Answer:
the correct one is C
Explanation:
For this exercise we must use the work definition
W = F. s
Where the bold characters indicate vectors and the point is the scalar producer
W = F s cos θ
Where θ is the angles between force and displacement.
Let us support this in our case. The cable creates an upward tension and with the elevator going down the angle between them is 180º so the work of the cable on the elevator is negative.
The evade has a downward force, its weight so the force goes down and the displacement goes down, as both are in the same direction the work is positive
When examining the statements the correct one is C
Final answer:
The correct statement for an elevator being lowered at constant speed is that the steel cable does negative work on the elevator, and the elevator does negative work on the cable, illustrating the principle that work can be negative if force and displacement are in opposite directions.
Explanation:
The question pertains to the work done by an elevator cable while lowering an elevator at constant speed. According to the principles of work and energy in physics, work done is defined as the force applied in the direction of motion times the distance moved. If an elevator is being lowered at a constant speed, the steel cable exerts an upward force to counteract gravity but the elevator moves downward. Therefore, the displacement of the elevator is in the opposite direction to the force exerted by the cable, resulting in the cable doing negative work on the elevator. Conversely, because the elevator is moving downwards (in the direction opposite to the force exerted by the cable), we can interpret this as the elevator doing negative work on the cable as well, due to the concept that positive work adds energy to a system while negative work removes it.
Thus, the correct statement is: D. The cable does negative work on the elevator, and the elevator does negative work on the cable. This illustrates the application of the definition of work in physics, particularly in scenarios involving opposite directions of force and motion.
A 7-m long wire with a mass of 40 g is under tension. A transverse wave for which the frequency is 630 Hz, the wavelength is 0.5 m, and the amplitude is 7.1 mm is propagating on the wire.
The maximum transverse acceleration of a point on a wire is closest to ____ m/s².
Answer:
111249.41m/s²
Explanation:
Mass of the wire m = 40g = 0.04 Kg
Length of the wire l = 7 m
Linear density of wire μ = m/l
= 0.04/7
= 0.0057 Kg/m
Frequency n = 630 Hz
Wavelength λ = 0.5 m
Amplitude A = 7.1mm = 0.0071m
Speed of the wave v = nλ
= 630*0.5
= 315 m/s
Angular speed ω = 2πn
= 2π*630
= 3958.40 rad/s
Maximum transeverse acceleration
a = ω^2 A
= 3958.40² × 0.0071m
= 111249.41m/s²
The goal in a command economy is economic ____,
Answer:
A- Equality
Explanation:
A 145 g block connected to a light spring with a force constant of k = 5 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 3 cm from equilibrium and released from rest. Find the period of its motion. (Recall that the period, T, and frequency, f, are inverses of each other.) s Determine the maximum acceleration of the block.
Answer:
T = 1.07 s
a = 1.034 m/s2
Explanation:
Metric unit conversion
m = 145 g = 0.145 kg
x = 3 cm = 0.03 m
Suppose this is a simple harmonic motion, then its period T can be calculated using the following equation
[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi\sqrt{\frac{0.145}{5}} = 1.07 s[/tex]
The maximum acceleration would occurs when spring is at maximum stretching length, aka 0.03m
The spring force at that point would be
[tex]F_s = kx = 5*0.03 = 0.15 N[/tex]
According to Newton's 2nd law, the acceleration at this point would be
[tex]a = F_s/m = 0.15 / 0.145 = 1.034 m/s^2[/tex]
The type of friction between the pavement and the tires on a moving vehicle is called.._____friction.
A) kinetic
B) fluid
C) static
D) inertial
Answer:
A) Kinetic
Explanation:
Kinetic friction is friction caused by motion. Kinetic energy is energy in motion.
A track consists of a frictionless arc XY, which is a quarter-circle of radius R, and a rough horizontal section YZ. Block A of mass M is released from rest at point X, slides down the curved section of the track, and collides instantaneously and inelastically with identical block B at point Y. The two blocks move together to the right, sliding past point P, which is a distance l from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is ì. Express your answers in terms of M, R, ì, R, and g.
The question is not complete, so i have attached an image with the complete question.
Answer:
A) speed of block a before it hits block B;v = √2gR
B) speed of combined blocks after collision;V_ab = ½√2gR
C) kinetic energy lost = ½MgR
D) temperature change; ∆t = ½gR/c
E) Additional thermal energy;W_f = 2μMgL
Explanation:
A) From conservation of energy, potential energy = kinetic energy.
Thus; P.E = K.E
So, mgr = ½mv²
m will cancel out to give;
gr = v²
So, v = √2gR
B) from conservation od momentum, momentum before collision = momentum after collision.
Thus;
M_a•V_a = M_ab•V_ab
Where;
M_a is the mass of block A
V_a is speed of block A
M_ab is the mass after collision
V_ab is speed after collision
So, making V_ab the subject, we have;
V_ab = M_a•V_a/M_ab
Now from answer in a above,
V_a = V = √2gR
Also, M_a = M and M_ab = 2M because it's the sum of 2 masses after collision.
