,
Fnet = 0
So, if he is applying a force F to push the box and the box is not moving then, there is an external force that is pushing the force back opposite the direction he his pushing and this force counterbalance is own force.
F—F' = 0
F' = F
So, F' is the counter balance force and it is equal to the force applied by Luis
Or it might be frictional force, because if the static friction is not overcome, then, the body will not leave it's state of rest. So if the fictional force is very high, then the box will not leave it rest position and we also know that frictional force opposes motion,
F—Fr=0
F = Fr
So using this explanation,.
The answer is B
B. There is another force acting on the box that balances Luis's force.
If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T west, what is the direction and the magnitude of the velocity?
Answer:
Explanation:
Given that,
Force is downward I.e negative y-axis
F = -2 × 10^-14 •j N
Magnetic field is westward, +x direction
B = 8.3 × 10^-2 •i T
Charge of an electron
q = 1.6 × 10^-19C
Velocity and it direction?
Force in a magnetic field is given as
F = q(V×B)
Angle between V and B is 270, check attachment
The cross product of velocity and magnetic field
F =qVB•Sin270
2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2
Then,
v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)
v = 1.51 × 10^6 m/s
Direction of the force
Let x be the direction of v
-F•j = v•x × B•i
From cross product
We know that
i×j = k, j×i = -k
j×k =i, k×j = -i
k×i = j, i×k = -j OR -k×i = -j
Comparing -k×i = -j to given problem
We notice that
-F•j = q ( -V•k × B×i)
So, the direction of V is negative z- direction
V = -1.51 × 10^6 •k m/s
The starting conditions of an oscillator are characterized by the
Answer:
The phase constant
Explanation:
The phase constant tells how much a signal is shifted along the x-axis. A phase constant of ϕ means that each value of the signal happens ϕ amount of time earlier. If the signal has a beginning, then a phase constant of ϕ means the signal occurs that much sooner.
The starting conditions of an oscillator are characterized by the phase constant. Option B is correct.
Phase Constant:
This represents the number of oscillation per cycle of wave. It is denoted by [tex]\bold {\phi}[/tex]. It is also known as Propagation constant. The phase constant can be calculated using the formula.
[tex]\bold {\phi = \dfrac {2\pi }{\lambda}}[/tex]
Where, lambda is wavelength.
Frequency is the number of waves passing from a point per unit time.Amplitude is the height of the wave from its mid point.Therefore, option B is correct. The starting conditions of an oscillator are characterized by the phase constant.
To know more about phase constant,
https://brainly.com/question/17095727
Identify which type of natural selection is described in each case below.
Results in the evolution of the species:
Results in new species:
Favors traits at the two extremes:
No increase in diversity:
One extreme trait is favored:
Diversity decreases:
Diversity increases:
Answer:
1.) A
2.) B
3.) B
4.) A
5.) A
6.) C
7.) B
Explanation:
Answer:
A directional selection
B disruptive selection
B disruptive selection
A directional selection
A directional selection
C stabilizing selection
B disruptive selection
Explanation:
__________ is a type of molding in which thermoplastics are heated above melting temperature and, using an extruder, are forced into a closed die to produce a molding.
a. Injection molding
b. Blow molding
c. Thermoforming
d. Extrusion
Answer: The correct and is
D. Extrusion
Explanation:
What is extrusion?
Extrusion is a manufacturing process used to make basically pipes and hoses.
The pvc granules are melt into a liquid which is forced through a die, forming a long 'tube like' shape. The shape of the die determines the shape of the tube.
The extrusion is then cooled and forms a solid shape
Extrusion moulding is used to create products with a consistent cross-section.
A fisherman fishing from a pier observes that the float on his line bobs up and down, taking 2.4 s to move from its highest to its lowest point. He also estimates that the distance between adjacent wave crests is 48 m. What is the speed of the waves going past the pier?
(A) 20 m/s
(B) 1.0 m/s
(C) 10 m/s
(D) 5.0 m/s
(E) 115 m/s
Answer:
(c) 10m/s
Explanation:
to find the speed of the waves you can use the following formula:
[tex]v=\frac{\lambda}{T}[/tex]
λ: wavelength of the wave
T: period
the wavelength is the distance between crests = 48m
the period is the time of a complete oscillation of the wave. In this case you have that the float takes 2.4 s to go from its highest to the lowest point. The period will be twice that time:
T = 2(2.4s)=4.8s
by replacing you obtain:
[tex]v=\frac{48m}{4.8s}=10\frac{m}{s}[/tex]
the answer is (c) 10m/s
Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 1515 grams and velocity of 77 centimeters per second has a kinetic energy of 147147 ergs. Find the kinetic energy for a mass of 1010 grams and velocity of 99 centimeters per second?
