Young's modulus for the rope 5.462 × [tex]10^{9}[/tex]N/m²
Mass is 70 kg
F = mg = 70 × 9.8N
F = 686N
ΔL = 8cm
A = πr² = 0.785 ×[tex]10^{-4}[/tex]
Y = F/A l/Δl = 686 × 50 / 0.785 ×[tex]10^{-4}[/tex] × 8 × [tex]10^{-2}[/tex]
Y = 5.462 × [tex]10^{9}[/tex]N/m²
Young's modulus (displaystyle E), also known as the Young modulus or the modulus of elasticity in tension or compression (i.e., negative tension), is a mechanical characteristic that gauges the stiffness of a solid material under tensile or compressive tension when a force is applied longitudinally. Using the formula, it quantifies the relationship between axial strain and tensile/compressive stress in the linear elastic zone of a material. Axial strain is measured as the proportionate deformation of the relationship between axial strain and tensile/compressive stress. Young's modulus was first introduced by Leonhard Euler in 1727, while being named after the 19th-century British scientist Thomas Young. Giordano Riccati, an Italian physicist, conducted the first experiments in 1782 that utilized the idea of Young's modulus in its current form, 25 years before Young's work.
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You measure the mass of your rubber band airplane and find that it is 15 grams (0.015 kg). You also measure the acceleration of the airplane with a motion detector and find that the airplane initially accelerates upward at a rate of 2 m/s 2. Calculate the net force on the airplane and the upward force from the lift.
An Atwood’s machine is set up by suspending two blocks connected by a string of negligible mass over a pulley, as shown above. The blocks are initially held at rest and then released. The acceleration of mass M1 is 4.9 m/s2 downward and M1 = 6 kg. Find the mass of M2.
The mass of M₂ = 12kg.
What is acceleration of mass?Newton's second law of motion states that an object's acceleration is equal to its mass divided by the net force acting on it, or a = F m. When an object's mass and the net force acting on it are known, this equation for acceleration can be used to determine the object's acceleration.
The formal formulation of Newton's second law is as follows: The acceleration of an object caused by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
F = ma
F = 6×4.9
F = 29.4N.
F = (M₂ - M₁)a
29.4 = (M₂- 6) 4.9
29.4/ 4.9 = M₂- 6
M₂ = 6+6
M₂ = 12kg.
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Problem 15:
Lava has an area of 0.80 m², and emissivity of 1.0, and a temperature of 1450 K. The
temperature of the surroundings is 300 K. What is the rate of heat transfer, from the lava to the
surroundings, due to radiation?
a. 1.0 x 105 W
c. 3.0 x 105 W
d. 4.0 x 105 W
b. 2.0 x 105 W
The heat transfer by an emitted radiation is calculated using Stefan-Boltzmann law of radiation. The rate of heat transfer is 2.0 × 10⁵ W.
What is Stefan-Boltzmann law of radiation?According to Stefan-Boltzmann law of radiation, the heat transfer by an emitted radiation Q/t is related to the absolute temperature T and the area A as:
Q/t = σe AT⁴
Where σ is the Stefan-Boltzmann constant e be the emissivity. σ has a value of 5.67 × 10⁻⁸ J/s m² K⁴.
Given that, T = 1450 K
e = 1.0, A = 0.80 m².
Now, heat transfer Q/t = 5.67 × 10⁻⁸ J/s m² K⁴ × 0.80 m² × (1450 K)⁴
= 2.0 × 10⁵ W.
Therefore, the heat transfer to the surrounding to the radiation is 2.0 × 10⁵ W.
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Figure shows an electric dipole. What are the (a) magnitude-axis) direction (relative to
the positive direction of the x axis) of the dipole’s electric field at point P, located at
distance r >> d?
The magnitude-axis direction relative to the positive direction of the x axis of the dipole’s electric field at point P, located at distance r >> d is 1 / 4λεο x qd/r³.