Thus;
V_ab = (M√2gR)/(2M)
M will cancel out to give;
V_ab = ½√2gR
C) From the work-kinetic energy theorem, the net work done on the object is equal to the change in the kinetic energy of the object.
Thus;
W_net = K_f - K_i = ½m_ab•v_ab²- ½m_a•v_a²) = ∆K.
We have seen that:
M_a = M and M_ab = 2M
V_a = √2gR and V_ab = ½√2gR
Thus;
∆K = ½(2M•(½√2gR)²) - ½(M•(√2gR)²)
∆K = ½MgR - MgE
∆K = -½MgR
Negative sign means loss of energy.
Thus kinetic energy lost = ½MgR
D) The formula for heat energy is given by;
Q = m•c•∆t
Thus, change in temperature is;
∆t = Q/Mc
Q is the heat energy, thus Q = ½MgR
Thus;∆t = ½MgR/(Mc)
M will cancel out to give;
∆t = ½gR/c
E) Additional thermal energy is gotten from;
Work done by friction;W_f = F_f x d
Where;
F_f is frictional force given by μF_n. F_n is normal force
d is distance moved
Thus;
W_f = μF_n*d
Mass after collision was 2M,thus, F_n = 2Mg
We are told to express distance in terms of L
Thus;
W_f = μ2Mg*L
W_f = 2μMgL
The problem involves principles of energy conservation and kinetic friction. As the block slides down, it converts potential energy to kinetic, collides with another block, and moves together along a rough surface. The work done against friction equals energy at the collision point yielding the expression for velocity after collision V' = sqrt(2µgl).
Explanation:The important concept here is conservation of energy. As block A slides down, it is converting potential energy into kinetic energy. At point X, the potential energy is maximum because it's at the highest point. The potential energy is given by M*g*R. By the time it reaches point Y, the potential energy has turned into kinetic energy which results in the block's speed (½MV²) at point Y.
After the collision, the two blocks move together with a new mass of 2M and new velocity V': MV = 2MV'. In the horizontal section YZ, due to kinetic friction, the blocks lose energy as work done against friction which causes them to stop.
The work done against friction is: F *d = µ*2M*g*l. Equating the energy at point Y after collision to work done against friction, we get: 2*½MV'² = µ*2M*g*l. Resolving gives the expression V' = sqrt(2µgl).
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A series RC circuit, which is made from a battery, a switch, a resistor, and a 4-μF capacitor, has a time constant of 8 ms. If an additional 7-μF is added in series to the 4-μF capacitor, what is the resulting time constant?
Answer:
The resulting time constant will be [tex]\bf{5.09~ms}[/tex].
Explanation:
Given:
the time constant of the RC circuit, [tex]\tau = 8~ms[/tex]
The value of the capacitor in the circuit, [tex]C_{1} = 4~\mu F[/tex]
The value of addition capacitor added to the circuit, [tex]C_{2} = 7~\mu F[/tex]
The value of the time constant for a series RC circuit is give by
[tex]\tau = RC[/tex]
So the value of the resistance in the circuit is
[tex]R &=& \dfrac{\tau}{C}\\&=& \dfrac{8 \times 10^{-3}~s}{4 \times 10^-6~F}\\&=& 2000~\Omega[/tex]
When the capacitor [tex]C_{2}[/tex] is added to the circuit, the net value of the capacitance in the circuit is
[tex]C &=& \dfrac{C_{1}C_{2}}{C_{1} + C_{2}}\\&=& \dfrac{4 \times 7}{4 + 7}~\mu F\\&=& \dfrac{28}{11}~\mu F[/tex]
So the new time constant will be
[tex]\tau_{n} &=& (2000~\Omega)(\dfrac{28}{11} \times 10^{-6})~s\\&=& 5.09~ms[/tex]
The Earth produces an inwardly directed electric field of magnitude 150V/m near its surface. (a) What is the potential of the Earth's surface relative to V=0 at r=[infinity]? (b) If the potential of the Earth is chosen to be zero, what is the potential at infinity? (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth's net charge; how would this affect your answer?)