Answer:
The second kinetic energy is 162 J.
Explanation:
Given that,
Mass, [tex]m_1=15\ g[/tex]
Velocity, [tex]v_1=7\ cm/s[/tex]
Kinetic energy, [tex]K_1=147\ ergs[/tex]
Mass, [tex]m_2=10\ g[/tex]
Velocity, [tex]v_2=9\ cm/s[/tex]
We need to find kinetic energy [tex]K_2[/tex]. Kinetic energy is given by :
[tex]K=\dfrac{1}{2}mv^2[/tex]
So,
[tex]\dfrac{K_1}{K_2}=\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}\\\\K_2=\dfrac{K_1}{\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}}\\\\K_2=\dfrac{147}{\dfrac{15}{10}\times \dfrac{7^2}{9^2}}\\\\K_2=162\ J[/tex]
So, the second kinetic energy is 162 J.
During which phase of the moon may a lunar eclipse occur?
Answer: New moon
Explanation: A solar eclipse is only possible during a new moon phase
why hot air balloon rises up in air n large ships Floyd in water
Answer:
Hot air is less dense than cool air; the heated air causes the balloon to rise simply because it is lighter than an equal volume of cold air.So hot air balloon rises up in air.
Large ships float in water as their density is lesser than the water so.
A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rotate. It is supported by a cable that makes an Angle of 53 degrees with the horizontal. If the person of weight 350 N stands 1.50 m from the wall. Find the magnitude of the tension T in the cable. (Show your workout instead of giving the final answer).
The magnitude of the tension T in the cable is approximately 468.164 N.
To find the tension T in the cable supporting the beam, we can analyze the forces acting on the beam in equilibrium.
First, let's consider the forces acting on the beam:
1. The weight of the beam, acting downward at its center (2.0 m from the wall).
2. The tension T in the cable, acting upward and at an angle of 53 degrees with the horizontal.
3. The reaction force at the pin connection, acting horizontally to the left.
4. The vertical force exerted by the person standing on the beam, which contributes to the vertical component of the tension in the cable.
The beam is in equilibrium, so the sum of the torques about any point must be zero. Let's take moments about the pin connection at the wall.
Clockwise torques:
- Weight of the beam: [tex]\(200 \, \text{N} \times 2.0 \, \text{m}\)[/tex]
- Vertical component of the tension: [tex]\(T \times \cos(53^\circ) \times 1.5 \, \text{m}\)[/tex]
Counterclockwise torques:
- Tension in the cable: [tex]\(T \times \sin(53^\circ) \times 1.5 \, \text{m}\)[/tex]
- Vertical force exerted by the person: [tex]\(350 \, \text{N} \times 1.5 \, \text{m}\)[/tex]
Since the beam is in equilibrium, these torques must balance:
[tex]\[ 200 \, \text{N} \times 2.0 \, \text{m} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 350 \, \text{N} \times 1.5 \, \text{m} \][/tex]
Now, we can solve for T:
[tex]\[ 400 \, \text{N} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ 400 \, \text{N} = T \times \sin(53^\circ) \times 1.5 \, \text{m} - T \times \cos(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ 400 \, \text{N} = T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} = 125 \, \text{N} \][/tex]
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
Now, we can calculate T:
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{(0.7986 - 0.6206) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{0.178 \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{0.267 \, \text{m}} \][/tex]
[tex]\[ T \approx 468.164 \, \text{N} \][/tex]
So, the magnitude of the tension T in the cable is approximately 468.164 N.
Write down the (real) electric and magnetic fields for a monochromatic plane wave of amplitude E0 , frequency w, and phase angle zero that is (a) traveling in the negative x direction and polarized in the z direction; (b) traveling in the direction from the origin to the point (1, 1, 1), with polarization parallel to the xz plane. In each case, sketch the wave, and give the explicit Cartesian components ofk and ii.
Answer:
a) the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
b) the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
Explanation:
a) the polarization the determined wave oscillates the electric field, which is the z axis
As the wave travels on the negative x-axis and the magnetic field is perpendicular, this field goes on the positive y-axis
the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
be) in the case of a polarization in the xi plane the magnetic field must go in the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
A sound wave (a periodic longitudinal wave) from a loudspeaker travels from air into water. The frequency of the wave does not change, because the loudspeaker producing the sound determines the frequency. The speed of sound in air is 343 m/s, whereas the speed in fresh water is 1482 m/s. When the sound wave enters the water, does its wavelength increase, decrease, or remain the same
Answer:
When the sound wave enters the water its wavelength increase.