What is electric field?Electric field is defined as the physical field that surrounds electrically charged particles and acts as an attractor or repellent on all other charged particles in the field. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges.
| E net | = 2 E sinθ = 2[1 / 4λεο x q / (d/2)² + 1³] d/2 / √(d/2)² + r ²
= [1 / 4λεο x qd / [(d/2)² + r²] ³/²
For r >> d [(d/2)² + r²] ³/² ≅ r
So the expression will be
| E net | = 1 / 4λεο x qd / r³
Thus, the magnitude-axis direction relative to the positive direction of the x axis of the dipole’s electric field at point P, located at distance r >> d is 1 / 4λεο x qd/r³.
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2. A student, Jay, makes himself as cylindrical as possible while sitting on a spinning chair.
Imagine the cylinder with its flat end in the chair and directly above the chair. His center of
mass is 0.10 m from the axis of rotation of the chair. His mass is 65 kg and the cylindrical shape
he is making has a circumference of 1.2 m. The chair's mass is 30 kg. Assume the friction is
directly proportional to the weight of the chair and whatever is on the chair, and the distance
from the axis of rotation to where the friction is applied is constant.
(a) What is the student's moment of inertia about his center of mass?
(b) What is his moment of inertia about the axis of rotation of the chair?
To determine the angular acceleration due to torque from friction, Andrea spins the chair while
the Jay is in his first position (the cylindrical position described above), while John times
certain rotations after releasing the chair. The first complete revolution while spinning takes 1.8
seconds, and the first three complete revolutions take a total of 5.9 seconds. Then Jay gets into
a second position, and this test happens again. John times that it takes 2.3 seconds for the first
revolution and 7.2 seconds for the first three revolutions. Next, Jay gets out of the chair, and
they spin the chair and calculate that the angular acceleration of the chair is -0.0700 rad/s2
.
[Assume that the angular acceleration for a specific situation is constant.]
(c) What is the angular acceleration of the chair while Jay is in cylindrical position?
(d) What is the angular acceleration of the chair while Jay is in second position?
(e) What is the moment of inertia of the chair?
(f) What is the torque due to friction when Jay is in the chair?
(g) What is Jay's moment of inertia in his second position?
Jay then gets into his cylindrical position again and sits as described earlier. Andrea spins him
very fast. After Jay goes around a few times, John says, “Start” and begins timing. After two
revolutions (3.2 seconds), John says, “Change”, Jay switches instantly into final position, and
John times Jay until the chair stops.
Disk 1 Disk 2 Disk 3
F⃗ F⃗ F⃗
(h) What was Jay's angular speed when the timing started (before the position change)?
(i) What was Jay's angular speed just before changing positions?
(j) What was Jay+Chair's angular momentum just before changing position?
(k) What was Jay+Chair's angular momentum just after changing position?
(l) What was Jay's angular speed just after changing position?
Answer: (a) To find the moment of inertia of Jay about his center of mass, we can use the formula I = mr^2, where m is the mass of the object and r is the distance from the center of mass to the point of rotation. Plugging in the values, we get I = 65 kg * (0.10 m)^2 = 0.65 kg*m^2.
(b) To find the moment of inertia of Jay about the axis of rotation of the chair, we need to use the formula I = Icm + md^2, where Icm is the moment of inertia about the center of mass and d is the distance from the center of mass to the axis of rotation. Plugging in the values, we get I = 0.65 kgm^2 + 65 kg * (0.10 m)^2 = 0.65 kgm^2 + 0.65 kgm^2 = 1.30 kgm^2.
(c) To find the angular acceleration of the chair while Jay is in his cylindrical position, we can use the formula alpha = T/I, where T is the torque and I is the moment of inertia. The angular acceleration is given as -0.0700 rad/s^2, so we can rearrange the formula to solve for T: T = I * alpha = 1.30 kg*m^2 * (-0.0700 rad/s^2).
(d) To find the angular acceleration of the chair while Jay is in his second position, we can use the same formula as before: alpha = T/I. The angular acceleration is not given, so we cannot solve for T directly.
(e) To find the moment of inertia of the chair, we can use the formula I = mr^2, where m is the mass of the object and r is the distance from the center of mass to the point of rotation. The mass of the chair is given as 30 kg and the distance from the center of mass to the axis of rotation is not given, so we cannot solve for I directly.