Answer:
a
The potential of the earth surface is [tex]V_E= - 9.6*10^9 V[/tex]
b
when the potential of the earth is zero the charge is choose to be negative this implies that the potential at infinity would be [tex]V_\infty=+9.6*10^9V[/tex] and (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth's net charge the potential at the surface increase and the electric field remains the same
Explanation:
From the question we are told that
The magnitude electric field is [tex]E = 150V/m[/tex]
The potential of the earth surface when V = 0 and [tex]r = \infty[/tex] is mathematically represented as
[tex]V_E = - \frac{q}{4 \pi \epsilon_o r} = - E *R[/tex]
Where q is the charge on the surface of the earth which is negative
R is the radius of the earth
[tex]V_E = 150 * 64 *10^6[/tex]
[tex]V_E= - 9.6*10^9 V[/tex]
when the potential of the earth is zero the charge is choose to be negative
Then the potential at infinity would be [tex]+9.6*10^9V[/tex]
Immediately after being struck by a hammer, the nail (mass of 50 g) has a velocity of 50 m/s. The total frictional force is 62.5 kM. How far does the nail move before it comes to a stop
Answer:
nail will stop after traveling 1000 m
Explanation:
We have given mass m = 50 gram = 0.05 kg
Frictional force which is used to stop the mass F = 62.5 kN
Initial velocity is given u = 50 m/sec
From newton's law force is equal to F = ma, here m is mass and a is acceleration
So [tex]a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2[/tex]
As finally nail stops so final velocity v = 0 m/sec
From third equation of motion [tex]v^2=u^2+2as[/tex]
So [tex]0^2=50^2-2\times 1.25\times 10^6\times s[/tex]
s = 1000 m
So nail will stop after traveling 1000 m
Two concentric circular loops lie in the same plane. One is connected to a source that supplies an increasing current; the other is a simple closed ring. Is the induced current in the ring in the same direction as the current in the loop connected to the source, or opposite? What if the current in the first loop is decreasing?
Answer:
Explanation:
We shall apply Lenz's law to solve the problem . This law states that direction of induced current is such that it opposes the change that creates it. Since current increases in the coil it creates increasing magnetic field in the other coil . So the current will be induced in it so that it opposes this increase . It can be done only if current in opposite direction is induced in it . Hence in the first case, current will be induced in opposite direction .
In this case, current is decreasing in the primary coil and current will be induced in the secondary coil. Decreasing current will create decreasing magnetic field . So induced current will try to increase it . In can be done if current in the same direction is induced in the secondary coil.
Hence in the second case , current will be induced in the same direction .
The two spheres are rigidly connected to the rod of negligible mass and are initially at rest on the smooth horizontal surface. A force F is suddenly applied to one sphere in the y-direction and imparts an impulse of 9.8 N·s during a negligibly short period of time. As the spheres pass the dashed position, calculate the magnitude of the velocity of each one.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The velocity of the both spheres is [tex]v=4.62 m/s[/tex]
Explanation:
The free body diagram of this question is shown on the second uploaded image
Looking at this diagram we can deduce that there is no impulse force along the horizontal direction so
The mathematical equation for the impulse force along the horizontal axis is
[tex]\int\limits {\sum F_s} \, dt = 0[/tex]
The mathematical equation for the impulse force along the horizontal axis is
[tex]\int\limits {\sum F_y} \, dt = \Delta I_y[/tex]
Where [tex]I_y[/tex] is the impulse momentum along the y-axis and this is mathematically given as
[tex]\Delta I_y = 2mv_y[/tex]
substituting 9.8 N.s for [tex]\Delta I_y[/tex] and 1.5kg for m (mass of sphere) and the making [tex]v_y[/tex] the subject
[tex]v_y = \frac{9.8}{2 *1.5}[/tex]
[tex]=3.267 m/s[/tex]
The sum of the moment about the point I is mathematically represented as
[tex]\int\limits {\sum M_I } \, dt =\Delta H_I[/tex]
From the free body diagram [tex]\int\limits {\sum M_I } \, dt = I * 0.3 = 9.8 *0.3=2.94N.m[/tex]
[tex]\Delta H_I[/tex] is the angular momentum along the horizontal axis given as
[tex]\Delta H_I[/tex] [tex]= 2mv_x(0.3)[/tex]
substituting parameters into above equation
[tex]2.94 = 2 * 1.5 * v_x * 0.3[/tex]
Making [tex]v_x[/tex] the subject
[tex]v_x = \frac{2.94}{2*1.5*0.3}[/tex]
[tex]=3.267 m/s[/tex]
the velocity of the sphere is mathematically represented as
[tex]v = \sqrt{(v_x)^2 + (v_y)^2}[/tex]
Now substituting values
[tex]v = \sqrt{3.267^2 + 3.267^2}[/tex]
[tex]v=4.62 m/s[/tex]
The physics question pertains to the concept of impulse and momentum. The velocity of each sphere after the impulse is imparted can be calculated by dividing the change in momentum by twice the mass of one sphere.
Explanation:The Physics question is relevant to impulse and momentum in the domain of mechanics. From the problem, we know that a force F acting on the sphere imparts an impulse of 9.8 N·s. Impulse, denoted as J, in physics, is the product of force and the time for which it is applied and is equivalent to the change in momentum of the body. Thus, the change in momentum, Δp is equal to the impulse imparted, which is 9.8 N·s.
Now considering the system of two spheres rigidly connected, it is an isolated system (external force F is not considered as it acts for a very short time) and therefore, the total momentum of the system remains conserved. Therefore the momentum imparted to one sphere is equally distributed to both the spheres. Therefore, the final velocity v of each sphere can be calculated by using the formula v = Δp / 2m (m being the mass of each sphere).
