Answer:
B) AC current can be used in transformers while DC current cannot.
Explanation:
AC current is used for long distance transportation of electrical energy with amplifying the voltage and reducing the current in the conductors so that there is minimum loss of energy in the form of heat.
The relation between heat energy and electrical current is given by Joule's law as:
[tex]Q=i^2.R.t[/tex]
where:
Q = amount of heat energy generatedi = current in the conductorR = resistance of the conductort = time for which the current flowsThe above mentioned variation of current and voltage in the transmission lines of AC is achieved by a device called transformer. It consist of a rectangular shaped steel core common to two of the insulated wire winding of which one side acts as input (primary coil) and the other one acts as secondary (output).
AC i.e. alternating current can only be used in the transformers because they have the continuously varying amplitude which in turn provides the change in flux with respect to the time inducing an emf in the nearby secondary coil of the transformer without any physically moving parts.
The alternating nature of AC current aloows changes in magnet fields to be used to transform electricity and make it more efficient to transport over long distances. AC current can be used in transfromers while DC current cannot.
Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle α can have without any light being refracted out of the prism at face AC if the prism is immersed in water (with index of refraction 1.33). Express your answer in degrees. Ignore any reflections from the surface BC.
To solve this problem it is necessary to use the concepts related to Snell's law.
Snell's law establishes that reflection is subject to
[tex]n_1sin\theta_1 = n_2sin\theta_2[/tex]
Where,
[tex]\theta =[/tex] Angle between the normal surface at the point of contact
n = Indices of refraction for corresponding media
The total internal reflection would then be given by
[tex]n_1 sin\theta_1 = n_2sin\theta_2[/tex]
[tex](1.54) sin\theta_1 = (1.33)sin(90)[/tex]
[tex]sin\theta_1 = \frac{1.33}{1.54}[/tex]
[tex]\theta = sin^{-1}(\frac{1.33}{1.54})[/tex]
[tex]\theta = 59.72\°[/tex]
Therefore the [tex]\alpha_{max}[/tex] would be equal to
[tex]\alpha = 90\°-\theta[/tex]
[tex]\alpha = 90-59.72[/tex]
[tex]\alpha = 30.27\°[/tex]
Therefore the largest value of the angle α is 30.27°
The largest value the angle α can have without any light being refracted out of the prism at face AC is 30.27°.
How to calculate the angle?From the information given, the total internal reflection would be:
1.54sinb = (1.33) sin90°
sin b = (1.33 sin90° / 1.54)
b = 59.72°
Therefore, the value of the angle will be:
= 90° - 59.72°
= 30.27°
In conclusion, the correct option is 30.27°.
Learn more about angles on:
https://brainly.com/question/25716982
A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker?
Answer:F=40.09 N
Explanation:
Given
weight of crate [tex]W=50 N[/tex]
Inclination [tex]\theta =37^{\circ}[/tex]
Frictional Force [tex]f=10 N[/tex]
as the crate is moving with constant velocity therefore net Force on crate is zero
[tex]F-50\sin (37)-f=0[/tex]
[tex]F=50\sin (37)+10[/tex]
[tex]F=30.09+10[/tex]
[tex]F=40.09 N[/tex]
A 191 191 kg sculpture hangs from a horizontal rod that serves as a pivot about which the sculpture can oscillate. The sculpture's moment of inertia with respect to the pivot is 17.2 17.2 kg·m2, and when it is swung at small amplitudes, it is found to oscillate at a frequency of 0.925 0.925 Hz. How far is the pivot from the sculpture's center of mass?
Answer:
r = 0.31 m
Explanation:
Given that,
Mass of the sculpture, m = 191 kg
The sculpture's moment of inertia with respect to the pivot is, [tex]I=17.2\ kg-m^2[/tex]
Frequency of oscillation, f = 0.925 Hz
Let r is the distance of the the pivot from the sculpture's center of mass. The frequency of oscillation is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{mgr}{I}}[/tex]
[tex]r=\dfrac{4\pi^2f^2I}{mg}[/tex]
[tex]r=\dfrac{4\pi^2\times 0.925^2\times 17.2}{191\times 9.8}[/tex]
r = 0.31 m
So, the pivot is 0.31 meters from the sculpture's center of mass. Hence, this is the required solution.
You are lying in bed and want to shut your bedroom door. You have a bouncy "superball" and a blob of clay, both with the same mass. Which one would be more effective to throw at your door to close it?
a. The superball.
b. Neither will work.
c. The blob of clay.
d. Both the same.
Answer:
Choice A. The bouncy "superball" will exert approximately twice as much force than the blob of clay when tossed out at the same speed.
Explanation:
The momentum of an object is equal to the product of its mass and its velocity. The momentum of the system is conserved even if the collision is inelastic.
When the bouncy ball or the blob of clay hits the door, their momentum changes. Since momentum is conserved, the door would move in the opposite direction such that the total momentum stays at zero.
Let the mass of each object be [tex]m[/tex]. Suppose that both objects hit the door at a speed of [tex]v[/tex] in the same direction.
The bouncy ball will likely bounce back at about the same speed (in the opposite direction, away from the door.) The magnitude of the change in velocity will be at most [tex]2v[/tex]. The magnitude of the change in momentum will be at most [tex]2 \,m\cdot v[/tex].However, when the blob of clay hits the door, its shape would change (and it would potentially stick to the door.) In other words, its velocity becomes almost [tex]0[/tex] after the collision. The magnitude of the change in its momentum will only be around [tex]m \cdot v[/tex], which is only about [tex]\dfrac{1}{2}[/tex] the momentum change due to the bouncy ball.By Newton's Second Law, the net force on the door is proportion to the rate of change in its momentum. Assume that the two objects are in contact with the door for the same amount of time, the bouncy ball would exert about twice as much force on the door as the clay would. Hence it is "more effective" for closing the door.
