Answer:
False
Explanation:
Friction is the resisting force of motion that converts kinetic energy into heat energy. Kinetic energy is the energy of movement. Removing kinetic energy makes the object slower.
Suppose that the resistive force of the air on a skydiver can be approximated by f = −bv2. If the terminal velocity of an 82.0 kg skydiver is 33.4 m/s, what is the value of b (in kg/m)?
Answer:
The value of b 0.7351 kg/m
Explanation:
Given that;
Mass of sky diver = 82 kg
Velocity = 33.4 m/s
f = −bv²
It is a resistance force, therefore the negative sign is ignored.
since; f = mg
∴ mg = bv²
b = mg / v ² ........ (1)
At terminal velocity a = 0
Put parameters in (1)
b = (82 × 10) / (33.4)²
b = 820 / 1,115.56
b = 0.7351
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.5 m above the river, whereas the opposite side is a mere 2.4 m above the river. The river itself is a raging torrent 57.0 m wide.
A) How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?
Answer:
(A) Velocity of car in order just to clear the river and land safely on the opposite side is [tex]29.68 \frac{m}{s}[/tex]
(B) The speed of the car is [tex]35.14 \frac{m}{s}[/tex]
Explanation:
Given:
Car distance from river [tex]y = 20.5[/tex] m
Mere distance from river [tex]y_{o} = 2.4[/tex] m
River length [tex]x= 57[/tex] m
(A)
For finding the velocity of car,
Using kinematics equation we find velocity of car
[tex]y - y_{o} = v_{o}t + \frac{1}{2} gt^{2}[/tex]
Where [tex]v_{o} = 0[/tex], [tex]g = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex] ( given in question )
[tex]20.5 - 2.4 = \frac{1}{2} \times 9.8 \times t^{2}[/tex]
[tex]t= 1.92[/tex] sec
The speed of the car before it lands safely on the opposite side of the river is given by,
[tex]v_{x} = \frac{x-x_{o} }{t}[/tex]
[tex]v_{x} = \frac{57-0}{1.92}[/tex] ( [tex]x_{o} = 0[/tex] )
[tex]v_{x} = 29.68[/tex] [tex]\frac{m}{s}[/tex]
(B)
For finding the speed,
The horizontal distance travel by car,
[tex]v_{y} = v_{o} + at[/tex]
Where [tex]a = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex]
[tex]v_{y} = 9.8 \times 1.92[/tex]
[tex]v_{y} = 18.81[/tex] [tex]\frac{m}{s}[/tex]
For finding the speed,
[tex]v = \sqrt{v_{x}^{2} + v_{y} ^{2} }[/tex]
[tex]v = \sqrt{(29.68)^{2}+ (18.81)^{2} }[/tex]
[tex]v = 35.14[/tex] [tex]\frac{m}{s}[/tex]
Therefore, the speed of the car is [tex]35.14 \frac{m}{s}[/tex]
An object with mass 3.7 kg is executing simple harmonic motion, attached to a spring with spring constant 320 N/m . When the object is 0.017 m from its equilibrium position, it is moving with a speed of 0.60 m/s . Part A Calculate the amplitude of the motion. Express your answer to two significant figures and include the appropriate units. A = nothing nothing
Answer:
The amplitude of the motion of the spring is 1 m.
Explanation:
From the conservation of energy;
[tex](PE_{spring})_{max}= PE+KE[/tex]
which is
[tex]\frac{1}{2} kx^2_{max}=\frac{1}{2} kx^2+\frac{1}{2} mv^2[/tex]
Make [tex]x_{max}[/tex] the subject of formula
[tex]x^2_{max}= \frac{kx^2\times mv^2}{k}[/tex]
[tex]x_{max}= \sqrt{\frac{kx^2\times mv^2}{k} }[/tex]
Substitute 320 N/m for k, 0.017 m for x, 3.7 kg for m, and 0.60 m/s for v.
[tex]x_{max}= \sqrt{\frac{(320)(0.017)^2+ (3.7)(0.60)^2}{320} } \\\\=1m[/tex]
Thus, The amplitude of the motion of the spring is 1 m.
Answer: 0.0667m
Explanation:
given
Mass of the object, m = 3.7kg
Force constant of the spring, k = 320 N/m
Speed of the object, v = 0.6 m/s
Distance from equilibrium position, x = 0.017 m
Using the law of conservation of energy. The total energy conserved is
= 1/2kx² + 1/2mv²
= 1/2 * 320 * 0.017² + 1/2 * 3.7 * 0.6²
= 0.046 + 0.666
= 0.712 J
When v = 0, the maximum deflection is Xmas. This Xmas, is also the amplitude. Such that,
Energy = 1/2kx²
0.712 = 1/2 * 320 * x²
1.424 = 320 x²
x² = 1.424 / 320
x² = 0.00445
x = 0.0667
x = 6.67*10^-2 m
Thus, the amplitude is 0.0667 m
The chart shows the voltage of four electric currents.