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A local fun house incorporates a gently curved, concave, spherical mirror into its display. When a child stands 1.2 m from the mirror, her reflection is upside down and appears to float in front of the mirror. When she stands about 0.8 m from the mirror, she sees only a blur reflected in the mirror, but when she stands about 0.5 m from the mirror, her reflection is right side up and appears to be behind the mirror. The approximate focal length of the mirror is
Options:
a) more than 0.8 m .
b) equal to 0.8 m .
c) between 0.5 m and 0.8 m .
d) less than 0.5 m .
Answer:
b) equal to 0.8 m .
Explanation:
Note:
An upside down image = Inverted Image
An image that appears in front of the mirror = Real image
An image that appears behind the mirror = Virtual image
Let the object distance from the pole of the mirror be u
When the child stands 1.2 m from the mirror:
u = 1.2 m ( Real and Inverted image of the child is formed)
When the child stands about 0.8 m from the mirror:
u = 0.8 m (Virtual, erect and magnified image of the child is formed)
When the child stands about 0.5 m from the mirror:
u = 0.5 m ( Virtual and erect image of the child is formed)
Note: All objects positioned behind the focal length of a concave mirror are always real. Objects start becoming virtual when they are placed on the focal length or in front of it (Close to the pole of the mirror), although objects placed on the focus has its image formed at infinity.
Since the nature of the image formed changed from real to virtual when the child stands about 0.8 m from the mirror, then the focal length is approximately equal to 0.8 m
The approximate focal length of the mirror will be "0.8 m". A complete solution is provided below.
According to the question,
When a child stands 1.2 m, the real as well as inverted image will be formed, then
u = 1.2 mWhen a child stands 0.8 m, the virtual, erect as well as magnified image will be formed, then
u = 0.8 mWhen a child stands 0.5 m, the virtual as well as erect image will be formed, then
u = 0.5 mSince,
Whenever the image is formed from real to virtual, the focal length will become 0.8 m.
Thus the answer above is right.
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g An object spins in place with no unbalanced forces or torques acting upon it, what do we expect this object to do? The object’s spin will slow and eventually reverse direction. The object will continue spinning as it has been. The object will spin faster and faster. The object will begin to roll. None of these answers. The object will slowly stop spinning.
Answer:
The object will continue spinning as it has been.
Explanation:
When an object is spinning in a closed system with no unbalance
forces and no external force or torques are applied to it, it will have no change in angular momentum.
According to the law of conservation of angular momentum which states that when no external torque acts on an object, no change of angular momentum will occur
If the change in angular momentum is zero, then the angular momentum is constant; therefore, the object will continue spinning as it has been.
The same scenario is applicable to the rotation and the spinning of the earth.
Answer: The object will continue spinning as it has been.
Explanation: Unbalanced forces refers a force that changes the position, speed or direction of the object to which it is applied. While torque refers to otational or twisting effect of a force; a moment of force, defined for measurement purposes as an equivalent straight line force multiplied by the distance from the axis of rotation.
Any object that spins in place with no unbalanced forces or torques acting upon it will continue spinning as it has been. An example can be seen in the rotation of planets.
An object is electrically charged if the amounts of positive and negative charge it contains are not _______.
equal
negative
positive
increased
Answer:oops
Explanation:
If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?
P=6w
I=3A
V = ?
Answer:2 volts
Explanation:
Power=current x voltage
6=3 x voltage
Divide both sides by 3
6/3=(3 x voltage)/3
2=voltage
Voltage=2volts
If part of an electric circuit dissipates energy at 6 Watts when it draws a current of 3 Amperes, then the voltage is impressed across it would be 2 volts .
What is power ?The rate of doing work is known as power. The Si unit of power is the watt .
The power dissipated in the circuit = work / time
As given in the problem we have to find out what voltage is impressed across the circuit If part of an electric circuit dissipates energy at 6 Watts when it draws a current of 3 Amperes,
The power dissipated across the circuit = Voltage × current
6 watts = Voltage × 3 ampere
Voltage = 6 / 3
Voltage = 2 volts
Thus , the voltage across the circuit would be 2 volts .
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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 31.5 kg. The child grabs and clings to a bar that is 1.25 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 51.0 rpm to 17.0 rpm . What is the moment of inertia of the merry‑go‑round with respect to its central axis?
Answer:
24.609375 kgm²
Explanation:
I = Moment of inertia of the merry go round
[tex]N_1[/tex] = Initial speed = 51 rpm
[tex]N_2[/tex] = Final speed = 17 rpm
m = Mass of child = 31.5 kg
r = Radius = 1.25 m
In this system the angular momentum is conserved
[tex]IN_1=(I+mr^2)N_2\\\Rightarrow I\times 51=(I+31.5\times 1.25^2)17\\\Rightarrow I\dfrac{51}{17}-I=31.5\times 1.25^2\\\Rightarrow 2I=49.21875\\\Rightarrow I=\dfrac{49.21875}{2}\\\Rightarrow I=24.609375\ kgm^2[/tex]
The moment of inertia of the merry go round is 24.609375 kgm²
Which are examples of short-term environmental change? Check all that apply.
tsunamis
El Niño
large asteroid and comet impacts
volcanic eruptions
global warming
i just took the test it's: tsunamis, El Nino, and volcanic eruptions.