(f) To find the torque due to friction when Jay is in the chair, we can use the formula T = Fd, where F is the force of friction and d is the distance from the axis of rotation to the point of application of the force. The force of friction is not given, so we cannot solve for T directly.
(g) To find the moment of inertia of Jay in his second position, we need to use the formula I = Icm + md^2, where Icm is the moment of inertia about the center of mass and d is the distance from the center of mass to the axis of rotation. The moment of inertia about the center of mass and the distance from the center of mass to the axis of rotation are not given, so we cannot solve for I directly.
(h) To find Jay's angular speed when the timing started, we need to know the time it took for Jay to complete one revolution and the radius of the circle he was rotating around. Both of these values are not given, so we cannot solve for the angular speed directly.
(i) To find Jay's angular speed just before changing positions, we need to know the time it took for Jay to complete one revolution and the radius of the circle he was rotating around. Both of these values are not given, so we cannot solve for the angular speed directly.
(j) To find Jay+Chair's angular momentum just before changing position, we need to know Jay's angular speed and the moment of inertia of Jay+Chair
A cart at the end of a spring undergoes simple harmonic motion of amplitude A = 10 cm and frequency 5.0 Hz. Assume that the cart is at x=−A when t=0.
a) Determine the period of vibration.
b) Write an expression for the cart's position as a function of time.
c) Determine the position of the cart at 0.050 s .
d) Determine the position of the cart at 0.100 s .
A cart at the end of a spring undergoes simple harmonic motion of amplitude A = 10 cm and frequency 5.0 Hz. Assume that the cart is at x = −A when t = 0.
(a) T = 0.2 sec
(b) x(t) = − (10 cm)cos(10π) s⁻¹ t
(c) x(0.050) = 0
(d) x(0.100) = 10 cm
What is Simple harmonic motion?Simple harmonic motion is described as the periodic movement of a point down a straight line with an acceleration that is always toward a fixed point on that line and is proportional to that point's distance from the moving point.
Given that,
frequency (f) = 5.0 Hz
Amplitude (A) = 10 cm
(a) As we know,
T = 1/f
or, T = 1/5 Hz
or, T = 0.2 sec
(b) Because the expression for the cart position is at x= −A when t=0 , this means that the function which describes the position of the cart is a cosine function because, cos(0)=1.
Notice that the form of the equation that describes the position of a vibrating object is as follows :
x(t) = A cos (2Π/T) × t
Substitute for the values of T=0.2s ; A=10 cm and because x = −A when t = 0,then the function will have a negative sign
x(t) = −(10 cm)cos(10π) s⁻¹ t
(c) the position of the cart at 0.050 s:
x(0.050) = −(10 cm)cos(10π) s⁻¹ × (0.050)
x(0.050) = −(10 cm) (cos π/2)
x(0.050) = 0
(d) the position of the cart at 0.100 s:
x(0.100) = −(10 cm)cos(10π) s⁻¹ × (0.100)
x(0.100) = −(10 cm) (cos π)
x(0.100) = 10 cm
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identfy type of reaction
Ca + 2HCl -----> CaCl2 + H2
Ca + 2HCl -----> CaCl2 + H2 is the single-replacement reactions
What is Single replacement reactions?A single-replacement reaction is defined as a chemical reaction in which one strong element replace the other weak element. This reaction can also be called as Exchange reaction.
The given chemical equation Ca + 2HCl -----> CaCl2 + H2 is an example of single-replacement reactions in which Calcium (Ca) is stronger than Hydrogen (H ) and replaces calcium to form bond with Chloride.
Hence, the correct answer for the reaction Ca + 2HCl -----> CaCl2 + H2 is the "single-replacement reactions".
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A roller coaster car travels along the track as shown in the image. At which two points does the car have the least amount of potential energy?
Points 1 and 5.
Points 1 and 4.
Points 2 and 4.
Points 2 and 5.
Points 2 and 5 is the point the car have the least amount of potential energy. Hence option d is correct.
What is potential energy?Potential energy is defined as the energy that an object stores as a result of its position in relation to a zero position. Any object that is raised from rest has energy that can be released at a later time; for this reason, it is referred to as potential energy.