Which is best supported by the data in the chart?
Voltage of Currents
Current
Volts (V)
19.0
1.5
W
X
Current w flows at a higher rate than Current Z.
Current Y flows at a lower rate than Current X.
© Current X has a lower potential difference than Current
3.0
Z
4.5
Current Z has a greater potential difference than Current
W.
The question is incomplete. The complete question is as follows.
The chart shows the voltage of four electric currents. A two-column table with 2 row titled Voltage of Currents. The first column labeled Current has entries W, X, Y, Z. The second column labeled Volts (volts) has entries 9.0, 1.5, 3.0, 4.5.
Which is best supported by the data in the chart?
A. Current W flows at a higher rate than Current Z.
B. Current Y flows at a lower rate than Current X.
C. Current X has a lower potential difference than Current Y.
D. Current Z has a greater potential difference than Current W.
Answer: C. Current X has a lower potential difference than Current Y.
Explanation: The table described is show below:
Current Volts (volts)
W 9.0
X 1.5
Y 3.0
Z 4.5
Current is defined by the flow of charge that passes through a material, called conductor in an amount of time:
I = [tex]\frac{Q}{t}[/tex]
where:
Q is charge in Coulomb
t is time in seconds
I is the current in Ampere: A = C/s
According to the question, the currents are not defined by charge or numbers, so it's difficult to determine its value. So, alternatives A and B aren't correct.
Voltage is the difference in electrical potential between two points. It is because of the voltage that the charges flow.
From the table, Current W has the biggest voltage of all the others. So, alternative D is incorrect.
Analysing the voltages of Current X and Y, we observe that they are 1.5 and 3.0, respectively. So, volts or the potential diference of current X is lower than Y, making alternative C the correct one.
Answer:
Current X has a lower potential difference than Current Y.
Explanation:
What is equilibrium????
Answer:
Explanation:
An equilibrium is a state in which opposing forces or influences are banned.
An example of equilibrium is in economics when supply and demand are equal. An example of equilibrium is when you are calm and steady. An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.
Answer:
a state in which opposing forces or influences are balanced.
Explanation:
In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.
Interactive LearningWare 22.2 reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are 0.46 m x 0.68 m. The magnetic field has a magnitude of 3.0 T and is inclined at an angle of 67o with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of 0.49 s, what is the magnitude of the average emf induced in the loop
Explanation:
The dimension of a single rectangular loop is 0.46 m x 0.68 m.
Magnetic field, B = 3 T
The loop is inclined at an angle of 67 degrees with respect to the normal to the plane of the loop.
It is required to find the magnitude of the average emf induced in the loop if the magnetic field decreases to zero in a time of 0.49 s.
Te induced emf in the loop is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=A\dfrac{dB}{dt}\cos\theta\\\\\epsilon=0.46\times 0.68\times \dfrac{3}{0.49}\times \cos(67)\\\\\epsilon=0.74\ V[/tex]
So, the magnitude of the average emf induced in the loop is 0.74 V.
Does light travel faster or slower when it has to pass through air or water
Answer:
Light travels fast in air but i dont know about water
Answer:
light slower in water than in Air
Explanation:
Air's refractive index is about 1.0003, while water's is about 1.3. This means that light is “slower” in water than air. This is because it's more likely to hit a molecule and then get re-emitted, lengthening the amount of time the light takes to get through a certain distance of the medium.
An 8 foot metal guy wire is attached to a broken stop sign to secure its position until repairs can be made. Attached to a stake in the ground, the guy wire makes an angle of 51º with the ground. How far from the foot of the stop sign is the stake, to the nearest tenth of a foot?
Answer:
Explanation:
The guy wire is making a right angled triangle with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.
base / hypotenuse = cos51
base = hypotenuse x cos51
= 8 x cos51
= 5.03 ft .
The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .
If given a device that has unknown circuitry, and you measure that the voltage across the device leads the voltage across a resistor at 500 Hz, can you know what type of elements are in the device?If so, how?
Final answer:
A leading voltage across a device at 500 Hz compared to a resistor suggests inductive elements within the device. Voltage leads current in inductors, which can be determined by comparing phase differences using a voltmeter parallel to the device.
Explanation:
When measuring the voltage across a device and comparing it to the voltage across a resistor at a specific frequency, such as 500 Hz, the phase difference provides information about the type of elements in the device. If the voltage across the device leads the voltage across the resistor, this indicates the presence of inductive reactance, suggesting that there are inductive elements within the unknown circuitry.