Answer: tsunamis,El Nino,and volcanic eruptions
Explanation:
the correct answer for e d g e n u i t y
The examples of short-term environmental change are Tsunamis, El Niño, and Volcanic eruptions.
Tsunamis are huge ocean waves that are triggered by undersea disturbances like earthquakes or landslides. They can inflict major damage to coastal regions, however they usually happen in a short period of time.
El Nio is a climatic trend characterised by higher-than-normal equatorial Pacific ocean temperatures.
It can cause changes in weather patterns, such as greater rainfall in some areas and droughts in others, although the impacts are usually transient and endure for a few months to a couple of years.
Volcanic eruptions spew lava, ash, and gases into the atmosphere, which can have direct effects on the environment, including as changes in air quality and local weather patterns.
Thus, these effects, however, are often short-lived and localized.
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A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J. Calculate the kinetic energy and speed of the ejected electron.
Explanation:
Given that,
Wavelength of the light, [tex]\lambda=4170\ A=4170\times 10^{-10}\ m[/tex]
Work function of sodium, [tex]W_o=4.41\times 10^{-19}\ J[/tex]
The kinetic energy of the ejected electron in terms of work function is given by :
[tex]KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J[/tex]
The formula of kinetic energy is given by :
[tex]KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s[/tex]
Hence, this is the required solution.
cylindrical water tank is 20.0 m tall, open to the atmosphere at the top, and is filled to the top. It is noticed that a small hole has occurred in the side at a point 16.5 m below the water level and that water is flowing out at the volume flow rate of 2.90 10-3 m3/min. Determine the following. (a) the speed in m/s at which water is ejected from the hole
Answer:
a
The velocity is [tex]v =17.98 \ m/s[/tex]
b
The diameter is [tex]d = 0.00184m[/tex]
Explanation:
The diagram of the set up is shown on the first uploaded image
From the question we are told that
The height of the water tank is [tex]h = 20.0 \ m[/tex]
The position of the hole [tex]p_h = 16.5m[/tex] below water level
The rate of water flow [tex]\r V = 2.90 *10^{-3} m^3 /min = \frac{2.90 *10^{-3}}{60} = 0.048*10^{-3} m^3/s[/tex]
According to Bernoulli's theorem position of the hole
[tex]\frac{P_o + h \rho g}{\rho} + \frac{1}{2} u^2 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]
Where u is the initial speed the water through the hole = 0 m/s
[tex]P_o[/tex] is the atmospheric pressure
[tex]\frac{P_o }{\rho} + \frac{ h \rho g}{\rho} + 0 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
Substituting value
[tex]v = \sqrt{2 * 9.8 * 16.5 }[/tex]
[tex]v =17.98 \ m/s[/tex]
The Volumetric flow rate is mathematically represented as
[tex]\r V = A * v[/tex]
Making A the subject
[tex]A = \frac{\r V}{v}[/tex]
substituting value
[tex]A = \frac{0.048 *10^{-3}}{17.98}[/tex]
[tex]= 2.66*10^{-6}m^2[/tex]
Area is mathematically represented as
[tex]A = \frac{\pi d^2}{4}[/tex]
making d the subject
[tex]d = \sqrt{\frac{4*A}{\pi} }[/tex]
Substituting values
[tex]d = \sqrt{\frac{4 * 2.67 *10^{-6}}{3.142} }[/tex]
[tex]d = 0.00184m[/tex]
Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 780 m diameter rotating about its central axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity?
Answer:1.513 rps
Explanation:
Given
Diameter of cylindrical space [tex]d=780\ m[/tex]
When the station rotates it creates centripetal acceleration which is given by
[tex]a_c=\omega ^2r[/tex]
Now it must create the effect of gravity so
[tex]g=\omega ^2r[/tex]
[tex]\omega =\sqrt{\frac{g}{r}}[/tex]
[tex]\omega =0.158\ rad/s[/tex]
and [tex]\omega =\frac{2\pi N}{60}[/tex]
Thus [tex]N=\frac{4.755}{3.142}[/tex]
[tex]N=1.513\ rps[/tex]
A color television tube generates some X-rays when its electron beam strikes the screen. What is the shortest wavelength of these X-rays, in meters, if a 31 kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these X-rays from reaching viewers.)
Answer: 4.0024 x 10^ -11 m or 0.040024 nm
Explanation:
λ = h c/ΔE
λ = wave lenght
h = 6.626 x 10 ^ -34 m² kg /s = planck constant
ΔE = 31 keV potential ( 1 keV = 1.6021 x 10^-16J)
c = velocity of light = 3 x 10⁸ m/s
substitute gives
λ = 6.626 x 10 ^ -34 m² kg /s x 3 x 10⁸ m/s = 4.0024 x 10^ -11 m
31 x 1.6021x10^-16 J
Kinetic energy caused by the vibration of particles in a medium such as steel water or air
Final answer:
Kinetic energy is the energy caused by the vibration of particles in a medium. It is calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.