The clay is more effective than the superball for shutting the door because upon impact the clay sticks to the door and transfers all of its momentum to it, while the superball bounces back and doesn't transfer all of its momentum.
Explanation:The most effective object to throw at your door to shut it would be c. The blob of clay. This is because of the principle of conservation of momentum. When the clay hits the door, it sticks to it, thus transferring all of its momentum to the door. This momentum is what causes the door to move and potentially close. On the other hand, when the superball hits the door, it bounces back, meaning not all of its momentum is transferred to the door. Hence, the clay, by virtue of its ability to stick and transfer its full momentum, would be more effective.
Learn more about Conservation of Momentum here:https://brainly.com/question/33316833
#SPJ11
A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a speed of 32.2 m > s. The motorcycle has a speed of 14.8 m > s. What frequency does the motorcyclist hear?
To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.
It can be described as
[tex]f = \frac{c\pm v_r}{c\pm v_s}f_0[/tex]
c = Propagation speed of waves in the medium
[tex]v_r[/tex]= Speed of the receiver relative to the medium
[tex]v_s[/tex]= Speed of the source relative to the medium
[tex]f_0 =[/tex]Frequency emited by the source
The sign depends on whether the receiver or the source approach or move away from each other.
Our values are given by,
[tex]v_s = 32.2m/s \rightarrow[/tex] Velocity of car
[tex]v_r = 14.8 m/s \rightarrow[/tex] velocity of motor
[tex]c = 343m/s \rightarrow[/tex] Velocity of sound
[tex]f_0 = 523Hz \rightarrow[/tex]Frequency emited by the source
Replacing we have that
[tex]f = \frac{c + v_r}{c - v_s}f_0[/tex]
[tex]f = \frac{343 + 14.8}{343 - 32}(523)[/tex]
[tex]f = 601.7Hz[/tex]
Therefore the frequency that hear the motorcyclist is 601.7Hz
Final answer:
To determine the frequency heard by the motorcyclist, we apply the Doppler shift formula using the given speeds of the motorcycle and police car and the emitted frequency of the police car's siren. The calculated frequency is the observed frequency by the motorcyclist due to the relative motion of the two vehicles. The final answer is about 602.1 Hz
Explanation:
The subject of the question is the Doppler Effect, which is related to the change in frequency of sound waves due to the relative motion of the source and the observer. To calculate the frequency the motorcyclist hears, we use the following Doppler shift formula for sound:
f' = f(v + vo) / (v - vs)
Where:
f' is the observed frequency by the motorcyclist,
f is the emitted frequency by the police car (523 Hz),
v is the speed of sound in air, which can be assumed to be approximately 343 m/s at room temperature,
vo is the speed of the observer (motorcyclist) towards the source (14.8 m/s),
vs is the speed of the source (police car) towards the observer (32.2 m/s).
Plugging in the given values, the equation becomes:
f' = 523 Hz (343 m/s + 14.8 m/s) / (343 m/s - 32.2 m/s) = 602.089 Hz
After performing the calculations, the frequency heard by the motorcyclist can be determined. This is an application of the Doppler effect as studied in high school physics.
A physics class conducting a research project on projectile motion constructs a device that can launch a cricket ball. The launching device is designed so that the ball can be launched at ground level with an initial velocity of 28 m/ s at an angle of 30° to the horizontal. Calculate the horizontal component of the velocity of the ball:a. initiallyb. after 1.0 sc. after 2.0 s
Answer:
Explanation:
initial velocity, u = 28 m/s
Angle of projection, θ = 30°
The acceleration in horizontal direction is zero, so the horizontal component of velocity is constant.
Horizontal component of velocity, u cos θ = 28 x Cos 30 = 24.25 m/s
At t = 2 sec, the horizontal component of velocity = 24.25 m/s
At t = 3 sec, the horizontal component of velocity = 24.25 m/s
In this exercise we have to use the knowledge about oblique launch to calculate the components of velocity for each case, so we have that:
For all times we will find a velocity equal to 24.25 m/s
organizing the information given in the statement we have that:
initial velocity, u = 28 m/sAngle of projection, θ = 30°The acceleration in horizontal direction is zerovelocity is constant.Knowing that the component can be written as:
[tex]u cos \theta = 28 * Cos 30 = 24.25 m/s[/tex]
So as the formula does not depend on time we have that for any value it will have a constant velocity.
See more about velocity at brainly.com/question/862972
A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings make an angle of 60° with the vertical. The length of the pendulum is 2.3 m. Find the speed of the bullet.
Answer:
217.43298 m/s
Explanation:
[tex]m_1[/tex] = Mass of bullet = 19 g
[tex]m_2[/tex] = Mass of bob = 1.3 kg
L = Length of pendulum = 2.3 m
[tex]\theta[/tex] = Angle of deflection = 60°
u = Velocity of bullet
Combined velocity of bullet and bob is given by
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s[/tex]
As the momentum is conserved
[tex]m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s[/tex]
The speed of the bullet is 217.43298 m/s
The problem involves conservation of momentum and energy principles. Initially, bullet's momentum equals the final momentum of the system. The bullet's speed can be found by solving these equations, using the provided values.
Explanation:This problem can be solved using principles from both conservation of momentum and conservation of energy. To find the speed of the bullet, we need to consider two scenarios: the before and after the bullet is fired into the bob. Initial momentum is the mass of the bullet multiplied by its velocity and final momentum is the combined mass of the bullet and bob at their highest point. Assuming there's no external force acting, we can have:
m_bullet * v_bullet = (m_bullet + m_bob) * v_final.
The final velocity here is the vertical component of the velocity when the pendulum reach its highest point. This can be calculated by:
v_final = sqrt(2*gravity*height).