Reactance and impedance play key roles in AC circuits. The voltage drop across a resistor is in phase with the current, meaning it neither leads nor lags the current. However, in an inductor, the voltage leads the current, and in a capacitor, the voltage lags the current. Therefore, measuring a leading voltage implies inductive characteristics. To evaluate this, a voltmeter would be placed in parallel with the device, ensuring minimal disturbance to the circuit as high resistance in the voltmeter minimizes current flow through it.
For components like inductors and capacitors, the relationship between voltage and current is not linear; thus, they are not ohmic devices like resistors. This is evident in an RLC series circuit, where the impedance at a given frequency depends on the resistance (R), inductance (L), and capacitance (C) of the circuit.
The voltage leading the resistor's voltage at 500 Hz suggests the device contains capacitive elements. A phase difference measurement can confirm this. Capacitors cause the voltage to lead the current in AC circuits.
If you observe that the voltage across a device leads the voltage across a resistor at 500 Hz, it is likely that the device includes a capacitive element. In AC circuits, capacitors cause the voltage to lead the current, which means that the voltage across the capacitor will peak before the voltage across a purely resistive element.
To confirm the presence of a capacitive component, you can perform a more detailed analysis:
Measure the phase difference between the voltage across the device and the voltage across the resistor.If the phase difference is close to 90 degrees, it indicates the presence of a pure capacitor.If the phase difference is less than 90 degrees but still significant, it may indicate a combination of resistive and capacitive elements, known as an RC circuit.This phase difference can be analyzed using an oscilloscope to visualize the waveform and confirm the capacitive nature of the device.
A helium balloon can just lift a load of 790 N. The skin of the balloon has a mass of 1.80 kg. What is the volume of the balloon?(the density of air is 1.29 kg/m³ and the density of helium is 0.179 kg/m³)
Answer:
74.10 cubic meters
Explanation:
The weught of load, balloon and heilium equals the weight of air displaced.
We know that density is mass per unit volume hence mass is product of density and volume, m=dv where d is density and v is volume
Weight is product of mass and acceleration due to gravity, W=mg=mgv where g is acceleration due to gravity and m is mass. We take g to be 9.81 m/s2
Weight of load=790 N
Weight of ballon= mg= 1.8*9.81
Weight of helium=0.179*9.81*v
Weight of air displaced =1.29*9.81*v
790+(1.8*9.81)+(9.81*0.179*v)=1.29*9.81*v
(1.29-0.179)v*9.81=807.658
10.89891v=807.658
V=74.104474667650
Rounding off, the volume is 74.10 cubic meters
A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation rate) ωi, and a stationary outer cylinder of inside radius Ro. In the tiny gap between the two cylinders is the fluid of viscosity μ. The length of the cylinders (into the page) is L. L is large such that end effects are negligible (we can treat this as a two-dimensional problem). Torque (T) is required to rotate the inner cylinder at a constant speed. (a) Showing all of your work and algebra, generate an approximate expression for T as a function of the other variables.
(b) Explain why your solution is only an approximation. In particular, do you expect the velocity prole in the gap to remain linear as the gap becomes larger and larger (i.e., if the outer radiusR0 were to increase, all else staying the same)?
Answer:
b) the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.
Explanation:
check attachment for solution to A
The torque required to rotate the inner cylinder of a rotating viscometer with a fluid of viscosity μ can be calculated using fluid mechanics. However, this solution is approximated and based on a linear velocity profile in the gap, which can change with a significant increase in the gap.
Explanation:In a rotating viscometer with a fluid of viscosity μ between two concentric cylinders, the torque required to rotate the inner cylinder at a constant speed can be computed using fluid mechanics principles. The shear stress on the fluid due to the rotating cylinder is the product of the viscosity and the velocity gradient across the gap, given by (ωi*ri)/ (ro - ri). This shear stress relates to the force acting on a unit area of the fluid as F/A, and the torque T is the product of the force and the radius of the inner cylinder. Following the equations, we'll obtain T = 2π * L * μ * ωi * (((ro^2 - ri^2) / ln (ro/ri)).
This solution is an approximation as it's based on the assumption that the velocity profile in the gap between the cylinders is linear, which holds true when the gap is thin. If the gap increases disproportionately (i.e., the outer radius R0 were to increase), the linear approximation of the velocity profile may no longer hold. In this scenario, edge effects and non-linear velocity profiles have to be considered, potentially modifying the expected torque value.
Learn more about Viscosity and Torque here:https://brainly.com/question/34645202
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Suppose we take a 1 m long uniform bar and support it at the 21 cm mark. Hanging a 0.40 kg mass on the short end of the beam results in the system being in balance. Find the mass of the beam.
Answer:
The mass of the beam is = 29 kg.