Explanation:
Kinetic energy is the energy an object has because of its motion. It is caused by the vibration of particles in a medium such as steel, water, or air. Kinetic energy is calculated as one-half the product of the mass of the particle and the square of its speed.
For example, when a rock is thrown into a pond or when a swimmer splashes the water's surface, the kinetic energy of the vibrating water particles is generated. Similarly, mechanical sound waves also have kinetic energy due to the movement of air particles and the potential energy caused by the elasticity of the material through which the sound propagates.
In summary, kinetic energy arises from the motion of particles in a medium and can be calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.
A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much work is done in stretching the spring 5.5 m from its equilibrium position? b. How much work is done in compressing the spring 1.5 m from its equilibrium position?
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer:
a) W = 1815 J
b) W = 135 J
Explanation:
We need to model the force using Hooke's law
We first get the elastic constant, k
A force of 60 N causes an extension of 0.5 m
F = 60 N, e = 0.5 m
F = k * e
60 = 0.5 k
k = 60/0.5
k = 120 N/m
Therefore the force in terms of the extension, x is:
F(x) = 120x
a) Work done in stretching the spring 5.5 m from its equilibrium position
b = 0 m, a = 5.5 m
[tex]W = \int\limits^a_b {F(x)} \, dx[/tex]
[tex]W = \int\limits^a_b {120x} \, dx \\a =0, b = 5.5[/tex]
[tex]W = 60x^{2} \\W = 60 [5.5^{2} -0^{2} ][/tex]
[tex]W = 1815 J[/tex]
b) Work required to compress the spring 1.5 m from its equilibrium position
[tex]W = \int\limits^a_b {F(x)} \, dx[/tex] b = 0, a = -1.5
[tex]W = 60x^{2} \\W = 60 [(-1.5)^{2} -0^{2} ]\\[/tex]
W = 135 J
The work done to compress the spring 1.5 m from its equilibrium position is 135 J
A string under tension τi oscillates in the third harmonic at frequency f3, and the waves on the string have wavelength λ3. If the tension is increased to τf = 5.0τi and the string is again made to oscillate in the third harmonic, what then are (a) the ratio of frequency of oscillation to f3 and (b) the ratio of the wavelength of the waves to λ3?
Answer:
(a). [tex]\dfrac{f_3'}{f_3} =\sqrt{5}.[/tex]
(b). The wavelength remains unchanged.
Explanation:
The speed [tex]v[/tex] of the waves on the string with tension [tex]T_i[/tex] is given by
[tex]v = \sqrt{\dfrac{T_iL}{m} }[/tex]
And if the string is vibrating, its fundamental wavelength is [tex]2L[/tex], and since the frequency [tex]f[/tex] is related to the wave speed and wavelength by
[tex]f = v/\lambda[/tex]
the fundamental frequency [tex]f_1[/tex] is
[tex]f_1 =\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}[/tex]
and since the frequency of the third harmonic is
[tex]f_3 = 3f_1[/tex]
[tex]f_3 = 3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L},[/tex]
and the wavelength is
[tex]\lambda_3 = \dfrac{2L}{3}.[/tex]
(a).
Now, if we increase to the string tension to
[tex]T_f = 5.0T_i[/tex]
the third harmonic frequency becomes
[tex]f_3' = 3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L},[/tex]
The ratio of this new frequency to the old frequency is
[tex]\dfrac{f_3'}{f_3} = \dfrac{3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L}}{3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}}[/tex]
[tex]\boxed{\dfrac{f_3'}{f_3} =\sqrt{5}.}[/tex]
(b).
The wavelength of the third harmonic remains unchanged because [tex]\lambda_3 = \dfrac{2L}{3}.[/tex] depends only on the length of the string
Which of the following energy sources is no way derived from the sun is it nuclear energy title energy when energy or petroleum
Answer:nuclear energy is not derived from the sun
Explanation:
Nuclear energy is not derived from the sun.nuclear energy comes from the energy released when atoms are split apart and some mass is converted to energy
At the instant when the speed of the loop is 3.00 m/sm/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
Complete question:
A rectangular loop of wire with dimensions 2.0 cm by 10.0 cm and resistance 1.0 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude of 2.0 T and is directed into the plane. At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
Check the image uploaded
Answer:
The magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N
Explanation:
Given;
resistance of the wire; R = 1.0 Ω
magnitude of magnetic field strength, B = 2.0 T
speed of the loop, v = 3.00 m/s
Induced emf is given as;
ε = IR
[tex]I = \frac{emf}{R} = \frac{VBL}{R}[/tex]
magnitude of the force that the magnetic field exerts on the loop:
F = BIL
Substitute in the value of I
[tex]F = \frac{VB^2L^2}{R}[/tex]
where;
L is the displacement vector between the initial and final end of the portion of the wire inside the field = 2.0 cm
Substitute the given values and solve for F
[tex]F = \frac{3*2^2*(2*10^{-2})^2}{1} \\\\F = 4.8 *10^{-3} \ N[/tex]
Therefore, the magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N
Complete Question:
A rectangular loop of wire with dimensions 1.50 cm by 8.00 cm and resistance 0.700 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.20 T and is directed into the plane of.