The object has a maximum amount of gravitational potential energy and no kinetic energy at the top of the roller coaster (assumed there is no velocity). The gravitational potential energy of the object decreases and the kinetic energy rises as it starts to descend to the bottom.
Thus, points 2 and 5 is the point the car have the least amount of potential energy. Hence option d is correct.
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10. IB challenge question: Two masses hang from a pulley as shown. Calculate the acceleration of the smaller mass. Use 10 for acceleration due to gravity instead of 9.81. (Hint: look at each mass separately, and create a system of equations) I 1kg 3kg
The acceleration on the block of mass 1 kg will be equal to the acceleration due to gravity that is 9.81 m/s².
What is Acceleration?
Acceleration of an object is the rate of change of the velocity of the object with respect to time.
As the string and pulley system is ideal so the tension will be constant throughout the string and thus both of the blocks will move with the same acceleration, suppose the tension is T and the common acceleration is a
Then, 3kg being heavier weight will move downward.
The tension always act towards a direction away from the block.
The net downward force on 3kg block will be (3g − T)
So, the equation of Force will be F = ma or (3g − T)= 3a ...(1)
Similarly, the net upward force on 1kg block will be (T − 1g)
So, the equation of Force will be F = ma or (T − 1g) = 1a ...(2)
By adding both these equations 1 and 2.
We get,
2g = 2a
a = g
Therefore, acceleration on the block of 1 kg will be equal to the acceleration due to gravity that is 9.81m/s².
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The force required to stretch a Hooke’s-law spring varies from 0 N to 42.8 N as we stretch the spring by moving one end 14.3 cm from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Find the work done in stretching the spring. Answer in units of J.
The force constant and the work done in stretching the spring with a force of 42.8 N acting on it is 299.3 N/m and 3.06 J respectively.
How to calculate the force constant of a spring?To calculate the force constant of the spring, we use the formula below.
Formula:
F = ke........ Equation 1Where:
F = Force on the springk = Force constant of the springe = Extension of the springFrom the question,
Given:
F = 42.8 Ne = 14.3 cm = 0.143 mSubstitute these values into equation 1 and solve for k
42.8 = 0.143kk = 42.8/0.143k = 299.3 N/mTo calculate the work done in stretching the spring, we use the form ula below.
Formula:
W = ke²/2W = 299.3×0.143²/2W = 3.06 JHence, the force constant and the work done in stretching the spring is 299.3 N/m and 3.06 J respectively.
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b. give a brief description of the following symbols given in the above wavepacket. what properties of the wavepacket does each represent?
A wave group is another name for a wave packet. The concept of superposition allows for this circumstance. According to, the total of both waves is a solution if any two beams are just a response to wave equation.
What does the term "wave packet" mean?A wave packet is a condensed train of (classical) waves with different wavelengths or momenta that is constrained to a specific area of space.
What causes a wave packet to form?a wave packet is created by the superposition of two waves with marginally differing frequencies: The wave packet gets increasingly spatially confined as that the amplitude of the wave rises. Note that the geometry of the wavepacket remains constant.
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The position , in meters, of an object moving along the -axis is given as a function of time , in seconds, as ()=0.09814−0.1953+0.8552+3.15−7.49 Find the object's acceleration at time =3.41 s.
With the knowledge of differential calculus, the magnitude of the object's acceleration is 11.4 m/s²
What is Position ?Position is any point of coordinate in space or any region in which an object can be located.
Given that the position , in meters, of an object moving along the x-axis is given as a function of time t, in seconds, as
x(t) = 0.09814[tex]t^{4}[/tex] − 0.1953t³ + 0.8552t² + 3.15t − 7.49
The object's acceleration at time =3.41 s can be found with the knowledge of differential calculus.
The derivative of position = velocity
The derivative of velocity = acceleration
That is,
If x(t) = 0.09814[tex]t^{4}[/tex] − 0.1953t³ + 0.8552t² + 3.15t − 7.49
dx/dt = 0.39256t³ - 0.5859t² + 1.7104t + 3.15 = v
dv/dt = 1.17768t² - 1.1718t + 1.7104
when time t = 3.41 s
Substitute for t
dv/dt = a = 1.17768(3.41)² - 1.1718(3.41) + 1.7104
a = 13.69 - 4.00 + 1.7104
a = 11.4 m/s²
Therefore, the object's acceleration at time = 3.41 s, is 11.4 m/s² approximately.