Explanation:
A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.
Taking moment about Point S
40 × 21 = [tex]M_{beam}[/tex] × 29
[tex]M_{beam}[/tex] = 29 kg
Therefore the mass of the beam is = 29 kg.
g A 48.7 g clay, moving at 9.8 m/s, crashes into 1.1 kg brick that is sitting at rest on a frictionless surface. After the collision, the clay is stuck to the brick. This is an inelastic collision, where some kinetic energy (for the system of both masses) is lost. How much kinetic energy is lost (so your answer will be negative!)
Answer:
-2.2394181 J amount of kinetic energy is lost.
Explanation:
Let Vf be the final speed
(48.7) (9.8) = (48.7 + 1100) Vf
477.26 = 1148.7 Vf
=> Vf = 0.4155 m/s
KE initiali = 1/2 (0.0487) (9.8)²
KE initial = 1/2 (0.0487) (96.04)
KE initial = 1/2 (4.677148)
KE initial = 2.338574 J
KE final = 1/2 (1.1 + 0.0487) (0.4155)²
KE final = 1/2 (1.1487) (0.17264025)
KE final = 1/2 (0.198311855175)
KE final = 0.0991559 J
KE (lost) = KEf - KEi
= 0.0991559 J - 2.338574 J
= -2.2394181 J
Why are the peaks opposite in direction?
a. The peaks are opposite in direction because the change in magnetic field at one end of the coil is opposite to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field.
b. The peaks are opposite in direction because the change in magnetic field at one end of the coil has the same direction to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field.
c. The peaks are in the same direction because the change in magnetic field at one end of the coil has the same direction to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field
Answer:
A
Explanation:
Peaks are in opposite direction because change in magnetic field at one end of the coil is opposite to the change in magnetic field at other end of the coil. Faraday's law predict that the direction of induced voltage is dependent on the nature of change in magnetic field
A 1.3 kgkg block slides along a frictionless surface at 1.3 m/sm/s . A second block, sliding at a faster 5.0 m/sm/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/sm/s . What was the mass of the second block?
Answer:
0.624 kg
Explanation:
We are given that
Mass of one block,[tex]m_1=1.3 kg[/tex]
[tex]v_1=1.3 m/s[/tex]
[tex]v_2=5 m/s[/tex]
[tex]V=2.5 m/s[/tex]
We have to find the mass of second block.
Mass of second block,[tex]m_2=\frac{m_1V-m_1v_1}{v_2-V}[/tex]
Substitute the values
[tex]m_2=\frac{1.3\times 2.5-1.3\times 1.3}{5-2.5}[/tex]
[tex]m_2=0.624 kg[/tex]
Hence, the mass of second block=0.624Kg
Answer:
The mass of the second block is 0.624 kg.
Explanation:
Given that,
Mass of the block 1, m₁ = 1.3 kg
Speed of block 1, u₁ = 1.3 m/s
Speed of block 2, u₂ = 5 m/s
The final velocity of the combined blocks is 2.5 m/s, V = 2.5 m/s
It is a case of inelastic collision. Using the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\1.3\times 1.3+5m_2=(1.3+m_2)2.5\\\\1.69+5m_2=3.25+2.5m_2\\\\m_2=0.624\ kg[/tex]
So, the mass of the second block is 0.624 kg.
g A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.462 of the escape speed from Earth and (b) its initial kinetic energy is 0.462 of the kinetic energy required to escape Earth
Answer:
a.)r=4RE/4
b.)r=2RE
c.)ZERO
Explanation:
a.) given v= 0.462 which is V= 0.466Ve
Since the projectile is shot directly away from earth surface the speed it escape with;
v=√2GM/RE......................eqn(1)
M is the Earth's Mass
RE is the Radius
From the Law of conservation of Energy
K+U=0...............................eqn(2)
K₁+U₁=K₂+U₂....................eqn(3)
where K₁ and U₁ is initial kinetic and potential energy
K₂ and U₂ are final kinetic and potential energy
Kinetic Energy (K.E) decrease with time as the projectile moves up and there is decrease in Potential Energy (P.E) , it will let to a point where K.E will turn to zero i.e K₂=0
U₂=K₁U₂ ..........................eqn(3)
From Gravitational Law
U₁= -GMm/RE ..................(5)
U₂= -GMm/r .....................(6)
Where "r" is the distance
v= 0.462√2GM/RE
v= √GM/2RE
GM/4RE - GM/RE = -GM/r
r= 4RE/3
b.) "r" is calculated by this equation;
K₁=0.466
K₁= 1/4MVe².............................eqn(9)
substitute eqn(1) into eqn (9) then
1/4m2GM/RE=0.466GM/2Re
GM/2RE - GM/RE =-GM/r
r=2RE
c.)The potential energy and kinetic energy is the same in terms of their size both in different directions, while the potential energy face outward, the kinetic energy face inward therefore the least initial mechanical energy
required at launch if the projectle is to escape is ZERO
A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.350-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 10.00 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
Answer:
a. The energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b. The total number of revolutions the particle makes before stopping is 1.43 revolutions
Explanation:
a.