At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
Answer:
F = 0.133 N
Explanation:
Magnitude of the magnetic field, B = 2.20 T
Length of the loop = 1.5 cm = 0.015 m
The speed of the loop, v = 3.00 m/s
The emf induced in the loop , e = Blv
e = 2.20 * 0.015 * 3
e = 0.099 V
Current induced in the loop, I = e/R
I = 0.099/0.7
I = 0.1414 A
The magnitude of the force is given by, F = I *l *B sin90
F = 0.1414 * 0.015 * 2.20
F = 0.00467 N
At any angular speed, a uniform solid sphere of diameter D has the same rotational kinetic energy as a uniform hollow sphere of the same diameter when both are spinning about an axis through their centers. The moment of inertia of a solid sphere is :
Answer:
m = (3/5)*M
Explanation:
Given:-
- The angular speed of both hollow and solid sphere = w
- The diameter of solid & hollow sphere = D
- The mass of the solid sphere = M
- Both rotate about their common axis with similar rotational kinetic energy.
Find:-
The mass of hollow sphere (m) ?
Solution:-
- The formula for rotational kinetic energy (K.E) of any rigid body is:
K.E = 0.5*I*w^2
Where,
I : Moment of inertia of rigid body
- The rotational kinetic energies of both hollow sphere and solid sphere are same:
0.5*I_solid*w^2 = 0.5*I_shell*w^2
I_solid = I_shell
0.4*M*D^2 / 4 = (2/3)*m*D^2 / 4
(2/5)*M = (2/3)*m
m = (3/5)*M
A 62.0 kgkg skier is moving at 6.90 m/sm/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 mm long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 mm high.How fast is the skier moving when she gets to the bottom of the hill?
How much internal energy was generated in crossing the rough patch?
Answer:
the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.
the internal energy generated in crossing the rough patch is 820.26 J
Explanation:
Given that,
Mass, m = 62 kg,
Initial speed, = 6.90 m/s
Length of rough patch, L = 4.50 m,
coefficient of friction, = 0.3
Height of inclined plane, h = 2.50 m
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
According to energy conservation equation,
Part (b)
The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.
The magnitude of the friction force is:
[tex]F_f=\mu mg[/tex]
Therefore, the work done by friction is:
[tex]W=-F_f d =-\mu mg d[/tex]
The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier
Substituting,
[tex]W=-(0.300)(62.0)(9.8)(4.50)=-820.26 J[/tex]
So, the internal energy generated in crossing the rough patch is 820.26 J
Part (a)
If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:
[tex]E=K_i +U_i = \frac{1}{2}mu^2 + mgh[/tex]
After crossing the rough patch, the new mechanical energy is
[tex]E'=E+W[/tex]
where
W = -820.26 J is the work done by friction
At the bottom of the hill, the final energy is just kinetic energy,
[tex]E' = K_f = \frac{1}{2}mv^2[/tex]
where v is the final speed.
According to the law of conservation of energy, we can write:
[tex]E+W=E'[/tex]
So we find v:
[tex]\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\\\v=\sqrt{u^2+2gh+\frac{2W}{m}}\\\\=\sqrt{6.9^2+2(9.8)(2.50)+\frac{2(-820.26)}{62.0}}\\\\=8.38 m/s[/tex]
Thus, the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.
The question repeated the units attached the the values
So in order of their appearance in the question, the correct units are;
62kg, 6.9m/s, 4.5m, 0.30, 2.5m
Answer:
A) Speed at bottom of hill = 8.38 m/s
B) Internal energy generated in crossing the patch = 820.26J
Explanation:
From the question, it's clear that there are 2 distinct stages of of her skis. The first stage is when she is on the rough patch while the second stage is when she skis down the hill.
A) For the first stage when she is on the rough patch;
If we apply Newton's second law of motion to the skis along the vertical direction, we obtain;
ΣF_y = N - mg = 0
Thus,N = mg
m = 62kg,thus,N = 62 x 9.8 = 607.6N
Now, let's find the kinetic friction. It's given by; f_k = μN
Where μ is coefficient of kinetic friction
Thus, f_k = 0.3 x 607.6 = 182.28N
Now, let's calculate the work done by this frictional force,
So, W_fk = f_k x distance = 182.28 x
4.5 = 820.26J
This work is done in a direction opposite to the displacement and it will have a negative sign. Thus,
W_fk = - 820.26J
Now, since the skier skies horizontally and perpendicular to the gravitational force, the work done due to gravity is zero. Thus,
W_grav = 0
Let's calculate the kinetic energy at the beginning of the rough patch.