The height can be calculated using trigonometry:
height = length - length * cos(60).
Filling all the given values into the equations will give the speed of the bullet.
Learn more about Bullet Speed here:
https://brainly.com/question/30778897
#SPJ3
Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 1100 N exerted at the top would. How to the side does the top of the pole flex?
Answer:
The deformation in the pole due to force is 0.70 mm.
Explanation:
Given that,
Height = 20.0 m
Diameter = 4.00 cm
Force = 1100 N
We need to calculate the area
Using formula of area
[tex]A=\pi\times r^2[/tex]
[tex]A=\pi\times(2.00\times10^{-2})^2[/tex]
[tex]A=0.00125\ m^2[/tex]
[tex]A=1.25\times10^{-3}\ m^2[/tex]
We need to calculate the deformation
Using formula of deformation
[tex]\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)[/tex]
Where, s = shear modulus
F = force
l = length
A = area
Put the value into the formula
[tex]\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)[/tex]
[tex]\Delta x=0.000704\ m[/tex]
[tex]\Delta x=7.04\times10^{-4}\ m[/tex]
[tex]\DElta x=0.70\ mm[/tex]
Hence, The deformation in the pole due to force is 0.70 mm.
Suppose that a simple pendulum consists of a small 81 g bob at the end of a cord of negligible mass. If the angle θ between the cord and the vertical is given by θ = (0.068 rad) cos[(9.2 rad/s) t + φ], what are (a) the pendulum's length and (b) its maximum kinetic energy?
Answer:
(a) 0.115 m
(b) 2.08 x 10^-5 J
Explanation:
mass of bob, m = 81 g = 0.081 kg
The equation of oscillation is given by
θ = 0.068 Cos {9.2 t + Ф}
Now by comparison
The angular velocity
ω = 9.2 rad/s
(a) [tex]\omega^{2} =\frac{g}{L}[/tex]
where, L be the length of the pendulum
[tex]L =\frac{g}{\omega ^{2}}[/tex]
[tex]L =\frac{9.8}{9.2 \times 9.2}[/tex]
L = 0.115 m
(b) A = L Sinθ
A = 0.115 x Sin 0.068
A = 7.8 x 10^-3 m
Maximum kinetic energy
K = 0.5 x mω²A²
K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6
K = 2.08 x 10^-5 J
A glass flask whose volume is 1000 cm3 at a temperature of 0.300 ∘C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0 ∘C , a volume of 8.25 cm3 of mercury overflows the flask.
Final answer:
When the glass flask and mercury are warmed together, the mercury expands and overflows the flask. To calculate the initial volume of the mercury, use the equation: Volume of mercury = Volume of flask + Volume overflowed.
Explanation:
When the glass flask and mercury are warmed together, both substances expand due to the increase in temperature. As a result, some of the mercury overflows the flask. To calculate the initial volume of the mercury, we can use the equation:
Volume of mercury = Volume of flask + Volume overflowed
So, the initial volume of mercury is 1000 cm³ + 8.25 cm³ = 1008.25 cm³.
Sometime around 2022, astronomers at the European Southern Observatory hope to begin using the E-ELT(European Extremely Large Telescope), which is planned to have a primary mirror 42 m in diameter. Let us assume that the light it focuses has a wavelength of 600 nm. (1 light-year = 9.461×1015 m) Note: Jupiter's Diameter dj=1.43×108 m 1) What is the most distant Jupiter-sized planet the telescope could resolve, assuming it operates at the diffraction limit? (Express your answer to two significant figures.)
Answer:
[tex]8.2\times 10^{15}\ m[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength = 600 nm
d = Diameter of mirror = 42 m
D = Distance of object
x = Diameter of Jupiter = [tex]1.43\times 10^8\ m[/tex]
Angular resoulution is given by
[tex]\Delta\theta=1.22\frac{\lambda}{d}\\\Rightarrow \Delta\theta=1.22\frac{600\times 10^{-9}}{42}\\\Rightarrow \Delta\theta=1.74286\times 10^{-8}\ rad[/tex]
We also have the relation
[tex]\Delta\theta\approx=\frac{x}{D}\\\Rightarrow D\approx\frac{x}{\Delta\theta}\\\Rightarrow D\approx\frac{1.43\times 10^8}{1.74286\times 10^{-8}}\\\Rightarrow D\approx 8.2049\times 10^{15}\ m[/tex]
The most distant Jupiter-sized planet the telescope could resolve is [tex]8.2\times 10^{15}\ m[/tex]
A rigid cube (each side is 0.10 m) is filled with water and frozen solid. When water freezes its volume expands about 9%. How much pressure is exerted on the sides of the cube? Hint: Imagine trying to squeeze the block of ice back into the original cube.
Answer:
P = 1.89 10⁸ N / m²
Explanation:
To solve this problem we can use the definition of bulk modules
B = - P / (ΔV/V)
The negative sign is entered for the volume module to be positive, P is the pressure and ΔV/V is the volume change fraction
In this case the volume change is 9% this is
ΔV / V 100 = 9%
ΔV / V = 0.09
P = B ΔV / V
The bulk modulus value is that of water since it is in a liquid state and then freezes
B = 0.21 101¹⁰ N / m²
let's calculate
P = 0.21 10¹⁰ 0.09
P = 1.89 10⁸ N / m²
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . Suppose that the quarterback takes 0.30 s to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?