Given
m = mass of particle = 0.350-kg
u = initial speed of 10.00 m/s
v = final speed = 5.50 m/s
r = radius = 0.600 m
We assume that the floor is horizontal;
This means that F = mg.
We also assume the rotational kinetic energy to be negligable.
Having listed the assumptions, we proceed as follows;
Let ∆E represent The energy transformed from mechanical to internal in the particle hoop.
This is given by
∆E = KE1 - KE2
Where KE1 = ½mu²
KE2 = ½mv²
So, ∆E = KE1 - KE2 becomes
∆E = ½mu² - ½mv²
∆E = ½m(u² - v²)
∆E = ½ * 0.350 * (10² - 5.5²)
∆E = 12.20625
∆E = 12.21J (Approximated)
Hence, the energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b.
Let n = number of revolutions
The relationship between n and the energy is
1/n = (KE1 - KE2)/KE1
Make n the subject of formula
n = KE1 / (KE1 - KE2)
n = ½mu² / (½mu² - ½mv²) --- Simplify
n = ½mu² / (½m(u² - v²)) ---- Divide through by ½m
n = u² / (u² - v²)
n = 10² / (10² - 5.5²)
n = 1.433691756272401
n = 1.43 rev
Hence, the total number of revolutions the particle makes before stopping is 1.43 revolutions
Final answer:
The energy lost to friction in one revolution is 11.9875 J, and the particle makes a total of 2 revolutions before coming to a stop.
Explanation:
Energy Loss Due to Friction
The energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution (a) is determined by the change in kinetic energy of the particle. The initial kinetic energy is given by ½ mv², where ‘m’ is the mass of the particle and ‘v’ is the initial velocity. The final kinetic energy is likewise given by ½ mv² using the final velocity. The difference in these energies gives the energy lost to friction.
Kinetic Energy Initial = ⅓ (0.350 kg)(10.00 m/s)² = 17.5 J
Kinetic Energy Final = ⅓ (0.350 kg)(5.50 m/s)² = 5.5125 J
Energy Transformed = Kinetic Energy Initial - Kinetic Energy Final = 17.5 J - 5.5125 J = 11.9875 J
Total Number of Revolutions
For part (b), to find the total number of revolutions the particle makes before stopping, assuming a constant frictional force, we can use a deceleration approach or an energy approach. The work done against friction for each subsequent revolution will be the same; thus each revolution results in the same amount of energy loss until the particle comes to a stop. By dividing the total initial kinetic energy by the energy lost per revolution, we find the number of revolutions.
Revolutions = Total Initial Kinetic Energy / Energy Transform per Revolution = 17.5 J / 11.9875 J ≈ 1.46 revolutions
Since it cannot complete a partial revolution after losing all kinetic energy, we round down to get 1 full revolution. Therefore, the particle makes a total of 1 + 1 = 2 revolutions before stopping.
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 6.80 m before stopping. How far does the lighter fragment slide
Answer:
Explanation:
Given that,
One fragment is 7 times heavier than the other
Let one fragment mass be M
Let this has a velocity v
And the other 7M
And this a velocity V
Initially the fragment is at rest u = 0
Applying conservation of momentum
Momentum is given as p=mv
Initial momentum = final momentum
Po = Pf
(M+7M) × 0 = 7M •V − Mv
0 = 7M•V - Mv
Divide both sides by M
0 = 7V -v
v = 7V
Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block
The workdone by the 7M mass is
Distance moved by 7M mass is 6.8m, Then, d =6.8m
W = fr × d
Where fr = µkN
When N=W =mg, where m=7M
N= 7Mg
fr = −µk × 7mg
Then, W(7m) = −7µk•Mg×d
W(7m) = −7µk•Mg×6.8
W(7m) = −47.6 µk•Mg
Then, same procedure,
Let distance move by the small mass be m
Work done by M mass
W(m) = −µk•Mg×d'
Since it is a wordone by friction, that is why we have a negative sign.