K1 = (1/2)m(v1)²
K1 = (1/2) x 62 x (6.9)²= 1475.91J
Also,let's calculate the kinetic energy at end of rough patch.
K2 = (1/2)m(v2)²
where v2 is final velocity at end of rough patch
K2 = (1/2)(62)(v2)² = 31(v2)²
Now, the total work done when other forces other than gravity do work, is given by ;
W_total = W_other + W_grav = K2 - K1
In this case, W_other is W_fk
Thus,
- 820.26J + 0 = 31(v2)² - 1475.91J
31(v2)² = 1475.91J - 820.26J
31(v2)² = 655.65
(v2)² = 655.65/31
v2 =√21.15 = 4.6 m/s
Now, for the second stage when she skis down the hill;
In this case the only force acting is gravity, thus, W_other = 0
Work done by gravity = mgh
W_grav = 62 x 9.8 x 2.5 = 1519 J
Now K2 will be kinetic energy at top of hill while K3 will be kinetic energy at bottom of hill.
Thus,
K2 = (1/2)m(v2)²
K2 = (1/2)(62)(4.6)²
K2 = 655.96J
Similarly,
K3 = (1/2)m(v3)²
Where v3 is velocity at bottom of hill.
Thus,
K3 = (1/2)(62)(v3)²
K3 = 31(v3)²
Again, the total work done when other forces other than gravity do work, is given by ;
W_total = W_other + W_grav = K3 - K1
So,
W_other is zero.
Thus,
1519J = 31(v3)² - 655.96J
31(v3)² = 1519J + 655.96J
31(v3)² = 2174.96J
(v3)² = 2174.96/31
v3 = √70.16
v3 = 8.38 m/s
B) Workdone by non-conservative forces manifests itself as changes in the internal energy of bodies. Now, frictional force experienced in the first stage in crossing the patch is a non - conservative force. Thus,
Internal energy = W_fk = 820.26J
The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit. The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit. twice / twice twice / half half / half the same as / half half / twice half / the same as
Answer:
Current is half
Voltage is half
Explanation:
according to Ohms law V = IR
Case 1
I₁ = V / R
I₂ = V / (R+R) = V / 2R
∴ I₁ / I₂ = (V/R) / (V/2R)
=> I₁ / I₂ = (V/R) * (2R/V)
=> I₁ / I₂ = (2VR / RV)
I₁ / I₂ = 2
I₂ = I₁ / 2
The current through each resistor in the two-resistor circuit is half the current through the resistor in the one-resistor circuit (the circuit in Part A).
The voltage across each resistor in the two-resistor circuit is half the voltage across the resistor in the one-resistor circuit.
The current through each resistor in a two-resistor series circuit is the same as the current through a one-resistor circuit. However, the voltage across each resistor in the two-resistor circuit is half of the voltage across the single resistor due to how voltages distribute in series circuits.
Explanation:In a series circuit, such as the two-resistor circuit you have mentioned, the current through each resistor is the same as the current through the single resistor in the one-resistor circuit. This is due to Kirchhoff's Current Law which states that the current entering a junction or a node must equal the current leaving it. However, the voltage accross each resistor in the two-resistor circuit is half of the voltage across the single resistor in the one-resistor circuit. This is because in a series circuit, the total voltage is the sum of the individual voltage drops across each resistor, according to Ohm’s law.
Learn more about Series Circuit here:https://brainly.com/question/34427175
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To make things easier, how do people design a circuit and put it on paper ?
Explanation:
A paper circuit is a functioning electronic circuit built on a paper surface. Projects involving paper circuits are unique because of the use of traditional art techniques and some unique materials to create a circuit that combines aesthetics and functionality.
Paper circuit are sometimes carefully designed and then transfered on the paper, and in some cases, they are designed directly on the paper (usually by experts that already know what they're doing).
To make designing circuit on paper easier, there are three commonly used materials for the circuitry and they are, conductive paints, conductive tapes and conductive inks.
Conductive tapes are made from metal strip (usually copper) that are taped to the paper. They are good to work with since they allow components to be soldered on them creating a stronger and more reliable joint.
Conductive paints are special paints that can be used to outline circuit path and also serve as the circuit. The only problem with conducting paints is that it can be messy, and needs time to dry.
Conductive inks are easier to use as they need no drying time. They are far less messy and allows the drawing of elaborate and more intricate circuit on the paper.
There are also components that have been modified (like led lights etc) and available to make paper circuit design easier.
A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, compressing it 5.0 cm. How hard is the athlete pushing?
Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
Answer:
37.5N
Explanation:
According to Hooke's law, the load or force, F, applied on an elastic material (e.g a spring) is directly proportional to the extension or compression, e, caused by the load. i.e
F ∝ e
F = k x e -------------------------(i)
where;
k = proportionality constant known as the spring constant.
From the question;
k = 750N/m
e = 5.0cm = 0.05m
Substitute these values into equation (i) as follows;
F = 750 x 0.05
F = 37.5N
Therefore, the load applied which is a measure of how hard the athlete is pushing is 37.5N