Answer:
0.0241875 m
Explanation:
[tex]m_1[/tex] = Mass of quarterback = 80 kg
[tex]m_2[/tex] = Mass of football = 0.43 kg
[tex]v_1[/tex] = Velocity of quarterback
[tex]v_2[/tex] = Velocity of football = 15 m/s
Time taken = 0.3 seconds
In this system as the linear momentum is conserved
[tex]m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.43\times 15}{80}\\\Rightarrow v_1=0.080625\ m/s[/tex]
Assuming this velocity is constant
[tex]Distance=Velocity\times Time\\\Rightarrow Distance=0.080625\times 0.3\\\Rightarrow Distance=0.0241875\ m[/tex]
The distance the quarterback will move in the horizontal direction is 0.0241875 m
Final answer:
The question is a physics problem regarding kinematics and requires calculating the horizontal distance an 80-kg quarterback moves while in the air after throwing a football. However, since the quarterback jumps vertically with zero initial horizontal speed and no horizontal force is acting after the throw, the resulting horizontal distance is zero.
Explanation:
The student is asking a question related to kinematics and the conservation of momentum in physics. Specifically, it concerns an 80-kg quarterback who jumps vertically and then throws a football horizontally. The core concept here is determining how far the quarterback will move horizontally during the time he is in the air, assuming a constant horizontal speed.
To calculate the distance (d) the quarterback moves horizontally, we can apply the equation for constant velocity motion: d = v * t. However, the information given does not directly provide the quarterback's horizontal speed after the throw; thus, we must assume that the horizontal speed of the quarterback is zero as the ball is thrown horizontally and no horizontal force on the quarterback has been mentioned.
Therefore, under the assumption that the quarterback's horizontal speed remains zero, the distance he moves horizontally d would also be zero, since he moves vertically up and down with no horizontal velocity component. If the problem had given a horizontal speed for the quarterback post-throw, we would use that given speed to calculate the distance using the equation.
The auditory canal behaves like a resonant tube to aid in hearing. One end terminated at the eardrum, while the other opens to the outside. Typically, the canal is about 2.4 cm long. At which frequency would it resonate in its first harmonic?
4.2 kHz
3.6 kHz
2.9 kHz
5.7 kHz
Answer:
3.6 kHz
Explanation:
The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.
The first harmonic will be as seen in the figure attached.
The length of the first harmonic will be λ/4.
λ/4=2.4 cm
λ=2.4 * 4=9.6 cm 0.096 m
Speed of Sound- 344 m/s(in air)
velocity(v) * Time Period(T) = Wavelength (λ)
Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}
\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency
Plugging in the values into the equation,
Frequency = [tex]\frac{344}{0.096}[/tex] Hz
= 3583.3 Hz≈3600 Hz= 3.6 kHz
Frequency= 3.6 kHz
Answer:
3.6 kHz
Explanation:
The auditory canal is a closed pipe because it has one closed end, the end terminated at the eardrum.
The length of the first harmonic of a closed pipe is given as;
L = λ/4 -------------------- (i)
where L = Length
and λ = wavelength
2.4 = λ/4
λ = 2.4 x 4 = 9.6 cm
Also, v = fλ ------------------ (ii)
where v = speed of sound in air = 344 m/s
f = frequency of wave in Hertz
f = v/λ ------------------ (iii)
convert 9.6 cm to m = 0.096 m
substitute for λ and v in (iii)
[tex]f = \frac{344}{0.096} = 3583.33[/tex]
3583.33 Hz = 3600 Hz = 3.6 kHz
Consider the amideanion.What is the central atom? Enter its chemical symbol.How many lone pairs are around the central atom?What is the ideal angle between the nitrogen-hydrogen bonds?Compared to the ideal angle, you would expect the actual angle between the nitrogen-hydrogen bonds to be ...
Answer:
a) The central atom of the amide ion is nitrogen
b) (NH2)-
c) There are two lone pairs around the nitrogen atom
d) The ideal angle between nitrogen hydrogen bonds is 109.5° C, typical of a tetrahedral electron density arrangement
e) I would expect the nitrogen-hydrogen bond angles in the amide ion to be less than 109.5 ° because unbonded pairs repel bonded pairs much more than other bonded pairs do. So the bonds in this ion would be pushed closer together than normal
Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).
The length of A is twice that of B, and the radius of A is one-half that of B.
The pressure across both tubes is the same.
To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.
It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.
The mathematical equation that expresses this concept is
[tex]\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}[/tex]
Where
P = Pressure at each point
r = Radius
[tex]\eta =[/tex] Viscosity
l = Length
Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]
From the problem two terms are given
[tex]R_A = \frac{R_B}{2}[/tex]
[tex]L_A = 2L_B[/tex]
Replacing we have to
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}[/tex]
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}[/tex]
Therefore the ratio of the flow rate through capillary tubes A and B is 1/32
An applied force of 122 N is exerted horizontally on an 28 kg box of dogs in Yulin. The coefficient of kinetic friction between the floor and the box is 0.35. If the box is initially at rest, what is the final kinetic energy after it has been moved 9.46 m across the floor?
Answer:
The final kinetic energy is 242 J.
Explanation:
Hi there!
According to the work-energy theorem, the total work done on the box is equal to its change in kinetic energy:
W = ΔKE
Where:
W = work done on the box.
ΔKE = change in kinetic energy (final KE - initial KE).
The only forces that do work in this case are the applied force and the friction force because the box moves only horizontally.
The equation of work is the following:
W = F · s
Where:
F = force.
s = traveled distance.
Then, the work done by the applied force is:
W = 122 N · 9.46 m = 1.15 × 10³ J
To calculate the work done by friction, we have to find the friction force:
Fr = N · μ
Where:
Fr = friction force.
N = normal force.
μ = coefficient of kinetic friction.
The box does not have a net vertical acceleration. It means that the sum of the vertical forces acting on the box is zero:
∑Fy = 0
In this case, the only vertical forces are the weight of the box and the normal force. Then:
Weight + N = 0
N = - Weight
The weight of the box is calculated as follows:
Weight = m · g
Where:
m = mass of the box.
g = acceleration due to gravity.