Using conservation of energy
Work done by 7M mass is equal to work done by M mass
W(7m) = W(m)
−47.6 µk•Mg = −µk•Mg×d
Then, M, g and µk cancels out
We are left with
-46.7 = -d
Then, d = 46.7m
Answer:
Distance the lighter fragment slides before stopping= 333.2 m
Explanation:
Gravitational acceleration = g
Mass of the lighter fragment = m
Mass of the heavier fragment = 7m
Velocity of the lighter fragment after the explosion = V1
Velocity of the heavier fragment after the explosion = V2
The object is at rest before the explosion hence the total momentum of the system is zero.
mV1 + 7mV2 = 0
V1 = -7V2
Coefficient of friction = [tex]\mu[/tex]
Friction force on the lighter fragment = f1 =[tex]\mu mg[/tex]
Friction force on the heavier fragment = f2 =[tex]7\mu mg[/tex]
Deceleration of the lighter fragment due to friction = a1
ma1 = -f1
ma1 = [tex]-\mu mg[/tex]
a1 =[tex]-\mu g[/tex]
Deceleration of the heavier fragment due to friction = a2
7ma2 = -f2
7ma2 = -7[tex]\mu mg[/tex]
a2 = [tex]-\mu g[/tex]
Final velocity of the lighter fragment = V3 = 0 m/s
Final velocity of the heavier fragment = V4 = 0 m/s
Distance traveled by the lighter fragment before coming to rest = d1
Distance traveled by the heavier fragment before coming to rest= d2 =6.8 m
V4² = V2² + 2a₂d₂
[tex],[/tex]
[tex](0)^2 = V_2^2 + (2)(-\mu g)d_2[/tex]
[tex]d_2 = \frac{V_2^2}{2 \mu g}[/tex]
V₃² = V₁² + 2a₁d₁
[tex](0) = (-7V_2)^2 + 2(-\mu g)d_1[/tex]
[tex]d_1 = \frac{49V_2^2}{2 \mu g}[/tex]
d₁ = 49d₂
d₁ = (49)(6.8)
d₁ = 333.2 m
Distance the lighter fragment slides before stopping = 333.2 m
Newton's law of gravity and Coulomb's law are both inverse-square laws. Consequently, there should be a "Gauss's law for gravity." The electric field was defined as E⃗ =F⃗ onq/q , and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g⃗ of a point mass? Write your answer using the unit vector r^ , but be careful with signs; the gravitational force between two "like masses" is attractive, not repulsive. Express your answer in terms of the variables G , M , r , and r^ .
The correct answer for the gravitational field of a point mass is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
The gravitational force between two masses is given by Newton's law of gravitation:
[tex]F = \dfrac{GM_1M_2}{r^2}[/tex]
[tex]F[/tex] is the gravitational force between [tex]M_1[/tex] and [tex]M_2[/tex]
[tex]G[/tex] is the gravitational constant,
[tex]r^2[/tex] is the square of the distance between two masses.
The gravitational force is a vector quantity, and it is directed along the line connecting the two masses.
[tex]g= \dfrac{F}{m}[/tex]
[tex]g[/tex] is the gravitational field.
[tex]F[/tex] is the gravitational force.
Let [tex]m[/tex] be the mass test object.
Substitute the value of gravitational force from Newton's law of gravity:
[tex]g = \dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
[tex]r[/tex]is the unit vector pointing from the mass M to the test mass, which represents the direction of the gravitational field.
[tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
The negative sign indicates the direction of the gravitational force is attractive.
The gravitational field is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
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The gravitational field of a point mass is given by -GM/(r^2)*r^, where the negative sign indicates the attractive nature (opposite direction to r^) of gravity.
Explanation:The gravitational field g of a point mass using the same reasoning as for the electric field would be analogous but with some slight modifications to account for the attractive nature of gravity, unlike the repulsive nature of electric charges of the same sign. The formula is given by -GM/(r^2)*r^, where G is the gravitational constant, M is the mass of the object, and r is the distance to the object. The negative sign indicates that the force is attractive and acts in the direction opposite to r (the vector pointing directly away from the mass).
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By considering the magnetic force in the second region, develop a mathematical expression that relates the mass of the particle to the other variables. Do not include the velocity in your expression. You can use the condition that the particle passed through the region of electric and magnetic fields undeflected to eliminate v from your expression. Your expression will also contain the radius of the circular path. i. Your expression for m should depend on B, E, r, and q
Answer:
M = [tex]\frac{qrB^{2} }{E}[/tex]
Explanation:
considering the magnetic force in the second region derive a mathematical expression that equates the mass of the particle to other variables
In a magnetic field
q = charge, M = mass of particle, E = electric field,B= magnetic field
qvb = [tex]\frac{mv^{2} }{r}[/tex] = therefore m = [tex]\frac{qrb}{v}[/tex] (equation 1)
note : the particle passes through the region undeflected
therefore : qvb = qE therefore (E = VB)
hence v = [tex]\frac{E}{B}[/tex] ( equation 2 )
insert equation 2 into equation 1
m = [tex]\frac{qrB^{2} }{E}[/tex]
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of north. The resultant vector A + B is given by:
(a) 9.8 m at an angle of 26° north of east,
(b) 3.3 m at an angle of 26° north of east,
(c) 7.2 m at an angle of 26° east of north,
(d) 3.3 m at an angle of 64° east of north or
(e) 9.8 m at an angle of 64° east of north
Answer:
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
C
Explanation:
Resolving the vectors to vertical and horizontal component;
Vertical;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical = 6sin30 + 4sin60 = 6.464m
Horizontal;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal = 6cos30 - 4cos60 = 3.196m
Resultant R = √(6.464^2 + 3.196^2) = 7.2m
Tanθ = 6.464/3.196
θ = taninverse (6.464/3.196) BN
θ = 64° north of East.