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A large truck is ahead of you and is turning right onto a street with two lanes in each direction. The truck:
A. May complete its turn in either of the two lanes.
B. May have to swing wide to complete the right turn.
C. Must stay in the right lane at all times while turning.
We can see a big truck turning right in front of you onto a roadway with two lanes going in either direction. When turning, the truck must always stay in the right lane.
What's mean by direction?Direction is characterized as the course that anything follows, the route that must be taken to reach a particular location, the direction in which something is beginning to take shape, or the direction you are facing. When you make a turn instead of left, that is an illustration of direction.
What is compass direction?Despite taking into account compass error, the direction as shown by the instrument. The true direction referred to an Earthly meridian may vary markedly from the direction represented by a magnetic compass.
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Calculate Q for each of the three processes. Express your answers in joules to three significant figures, separated by commas. A heat engine takes 0.350 mol of an ideal diatomic gas around the cycle shown in the pV diagram of (Figure 1). Process 1 + 2 is at constant volume, process 2 + 3 is adiabatic, and process 3 + 1 is at a constant pressure of 1.00 atm. The value of y for this gas is 1.40. The magnitude of the change in internal energy for each process is AU1–2= 2180 J, AU2-31 = 785 J, and |AU3_1] = 1396 J. Express your answers in joules to three significant figures, separated by commas. Q1+2, Q2+3, Q3+1 = ___ J, J, J Express your answers in joules to three significant figures, separated by commas.
W1+2, W2+3, W3+1 = ___ J, J, J
The three significant figures are ΔV2-3 = -785 J, W2-3 = 785.7 J and W3-1 = -558.72 J
Number of moles n=0.350
r = 1.40
| ΔV1-2 | = 2180 J | ΔV2-3| = 785 J
| Δ V3-1 | = 1396 J
From first law of thermodynamic
Q = ΔV + W
Internal Energy increases in this case as the temperature is increasing ΔV is positive and work done in a constant volume process is 0 as
W1-2 = P ΔV = P (O) = O
Q1-2 = Δ V1-2 + W1-2 = ΔV1-2 + O
Q1-2 = 2180 J
Change in internal energy is negative since temperature is decreasing as
R = 8.314 J
Δv2-3 = -785 J
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A car of mass 900kg starting from rest has a constant acceleration of 3.5m/s² . calculate it's momentum after it has travelled a distance of 40m
The momentum of the car is 15030kgm/s
What is momentum?The momentum of a body is the product of the mass of the body and it's velocity. It is a vector quantity and measured in kgm/s.
Momentum is expressed as :
momentum = mass× velocity
using the relation v² = u²+2as to get the velocity of the car.
v² = 0²+2×3.5× 40
v² = 280
v = √280
v = 16.7m/s
from momentum = mass× velocity
= 900× 16.7
= 15030kgm/s
therefore the momentum of the car is 15030kgm/s
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A 54 kg pole vaulter falls from rest from a
height of 4.7 m onto a foam rubber pad. The
pole vaulter comes to rest 0.40 s after landing
on the pad.
a) Calculate the athlete’s velocity just before reaching the pad.
Answer in units of m/s
Answer:
9.6
Explanation:
v² = u² + 2as
v² = 0 + 2(9.81)(4.7)
v² = 92.214
v = 9.6 m/s
use dimensional analysis to determine which of the following expressions gives the area of a circle: is it qrr2 or 2ttr? explain.
The dimensional analysis cannot determine whether the area of a circle is πr2 or 2πr2 because the constant 2 is dimensionless.
Dimensional Analysis :
Dimensional analysis is used to measure objects' dimensions and forms. It aids in our mathematical knowledge of object nature. Along with lengths and angles, it also includes geometrical traits like flatness and straightness. Similarly, only if two physical quantities have the same dimensions can they be equal. Dimensional analysis is the study of the relationship between physical quantities using dimensions and units of measurement. Because it maintains the units' consistency, dimensional analysis is crucial for facilitating accurate mathematical computations.