Then:
-Weight = N = 28 kg · 9.8 m/s² = 274.4 N
Now, we can calculate the friction force:
Fr = N · μ
Fr = 274.4 N · 0.35 = 96 N
The work done by the friction force will be:
W = Fr · s
W = 96 N · 9.46 m = 908 J
Since the work done by friction opposes to the sense of movement, the work is negative.
Now, we can calculate the total work done on the box:
W total = W applied forece + W friction force
W = 1.15 × 10³ J - 908 J = 242 J
Applying the work-energy theorem:
W = final KE - initial KE
Since the box is initially at rest the initial kinetic energy is zero. Then:
W = final KE - 0
W = final KE
Final KE = 242 J
The final kinetic energy is 242 J.
A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.
a) Calculate the magnitude of its magnetic moment
b) Suppose a uniform magnetic field of magnitude of 0.350 T isapplied parallel to the plane of the loop. What is the magnitude ofthe torque acting on the loop?
Answer:
(a) Magnetic moment will be [tex]17.212\times 10^{-4}A-m^2[/tex]
(b) Torque will be [tex]6.024\times 10^{-4}N-m[/tex]
Explanation:
We have given dimension of the rectangular 5.4 cm × 8.5 cm
So area of the rectangular coil [tex]A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2[/tex]
Current is given as [tex]i=15mA=15\times 10^{-3}A[/tex]
Number of turns N = 25
(A) We know that magnetic moment is given by [tex]magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2[/tex]
(b) Magnetic field is given as B = 0.350 T
We know that torque is given by [tex]\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m[/tex]
Final answer:
a) The magnitude of the magnetic moment is 0.0172 A·m². b) The magnitude of the torque on the loop is 0 N·m.
Explanation:
a) To calculate the magnitude of the magnetic moment, we can use the formula μ = nIA, where μ is the magnetic moment, n is the number of turns, I is the current, and A is the area of the coil. In this case, n = 25, I = 15.0 mA = 0.015 A, and A = (5.40 cm)(8.50 cm) = 45.9 cm² = 0.00459 m². Plugging these values into the formula, we get μ = (25)(0.015 A)(0.00459 m²) = 0.0172 A·m².
b) To calculate the magnitude of the torque, we can use the formula τ = μBsinθ, where τ is the torque, μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between μ and B. In this case, μ = 0.0172 A·m² (calculated in part a),
B = 0.350 T, and since the magnetic field is applied parallel to the plane of the loop, θ = 0°.
Plugging these values into the formula, we get τ = (0.0172 A·m²)(0.350 T)(sin 0°)
= 0 N·m.
A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force constant of 8.07 N/m. When the cannon is fired, the ball moves 16.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.033 0 N on the ball.
(a) With what speed does the projectile leave the barrel of the cannon? m/s
(b) At what point does the ball have maximum speed? cm (from its original position)
(c) What is this maximum speed? m/s
Answer:
a) the velocity is v=1.385 m/s
b) the ball has its maximum speed at 4.68 cm away from its compressed position
c) the maximum speed is 1.78 m/s
Explanation:
if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore
we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball
W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki
therefore
Ui + Ki = Uf + Kf + W fr
where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.
since
U= 1/2 k x² , K= 1/2 m v² , W fr = F*L
X= compression length , L= horizontal distance covered
therefore
Ui + Ki = Uf + Kf + W fr
1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L
a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0
1/2 k X² + 0 = 0 + 1/2 m v² + F*L
1/2 m v² = 1/2 k X² - F*L
v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s
b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.
1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]
1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d
taking the derivative
dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm
at x m = -0.4 cm the velocity is maximum
therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position
c) the maximum speed is
1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0
v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s
A conducting sphere of radius r1= 0.38m has a total charge of Q=.75 uC. A second uncharged conducting sphere of radius r2= 0.28m is then connected to the first by a thin conducting wire. The spheres are separated by a very large distance to their size. What is the total charge on sphere two, Q2, in C?
Answer:
The Q₂ is 0.318 μC
Explanation:
The charge flows is the same on both, then:
[tex]V_{1} =\frac{kQ_{1} }{r_{1} } \\V_{2} =\frac{kQ_{2} }{r_{2} } \\Q_{2} =Q-Q_{1} \\\frac{kQ_{1} }{r_{1} } =\frac{k*(Q-Q_{1}) }{r_{2} } \\Q_{1} =\frac{\frac{Q}{r_{2} } }{(1/r_{1})+(1/r_{2}) }[/tex]
But:
[tex]\frac{1}{r_{1} } +\frac{1}{r_{2} }=\frac{1}{0.38} +\frac{1}{0.28} =6.2[/tex]
Q = = 0.75 μC
Replacing:
[tex]Q_{1} =\frac{\frac{0.75}{0.28} }{6.2} =0.432\mu C[/tex]
The Q₂ is equal:
Q₂ = 0.75 - 0.432 = 0.318 μC
The second conducting sphere, once connected and then disconnected from the first, ends up with the same charge as the first sphere, 0.75 µC or 0.75 x 10^-6 C.
Explanation:When two conducting spheres are connected by a thin conducting wire, the charge distributes itself evenly across both spheres, assuming both spheres are identical in size. In the case of spheres with different radii, the amount of charge on each sphere once they are separated again will still be proportional to the original amount. Since the two spheres in the question are described to be separated by a large distance relative to their size, the thin wire connecting them would create an equipotential surface, allowing the charge to redistribute.
The total charge Q is conserved in the system, thus the charge on the second sphere Q2 after disconnecting the wire would be the same as the initial total charge Q since it was not charged before. Therefore, the total charge on sphere two, Q2, will also be 0.75 μC or 0.75 x 10^{-6} C.
Learn more about Charge Distribution on Conducting Spheres here:https://brainly.com/question/14487188
#SPJ3
The colors of the stars in the sky range from red to blue. Assuming that the color indicates the frequency at which the star radiates the maximum amount of electromagnetic energy, estimate the surface temperature of red, yellow, white, and blue stars.