Or
26° east of north
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
Given data:
The magnitude of vector A is, A = 6.0 m.
The direction of vector A is, 30° north of east.
The magnitude of vector B is, B = 4.0 m.
The direction of vector B is, 30° west of north.
The quantity having both the magnitude as well as the magnitude are known as vector quantities.
Resolving the vectors to vertical and horizontal component;
Along the Vertical direction;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical vector = 6sin30 + 4sin60 = 6.464 m
Along the Horizontal direction;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal vector = 6cos30 - 4cos60 = 3.196 m
Now, the resultant vector is calculated as,
[tex]R=\sqrt{6.464^{2}+3.196^{2}}\\\\R = 7.2 \;\rm m[/tex]
And the resultant direction of vectors is,
[tex]tan \theta = 6.464/3.196\\\\\theta = tan^{-1} (6.464/3.196) \\\theta =64 ^{\circ}[/tex]( North of East)
θ = 64° north of East.
Or
26° east of north
Thus, we can conclude that the resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
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In a Millikan oil-drop experiment, the condenser plates are spaced 2.00 cm apart, the potential across the plates is 4000 V, the rise or fall distance is 4.00 mm, the density of the oil droplets is 0.800 g/cm3 , and the viscosity of the air is 1.81 105 kg m1 s 1 . The average time of fall in the absence of an electric field is 15.9 s. The following different rise times in seconds are observed when the field is turned on: 36.0, 17.3, 24.0, 11.4, 7.54. (a) Find the radius and mass of the drop used in this experiment. (b) Calculate the charge on each drop, and show that charge is quantized by considering both the size of each charge and the amount of charge gained (lost) when the rise time changes. (c) Determine the electronic charge from these data. You may assume that e lies between 1.5 and 2.0 1019 C. 7.
Answer:
Answer is in the following attachment
Explanation:
A bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4 of the experiment, some light that reflects off the apparatus or from neighboring stations strikes the photometer even when the direct beam is blocked. In addition, due to electronic drifts, the photometer does not generally read 0.0 mV even in a dark room. It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement. Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?
Answer:
0.685
Explanation:
the background-corrected light measurement at 29mm: 20.7mv- 5.1mv⇒15.6mv
at 32.5mm: 15.8mv- 5.1mv⇒ 10.7mv
So, Normalize the data to the maximum value probably means to set the peak value to unity and scale the remaining data by the same factors.
Therefore, the normalized corrected value at 32.5mm is ⇒ 10.7mv/15.6mv ⇒0.685
The diagram below shows a light ray from a pencil hitting a mirror. An upright pencil sits to the left of a mirror, which is convex facing left. An arrow from the tip of the pencil runs to the center of the mirror. Line 1 runs from the point reflection at an acute angle below the first arrow. Arrow 2 runs at an acute angle above the first arrow. Arrow 3 runs perpendicularly up from the point reflection. Arrow 4 runs through the mirror down at an obtuse angle pointing toward a distant point on a line running from the base of the pencil. Which ray shows the correct direction of the reflected ray? 1 2 3 4
the answer would be 2, I just took the test..hope this helps :))
The ray 2 shows the correct direction of the reflected ray. Therefore, the correct option is b.
What is reflected ray?The ray that depicts the light that is reflected by the surface is the reflected ray that corresponds to a certain incident ray. The angle of reflection is the angle formed between the surface normal and the reflected beam. According to the Law of Reflection, the angle of reflection for a specular surface is always the same as the angle of incidence.
In the given problem, the second ray is the ray that is depicted correctly as a reflected ray as it is bounced back after striking the mirror at a specified angle above the horizontal. Rest all the other rays are not depicted correctly as the ray 1 is reflected below normal, ray 3 is reflected horizontally and the ray 4 is refracted.Therefore, the correct option is b.