Explanation:
Given:
Speed of sound in air [tex]v = 343 \frac{m}{s}[/tex]
Speed of sound in fresh water [tex]v' = 1482 \frac{m}{s}[/tex]
We know that when wave travel into one medium to another medium frequency doesn't change.
Velocity is given by,
[tex]v = f \lambda[/tex]
In above equation frequency is constant,
So when sound wave enters the water wavelength increase because speed sound is increase in water as compare to air.
Therefore, when the sound wave enters the water its wavelength increase.
Will water flow more easily through a wide pipe or a narrow pipe? Will current flow more easily through a thick wire or a thin wire?
Answer:
Explained below.
Explanation:
The water will flow more easily through a wide pipe, because as we know that if the diameter of the pipe is wider than the flow of water will be more.
And same case will be applied with the electric current, if the is wire thicker then the flow of the current will be more easy and the charge will also flow easily.
Which statement describes Newton's law of universal gravitation?
Mass has hittle effect on gravity between objects.
Gravity pushes objects away from Earth's center,
Gravity does not act between Earth and the moon.
Every object in the universe attracts every other object
Final answer:
Newton's law of universal gravitation states that all objects in the universe attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
Explanation:
Newton's Law of Universal Gravitation:
Sir Isaac Newton's law of universal gravitation states that every object in the universe attracts every other object with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The gravitational force is universally attractive and only depends on mass and distance, following the equation F = G × (m1 × m2) / r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the centers of the two masses.
An ideal transformer has 50 turns in its primary and 250 turns in its secondary. 12 Vac is connected to the primary. Find: (a) volts ac available at the second- ary; (b) current in a 1O-ohm device connected to the secondary; and (c) power supplied to the primary.
Answer:
a. 60 V b. 6 A c. 360 W
Explanation:
a. Voltage in secondary
For an ideal transformer,
N₁/N₂ = I₂/I₁ = V₁/V₂ where N₁ = turns in primary = 50 turns, N₂ = turns in secondary = 250 turns, V₁ = voltage in primary = 12 V, V₂ = voltage in secondary = ? V
N₁/N₂ = V₁/V₂
V₂ = V₁N₂/N₁ = 250 × 12/50 = 60 V
b. Since V₂ = I₂R,
I₂ = V₂/R R = 10 Ω
I₂ = 60 V/10 Ω
I₂ = 6 A
c. We first calculate the current in the primary from
N₁/N₂ = I₂/I₁ where I₁ = primary current
I₁ = N₂I₂/N₁ = 250 × 6 A/50 = 30 A
The power supplied to the primary is thus
P = I₁V₁ = 30 A × 12 V = 360 W
(a) The voltage supply at the secondary coil is of 60 V.
(b) The current flow through the 10- ohm device is 6 A.
(c) The power supply to the primary coil is 360 W.
Given Data:
Number of turns of primary coil is, n1 = 50.
Number of turns of secondary coil is, n2 = 250.
The voltage supply to primary coil is, V1 = 12 V.
The resistance of device is, R = 10 ohm.
(a)
The first part of the given problem is based on the Transformer equation relating the number of turns of coils at primary and secondary and voltage supply at each coils.
Therefore,
n1 / n2 = V1 / V2
Here,
V2 is the voltage supply to the secondary coil.
Solving as,
50 / 250 = 12 / V2
V2 = 250 / 50 × 12
V2 = 60 V.
Thus, we can conclude that the voltage supply at the secondary coil is of 60 V.
(b)
Now in order to find the current flow through the 10 - ohm device, we can use the Ohm's law as,
V2 = I' × R
Solving as,
60 = I' × 10
I ' = 60 / 10
I' = 6 A
Thus, we can conclude that the current flow through the 10- ohm device is 6 A.
(c)
Now, we first calculate the current in the primary from
n1 / n2 = I' / I
here,
I is the primary current
Solving as,
I = n2 × I' / n1
I = 250 × 6 /50
I = 30 A
The power supplied to the primary is thus
P1 = I × V1
P1 = 30 × 12
P1 = 360 W
Thus, we can conclude that the power supply to the primary coil is 360 W.
Learn more about the transformer equation here:
https://brainly.com/question/19574363
A block of mass = 4.00 kg is supported by a spring scale with unit of measure of Newtons. This spring scale is attached to top of an elevator. The elevator accelerates upward with a = 3.00 m/sec2. What is the reading on the spring scale to 3 significant figures?