A) Estimate the surface temperature of red star. Assume that red color corresponds to wavelengths of approximately 650 nm.B) Estimate the surface temperature of yellow star. Assume that yellow color corresponds to wavelengths of approximately 570 nm.C) Estimate the surface temperature of blue star. Assume that blue color corresponds to wavelengths of approximately 470 nmD) Estimate the surface temperature of white star. Since green is in the middle of the visible spectrum, choose 520 nm wavelength for white star.
Answer:
a) 4458K b) 5048K, c) 6166K, d) 5573K
Explanation:
The temperature of the stars and many very hot objects can be estimated using the Wien displacement law
[tex]\lambda_{max}[/tex] T = 2,898 10⁻³ [m K]
T = 2,898 10⁻³ / [tex]\lambda_{max}[/tex]
a) indicate that the wavelength is
Lam = 650 nm (1 m / 109 nm) = 650 10⁻⁹ m
Lam = 6.50 10⁻⁷ m
T = 2,898 10⁻³ / 6.50 10⁻⁷
T = 4,458 10³ K
T = 4458K
b) lam = 570 nm = 5.70 10⁻⁷ m
T = 2,898 10⁻³ / 5.70 10⁻⁷
T = 5084K
c) lam = 470 nm = 4.70 10⁻⁷ m
T = 2,898 10⁻³ / 4.7 10⁻⁷
T = 6166K
d) lam = 520 nm = 5.20 10⁻⁷ m
T = 2,898 10⁻³ / 5.20 10⁻⁷
T = 5573K
A ) the surface temperature of red star is about 4500 K
B ) the surface temperature of yellow star is about 5100 K
C ) the surface temperature of blue star is about 6200 K
D ) the surface temperature of white star is about 5600 K
[tex]\texttt{ }[/tex]
Further explanationLet's recall the Wien's Displacement Law as follows:
[tex]\boxed {\lambda_{max}\ T = 2.898 \times 10^{-3} \texttt{ m.K}}[/tex]
where:
λ_max = the wavelength of the maximum radiation energy ( m )
T = surface temperature of the star ( K )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
wavelength of red light = λ_r = 650 nm = 650 × 10⁻⁹ m
wavelength of yellow light = λ_y = 570 nm = 570 × 10⁻⁹ m
wavelength of blue light = λ_b = 470 nm = 470 × 10⁻⁹ m
wavelength of white light = λ_w = 520 nm = 520 × 10⁻⁹ m
Asked:
A ) the surface temperature of red star = T_r = ?
B ) the surface temperature of yellow star = T_y = ?
C ) the surface temperature of blue star = T_b = ?
D ) the surface temperature of white star = T_w = ?
Solution:
Part A) :[tex]T_r = ( 2.898 \times 10^{-3} ) \div \lambda_r[/tex]
[tex]T_r = ( 2.898 \times 10^{-3} ) \div ( 650 \times 10^{-9} )[/tex]
[tex]\boxed {T_r \approx 4500 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
Part B) :[tex]T_y = ( 2.898 \times 10^{-3} ) \div \lambda_y[/tex]
[tex]T_y = ( 2.898 \times 10^{-3} ) \div ( 570 \times 10^{-9} )[/tex]
[tex]\boxed {T_y \approx 5100 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
Part C) :[tex]T_b = ( 2.898 \times 10^{-3} ) \div \lambda_b[/tex]
[tex]T_b = ( 2.898 \times 10^{-3} ) \div ( 470 \times 10^{-9} )[/tex]
[tex]\boxed {T_b \approx 6200 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
Part D) :[tex]T_w = ( 2.898 \times 10^{-3} ) \div \lambda_w[/tex]
[tex]T_w = ( 2.898 \times 10^{-3} ) \div ( 520 \times 10^{-9} )[/tex]
[tex]\boxed {T_w \approx 5600 \texttt{ K} }[/tex]
[tex]\texttt{ }[/tex]
Learn moreSpeed of Wave : https://brainly.com/question/9834706Energy of Microwave Radiation : https://brainly.com/question/9962873Acceleration : https://brainly.com/question/2283922The Speed of Car : https://brainly.com/question/568302[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Mathematics
Chapter: Energy
Moe is carrying and supporting a large crate by exerting 1000 N. He is walking at constant velocity across a level, horizontal floor. If he covers a distance of 10 meters, how much work does he do on the crate? Use g = 10m/s2 if needed. Ignore friction between Moe and the crate.
Answer:
he did not work on the create
Explanation:
Honestly I have no idea it makes a lot of sense to use the work equation so that was my first answer but for some reason on UT Quest it said that was the wrong and moe not doing any work on the create was the right answer
Final answer:
The work done by Moe on the crate is 10,000 J.
Explanation:
The work done by Moe on the crate can be calculated using the formula:
Work = Force x Distance x Cosine(theta)
Since Moe is exerting a force of 1000 N and the distance covered is 10 meters, we can plug these values into the formula:
Work = 1000 N x 10 m x Cosine(0°)
The angle theta is 0° because Moe is exerting a force in the same direction as the displacement.
Cosine(0°) is equal to 1, so the work done by Moe on the crate is:
Work = 1000 N x 10 m x 1 = 10,000 J
What impulse occurs when an average force of 7.0 N is exerted on a cart for 2.5 s ?
Express your answer to two significant figures and include the appropriate units.
Answer:[tex]1.7\times 10[/tex]
Explanation:
Given
Force [tex]F=7 N[/tex]
time interval [tex]t=2.5 s[/tex]
Impulse is given by [tex]=Force \times time\ interval\ for\ applied\ force [/tex]
[tex]Impulse=7\times 2.5=17.5[/tex]
For two significant Figure
[tex]Impulse=1.7\times 10[/tex]
An 15-cm-long bicycle crank arm, with a pedal at one end, is attached to a 21-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 62 rpm to 95 rpm in 12 s . A) What is the tangential acceleration of the pedal?