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The question is incomplete, but most probably the complete question is,
The diagram below shows a light ray from a pencil hitting a mirror. An upright pencil sits to the left of a mirror, which is convex facing left. An arrow from the tip of the pencil runs to the center of the mirror. Line 1 runs from the point of reflection at an acute angle below the first arrow. Arrow 2 runs at an acute angle above the first arrow. Arrow 3 runs perpendicularly up from the point reflection. Arrow 4 runs through the mirror down at an obtuse angle, pointing toward a distant point on a line running from the base of the pencil. Which ray shows the correct direction of the reflected ray?
a.1
b.2
c.3
d.4
For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. The distance the worker’s eyes must travel is 20 in. The perpendicular distance from her eyes to the line of travel is 24 in. No refocus is required. What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element?
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 1.80 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.23 cm, and it takes 0.320 s to move it into the field.
1. What average current is induced in the ring in A if its resistance is 0.0100 Ω?
2. What average power is dissipated in mW?
3. What magnetic field is induced at the center of the ring in T?
Given that,
Magnetic field strength is
B = 1.8T
The wedding ring has a diameter of
d = 2.23 cm = 0.023m
Time take t = 0.32 secs
A. Current induced
From ohms law
V= iR, given that R = 0.01Ω
So, we need to get the induced emf
Using
ε = -NdΦ / dt
Where Φ = BA
ε = -A ∆B / ∆t
ε = -¼πd²(B2-B1) / (t2-t1)
ε = -¼ × π × 0.023² × -1.8 / 0.32
ε = 0.0023371 V
Then
I = ε / R
I = 0.002337 / 0.01
I =0.2337 A
B. Power discippated?
Power is given as
P = iV
P = 0.2337 × 0.002337
P = 0.0005462 W
P = 0.56 mW
C. The magnetic field at the centre of the ring.
The electric field at the centre of the ring is zero because each part of the ring will cause a symmetrical opposite magnitude at every point,
Then, B = 0 T
A closed loop conductor that forms a circle with a radius of is located in a uniform but changing magnetic field. If the maximum emf induced in the loop is what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Final answer:
The maximum rate at which the magnetic field strength is changing in a closed loop conductor can be determined using Faraday's law. The rate is equal to the maximum induced emf divided by the area of the loop.
Explanation:
The maximum rate at which the magnetic field strength is changing can be determined using Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf (voltage) in a closed loop conductor is equal to the negative rate of change of magnetic flux through the loop. In this case, the magnetic field is perpendicular to the plane of the loop, so the magnetic flux is given by the product of the magnetic field strength and the area of the loop. Therefore, the maximum rate of change of magnetic field strength is equal to the maximum induced emf divided by the area of the loop.
So, the maximum rate at which the magnetic field strength is changing is given by: (d(B)/dt) = (Emax / A)
where (d(B)/dt) is the rate of change of magnetic field strength, Emax is the maximum induced emf, and A is the area of the loop.
In an interference pattern, the intensity is Group of answer choices smaller in regions of constructive interference than in regions of destructive interference. unchanged in regions of destructive interference but smaller in regions of constructive interference. the same in both the regions of constructive interference and the regions of destructive interference. unchanged in regions of destructive interference but greater in regions of constructive interference. greater in regions of constructive interference than in regions of destructive interference.
Answer:
The same in both the regions of constructive interference and the regions of destructive interference.
Explanation:
Interference is a phenomenon which occurs when two waves meet while moving along the same medium . The amplitude formed as a result of the interference could be greater, lower, or the same amplitude.
Constructive and destructive interference result from the interaction of waves that are correlated or coherent with each other. This is because arose from the same source or they have the same or nearly the same frequency.
The waves being coherent, arising from the same source and having the same frequency explains why it’s the same in both the regions of constructive interference and the regions of destructive interference.
Answer:
Option E (greater in regions of constructive interference than in regions of destructive interference)Explanation:
Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes. Whereas in destructive interference opposite is true.
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One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied perpendicularly to the end face (uniformly across the area) at the other end, pulling directly away from the vise. What is the stress on the rod?
Answer: 1.91*10^8 N/m²
Explanation:
Given
Radius of the steel, R = 10 mm = 0.01 m
Length of the steel, L = 80 cm = 0.8 m
Force applied on the steel, F = 60 kN
Stress on the rod, = ?
Area of the rod, A = πr²
A = 3.142 * 0.01²
A = 0.0003142
Stress = Force applied on the steel/Area of the steel
Stress = F/A
Stress = 60*10^3 / 0.0003142
Stress = 1.91*10^8 N/m²
From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²
If the vertical displacement from crest to trough is 0.50 cm, what is the amplitude?
cm
0.50
0.25
1.0
Answer:
0.25
Explanation:
Since there is a vertical displacement of 0.5cm from the crest to the trough, there is half of that displacement to the midline, which is also known as the amplitude. Therefore, the amplitude is 0.5/2=0.25cm. Hope this helps!