Answer:
118 N
Explanation:
Given mass of the block, m = 4.00kg.
The acceleration of the elevator, a = 3.0 m/s^2.
As elevotar attaced with spring scale and accelerating upward
(block and elevator), so total force
[tex]F_N-mg=ma[/tex]
Here, mg is the weight of the block downward direction.
or
[tex]F_N=ma+mg=m(g+a)[/tex]
substitute the given value, we get
[tex]F_N=4kg(9.8m/s^2+3m/s^2)[/tex]
= 117.6 N = 118 N.
Thus, the reading on the spring scale to 3 significant figures is 118 N.
A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the particle between t=0.00 seconds and t=5.00 seconds. find the speed v of the particle at t=5.00 seconds?
Answer:
43 m/s
Explanation:
Mass, m = 5 kg
Force, F(t) = 6t² - 4t + 3
To find the speed, we first need to get the acceleration, a.
Force is the product of mass and acceleration. It is given as:
F = ma
Therefore, acceleration is:
a(t) = F(t)/m
a(t) = (6t² - 4t + 3) / 5
a(t) = 1.2t² - 0.8t + 0.6
Acceleration is the differentiation of velocity with respect to time, t. Therefore, to get velocity, v, we integrate a(t):
a(t) = dv(t) / dt
=> v(t) = (1.2/3)t³ - (0.8/2)t² + 0.6t
v(t) = 0.4t³ - 0.4t² + 0.6t
Therefore, at time t = 5secs, velocity is:
v(5) = 0.4 * (5³) - 0.4 * (5²) + 0.6 * 5
v(5) = 50 - 10 + 3
v(5) = 43 m/s
The velocity at time, t = 5 secs is 43 m/s
During a redox reaction the molecule that gains an electron is
In a redox (reduction-oxidation) reaction, the molecule that gains an electron is reduced. Reduction and oxidation always occur together in such reactions. Reduced molecules often act as energy carriers in metabolic pathways.
Explanation:During a redox (reduction-oxidation) reaction, the molecule that gains an electron is referred to as being reduced. This process is accompanied by an energy transfer. When a molecule loses an electron (a process termed as oxidation), energy is released; this energy and the electron are then transferred to another molecule, resulting in reduction of that molecule. These two processes, oxidation and reduction, always occur together in a redox reaction.
For example, considering a metabolic pathway involving Nicotinamide adenine dinucleotide (NAD+), when it accepts a hydride ion (H-) from a hydrogen atom, it gets reduced to form NADH. Here, NAD+ is the molecule that gets reduced in this redox reaction.
In summary, in redox reactions, the molecule that gains an electron and is reduced can act as an energy carrier, an important function in metabolism, facilitating the controlled transfer of energy within the cell. The concepts of oxidation and reduction are integral to understanding energy extraction and utilization in cells.
Learn more about Redox Reactions here:https://brainly.com/question/13978139
#SPJ12
A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the same point at t = 55 ms. (a) How far below the release point is the center of mass of the two stones at t = 470 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?
Answer:
Part(a): At [tex]\bf{t = 470~ms}[/tex] the center of mass will travel [tex]\bf{0.903~m}[/tex].
Part(b): The velocity of the first stone is [tex]\bf{4.606~m.s^{-1}}[/tex] and the velocity of the second stone is [tex]\bf{4.067~m.s^{-1}}[/tex].
Explanation:
Given:
The first stone is dropped at [tex]t_{1}=0~s[/tex].
The second stone is dropped at, [tex]t_{2}=55~ms=0.055~s[/tex]
The mass of the second stone is 3 times the mass of the first.
Both the stones are dropped from the same point.
Consider the mass of the first stone be [tex]m[/tex]. So the mass of the second stone is [tex]3m[/tex].
(a)
The formula to calculate the distance traveled by each stone is given by
[tex]y = \dfrac{1}{2}gt^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by each stone.
Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{1}[/tex] for [tex]y[/tex] and [tex]470~ms[/tex] or [tex]0.47~s[/tex] for first stone in equation (1), we have
[tex]y_{1}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.47~s)^{2}\\~~~~&=& 1.08~m[/tex]
where [tex]y_{1}[/tex] is the distance traveled by the first stone.
Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{2}[/tex] for [tex]y[/tex] and [tex](470-55)~ms = 415~ms[/tex] or [tex]0.415~s[/tex] for second stone in equation (1), we have
[tex]y_{2}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.415~s)^{2}\\~~~~&=& 0.844~m[/tex]
The formula to calculate the distance traveled by the center of mass is given by
[tex]y_{c} &=& \dfrac{my_{1}+3my_{2}}{m+3m} \\~~~~&=& \dfrac{1.08m + 0.844(3m)}{4m}\\~~~~&=& 0.903~m[/tex]
(b)
The formula to calculate the velocity of each stone is given by
[tex]v=gt~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Substituting [tex]v_{1}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.47~s[/tex] for [tex]t[/tex] in equation (2), we have
[tex]v_{1} &=& (9.8~m.s^{-2})(0.47~s)\\~~~~&=& 4.606~m.s^{-1}[/tex]
where [tex]v_{1}[/tex] is the velocity of the first stone after [tex]470~ms[/tex].
Substituting [tex]v_{2}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.415~s[/tex] for [tex]t[/tex] in equation (2), we have
[tex]v_{2} &=& (9.8~m.s^{-2})(0.415~s)\\~~~~&=& 4.067~m.s^{-1}[/tex]
where [tex]v_{2}[/tex] is the velocity of the second stone after [tex]470~ms[/tex].
Compute the quantity of heat released by 25.0 g of steam initially at 100.0oC, when it is cooled to 34.0°C and by 25.0 g of water initially at 100.0 oC, when it is cooled to 34.0°C.
Answer:
For steam, heat released E = 15.26KJ
For water, heat released E2 = 6.91KJ
Explanation:
Given;
Mass(steam) ms = 25g
Mass (water) mw = 25g
Change in temperature of both steam and water ∆T = 100-34= 66°C
Specific heat of water C = 4.186 J/g.°C
Specific Latent heat L = 334J/g
For steam;
Heat released E = msL + msC∆T
E = (25×334) + (25×4.186×66)
E = 15256.9J
E = 15.26KJ
For water;
Heat released E2 = mwC∆T
E2 = 25×4.186×66
E2 = 6906.9J
E2 = 6.91KJ
Air pollution from a power plant is being monitored for levels of nitrogen dioxide and ground-level ozone. The levels are measured daily at the same time. What is the impact of a dark and cloudy day on the readings for the levels of air pollutants measured?
An increase in the level of nitrogen dioxide and a decrease in the level of ground-level ozone occurs.
Explanation:
Ozone gas is normally found in stratosphere, it protects us from solar radiation. Under certain circumstances, it can be formed on the ground level.
Conditions that have to be met in order for this to happen are existing of nitrogen oxides and volatile organic compounds and their reaction catalyzed by heat and light.
During dark, cold and cloudy days, due to the lack of heat and light from the Sun, formation of ozone will be decreased.
Also, this will enable building up of nitrogen dioxide, due to the same reason, leading to increased concentration.
A bat locates insects by emitting 25.3 kHz ultrasonic chirps and then listening for echoes from insects flying around it. Humans can hear sounds with frequencies up to 20 kHz. How fast would a bat have to fly and in which direction (positive away from and negative towards a stationary human listener) in order for a person to hear these chirps
Answer:
Speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.
Explanation:
Given:
Frequency of the bat that is source here, [tex]f_S[/tex] = 25.3 kHz = 25.3 * 10^3 Hz
Frequency of the listener (human), [tex]f_L[/tex] = 20 kHz = 20*10^3 Hz
We have to identify how fast the bat have to fly in order for a person to hear these chirps .
Let the velocity of bat that is source is "Vs" and "Vs" = "Vbat".
Doppler effects formulae :
When the source is receding (moving away) [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex] When the source is approaching [tex]f_L=(\frac{V+V_L}{V-V_S}) f_S[/tex] Speed of sound in [tex]V_S =433.895-343[/tex]the air (medium), [tex]V = 343\ ms^-^1[/tex]Using the above formula and considering that the bat is moving away so that the human can listen the chirps also [tex]V_L=0[/tex] as listener is stationary.
⇒ [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex] ⇒ [tex]f_L=(\frac{V+0}{V+V_S}) f_S[/tex]
Re-arranging in terms of Vs.
⇒ [tex]V+V_S =\frac{V\times f_S}{f_L}[/tex]
⇒ [tex]V_S =\frac{V\times f_S}{f_L}-V[/tex]
⇒ [tex]V_S =\frac{343\times 25.3\times 10^3}{20\times 10^3}-343[/tex]
⇒ [tex]V_S=90.895[/tex] m/s
The speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.