B)What length of chain passes over the top of the sprocket during this interval?
The tangential acceleration of the pedal is 0.259 m/s^2. The length of chain passing over the top of the sprocket depends on the number of teeth on the sprocket, which is not provided in the question.
Explanation:A) The tangential acceleration of the pedal can be determined using the equation:
at = r × α
where at is the tangential acceleration, r is the radius of the pedal (half the length of the crank arm), and α is the angular acceleration. To find the angular acceleration, we can use the equation:
α = Δω / Δt
where Δω is the change in angular velocity and Δt is the change in time. Plugging in the given values:
α = (95 rpm - 62 rpm) * 2π / 60 s = (33 * 2π) / 60 s ≈ 3.459 rad/s2
Substituting this value and the radius into the first equation, we get:
at = 0.075 m * 3.459 rad/s2 = 0.259 m/s2
Therefore, the tangential acceleration of the pedal is 0.259 m/s2.
B) The length of chain passing over the top of the sprocket during the given time interval can be calculated using the formula:
d = 2π * r * (ωf - ωi) * t / N
where d is the length of chain, r is the radius of the sprocket, ωi and ωf are the initial and final angular velocities respectively, t is the time interval, and N is the number of teeth on the sprocket. Plugging in the given values:
d = 2π * (21 cm / 2) * [(95 rpm - 62 rpm) * 2π / 60 s] * 12 s / N
The length of chain depends on the number of teeth on the sprocket. Since the number of teeth is not provided in the question, we cannot calculate the exact length of chain. However, we have all the necessary equation components to calculate it once the number of teeth is known.
Learn more about Tangential acceleration and length of chain on a bicycle here:https://brainly.com/question/34642585
#SPJ3
A 5 kg ball takes 13.3 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion?
Answer: 0.47 rad/sec
Explanation:
By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:
ω = ∆θ/ ∆t
Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:
ω = 2 π / 13.3 rad/sec = 0.47 rad/sec
Answer:
0.472rad/s
Explanation:
Angular velocity = 2πf where f = frequency and frequency is the number of revolution per second and 2π represent a cycle of revolution. The mass of the body was 5kg, the time taken to complete a cycle was 13.3 s.
Frequency = 1/period where period is the time it takes to complete a revolution.
F = 1/13.3 = 0.075hz
Angular velocity = 2* 3.142* 0.075 = 0.472rad/s
The specifications for a ceiling fan you have just bought indicate that the total moment of inertia of its blades is 0.23 kg · m2 and they experience a net torque of 2.8 N · m.
(a) What is the angular acceleration of the blades in rad/s2? rad/s2
(b) When the blades rotate at 205 rpm, what is the rotational kinetic energy, in joules? J
Answer:(a)12.17 rad/s
Explanation:
Given
Moment of Inertia [tex]I=0.23 kg.m^2[/tex]
Torque [tex]T=2.8 N-m[/tex]
(a)Torque is given by Product of Moment of inertia and angular acceleration
[tex]T=I\cdot \alpha [/tex]
[tex]2.8=0.23\cdot \alpha [/tex]
[tex]\alpha =\frac{2.8}{0.23}=12.17 rad/s[/tex]
(b)RPM of blades [tex]N=205 rpm [/tex]
angular velocity [tex]\omega =\frac{2\pi N}{60}[/tex]
[tex]\omega =\frac{2\pi 205}{60}=21.47 rad/s[/tex]
Rotational Kinetic Energy [tex]=\frac{I\omega ^2}{2}[/tex]
[tex]=\frac{0.23\times (21.47)^2}{2}=53.01 J[/tex]
The angular acceleration of the blades is approximately 12.1739 rad/s^2. When the blades rotate at 205 rpm, the rotational kinetic energy is approximately 0.0948 J.
Explanation:(a) To find the angular acceleration of the blades, we can use the formula:
torque = moment of inertia × angular acceleration
Plugging in the given values:
torque = 2.8 N · m
moment of inertia = 0.23 kg · m2
Rearranging the formula, we get:
angular acceleration = torque / moment of inertia
Substituting the values:
angular acceleration = 2.8 N · m / 0.23 kg · m2
Solving for angular acceleration gives us:
angular acceleration ≈ 12.1739 rad/s2
(b) To find the rotational kinetic energy, we can use the formula:
rotational kinetic energy = ½ × moment of inertia × (angular velocity)2
Plugging in the given values:
moment of inertia = 0.23 kg · m2
angular velocity = 205 rpm = 205 revolutions / 60 seconds = 3.4167 rev/s
Rearranging the formula, we get:
rotational kinetic energy = ½ × moment of inertia × (angular velocity)2
Substituting the values:
rotational kinetic energy = ½ × 0.23 kg · m2 × (3.4167 rev/s)2
Solving for rotational kinetic energy gives us:
rotational kinetic energy ≈ 0.0948 J
Learn more about Angular acceleration and rotational kinetic energy here:https://brainly.com/question/29614393
#SPJ3
At a certain location, Earth's magnetic field of 34 µT is horizontal and directed due north. Suppose the net field is zero exactly 8.1 cm above a long, straight, horizontal wire that carries a constant current. What are (a) the magnitude of the current and (b) the angle between the current direction and due north?
Answer:
a) I = 13.77 A
b) 0 ° or to the East
Explanation:
Part a
The magnetic field by properties would be 0 at the radius on this case r =8.1 cm.Analyzing the situation the wirde would produce a magnetic field equals in magnitude to the magnetic field on Earth by with the inverse direction.
The formula for the magnetic field due to a wire with current is:
[tex] B = \frac{\mu_0 I}{2 \pi r} [/tex]
In order to have a value of 0 for the magnetic field at the radius then we need to have this balance
B (r=8.1) = B (Earth)
Replacing:
[tex] B = \frac{\mu_0 I}{2 \pi r)}= B_{Earth} [/tex]
Solving from I, from the last equation we got:
[tex] I = \frac{2 \pi r B_{earth}}{\mu_0} [/tex]
[tex] I=\frac{2 \pi 0.081 m (34 x 10^{-6} T)}{4 \pi x 10^{-7} Tm/A}[/tex] = 13.77 A
Part b
We can use the right hand rule for this case.
The magnetic field of the wire would point to the South, because the magnetic field of the earth given points to the North. Based on this the current need's to flow from West to East in order to create a magnetic field pointing to the south, because the current would be perpendicular to the magnetic field created.
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
A) Determine the center of mass of the system.
B) A 6 kg mass is added to the system. This four-mass system has a new center of mass at the origin. What is the location of the 6 kg mass?
Answer:
a) C.M [tex]=(\bar x, \bar y)=(0.767,0.7)m[/tex]
b) [tex](x_4,y_4)=(-1.917,-1.75)m[/tex]
Explanation:
The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"
The center of mass on a two dimensional plane is defined with the following formulas:
[tex]\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}[/tex]
[tex]\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}[/tex]
Where M represent the sum of all the masses on the system.
And the center of mass C.M [tex]=(\bar x, \bar y)[/tex]
Part a
[tex]m_1= 3 kg, m_2=5kg,m_3=7kg[/tex] represent the masses.
[tex](x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)[/tex] represent the coordinates for the masses with the units on meters.
So we have everything in order to find the center of mass, if we begin with the x coordinate we have:
[tex]\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m[/tex]
[tex]\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m[/tex]
C.M [tex]=(\bar x, \bar y)=(0.767,0.7)m[/tex]
Part b
For this case we have an additional mass [tex]m_4=6kg[/tex] and we know that the resulting new center of mass it at the origin C.M [tex]=(\bar x, \bar y)=(0,0)m[/tex] and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)
[tex]\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m[/tex]
If we solve for a we got:
[tex](3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0[/tex]
[tex]a=-\frac{(5kg*2.3m)}{6kg}=-1.917m[/tex]
[tex]\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m[/tex]
[tex](3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0[/tex]
And solving for b we got:
[tex]b=-\frac{(7kg*1.5m)}{6kg}=-1.75m[/tex]
So the coordinates for this new particle are:
[tex](x_4,y_4)=(-1.917,-1.75)m[/tex]
A heating coil inside an electric kettle delivers 2.2 kW of electrical power to the water in the kettle. How long will it take to raise the temperature of 0.50 kg of water from 30.0° C to 72.0° C
Answer:
0.04s
Explanation:
Specific heat of water c = 4.186J/g.C
The heat energy it would take to heat up 0.5 kg of water from 30C to 72 C is
[tex]E = mc\Delta T = 0.5 * 4.186 * (72 - 30) = 87.9 J[/tex]
If the power of the electric kettle is 2.2kW (or 2200W) and suppose the work efficiency is 100%, then the time it takes to transfer that power is
[tex] t = \frac{E}{P} = \frac{87.9}{2200} = 0.04 s[/tex]
006 10.0 points Two pucks of equal mass collide. One puck (puck A) is stationary before the collision. After the collision the puck that was initially stationary moves along in the same direction in which the puck that hit it (puck B) was initially moving. Puck B Puck A Which of the following statements is correct if the total energy of the system is conserved during the collision (elastic collision)? 1. There isn't enough information to answer the question. 2. Both pucks slide past the point of colli- sion. 3. The puck B remains at the point of colli- sion. 4. The puck B rebounds.
Answer: 3. The puck B remains at the point of collision.
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.
The initial momentum is due only to puck B, as Puck A is at rest.
The final momentum is given by the sum of the momenta of both pucks, so we can write the following equation:
mA*viA = (mA * vfA) + (mB * vfB)
As mA = mB = m, we can simplify the former equation as follows:
viA = vfA + vfB (1)
Now, we also know, that the collision was an elastic collision, so total kinetic energy must be conserved too:
½ m viA²= ½ m vfA² + ½ m vfB²
Simplifying on both sides, we finally have:
viA² = vfA² + vfB² (2)
Now, if we square both sides of (1), we get:
viA² = (vfA + vfB)²= vfA² + 2* vfA * vfB +vfB² (3)
As the right side in (2) and (3) must be equal each other (as the left sides do), the only choice is that either vfA or vfB, be zero.
As we are told that puck A (initially stationary) after the collision, moves, the only possible choice is that puck B remain at rest in the point of collision, after the collision, exchanging his speed with puck A.
In an elastic collision where the total energy of the system is conserved, Puck B rebounds. This is because kinetic energy, along with momentum, is conserved in such a collision, causing the moving puck to rebound.
Explanation:Considering this scenario from the lens of physics, the principle of conservation of momentum and kinetic energy comes into play, which means the total momentum and kinetic energy before and after the collision should remain constant. This is consistent with the properties of an elastic collision. Both pucks have the same mass and initially, Puck B is in motion while Puck A is stationary. Hence, all of the initial energy and momentum are with Puck B.
Upon collision, the energy is transferred, causing Puck A to move in the same direction that Puck B was initially moving. The correct statement is that if the collision is elastic and the total energy of the system is conserved, 'Puck B rebounds'. This rebounding comes from the conservation of kinetic energy principle which holds in an elastic collision. That is, the total initial kinetic energy equals the total final kinetic energy.
Learn more about Elastic Collision here:https://brainly.com/question/33268757
#SPJ3