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A cube with edge length L = 0.29 m and density ρc = 0.89×103 kg/m3 floats in equilibrium in a liquid of density ρl = 1.5×103 kg/m3, with the top of the cube a distance d above the liquid’s surface
A; Write an expression to find d using the image
It is given that,
Cube Length (L) = 0.29m
Density of Cube (Pc) = 0.89 *103 Kg/m3
Density of Liquid (P1) = 1.5 *103 Kg/m3
Distance (d) =?
We know that,
d = {(P1 - Pc)/P1}*L
d = {(1.5*103–0.89*103)/1.5*103}* 0.29
d = {(1.5 – 0.89)/1.5} * 0.29
d = 0.118m
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The expression to find distance using the given quantities is:
d = ( ρl - ρc ) L / ρl
How to apply this expression to find distance?In this question, it is given that a cube with length L= 0.29 meter and the density of cube ρc = 0.89×103 kg/m3 and it floats in equilibrium in liquid of density ρl = 1.5×103 kg/m3 with the top of the cube a distance d above the liquid surface. So here is the liquid, and the cube will float on the liquid at a distance d. Thus, the expression in terms of the given quantities to find distance d can be written as-
d = ( ρl - ρc ) L/ρl
We are given that,
Density of cube, ρc = 0.89×103 kg/[tex]m^{3}[/tex]
Density of liquid, ρl = 1.5×103 kg/[tex]m^{3}[/tex]
Edge length of cube, L = 0.29 m
So, on substituting the given values in above expression, we get:
d = [(1.5×103 – 0.89×103) / 1.5×103] × 0.29
d = [(1.5 – 0.89) / 1.5] × 0.29
d = 0.118 m
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7.)Summarize Newton’s Third Law.
8.) When gravity pulls your book down onto your desktop with 4 N of force, how strong does your desk push back?
9.) Describe/explain an everyday example
Newton’s 3rd Law in action.
(Please help me I need to study for my Test T-T)
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A.
8)The force of gravity pulling down and the force of the table (FRICTION) pushing upwards on the book are of equal magnitude and opposite directions. These two forces balance each other.
9)A man walking on the ground, While walking, a person pushes the ground in the backward direction, and the ground in return pushes the person in the forward direction, thus making them walk.
How does Newton's 3rd law explain how a person walks across a floor?
Newton's third law states that every action has an equal and opposite reaction. This is relevant to walking because when you put your foot on the ground, you are applying force to it. In doing this, the ground also actually applies an equal force onto your foot, in the opposite direction, pushing you forward.
What is an example of Newton's 3rd law?
Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies an equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.
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A constant force of 10 N acts on a 2.5 kg object for 9 seconds. What is the velocity of the object?
Answer:
[tex]\huge\boxed{\sf v = 36 \ m/s}[/tex]
Explanation:
Given data:Force = F = 10 N
Mass = m = 2.5 kg
Time = t = ?
Required:Velocity = v = ?
Formula:[tex]\displaystyle F=\frac{mv}{t}[/tex]
Solution:Put the given data in the above formula.
[tex]\displaystyle 10=\frac{(2.5)v}{9} \\\\10 \times 9 = 2.5v\\\\90=2.5 v\\\\Divide \ 2.5 \ to \ both \ sides\\\\90/2.5=v\\\\36 \ m/s=v\\\\v = 36 \ m/s\\\\\rule[225]{225}{2}[/tex]
A 2,500-kilogram roller coaster car is just reaching the bottom of a 30.00-meter hill. Assuming friction is 0, what are the potential energy (PE) and kinetic energy (KE) of the car at a height of 0 meters?
PE = 367,500 J; KE = 367,500 J.
PE = 0 J; KE = 735,000 J.
PE = 735,000 J; KE = 0 J.
PE = 735,000 J; KE = 735,000 J.
All of the potential energy has been converted to kinetic energy at the base of hills, resulting in greater speed. Potential energy = 735000 J ; kinetic energy = 0 J (option -C) is correct answer.
How does the energy change when a roller coaster is at the base of a hill?At the bottom of hills, roller coaster cars always move the fastest. This is connected to the first idea in that at the base of hills, all potential energy has been transformed into kinetic energy, leading to increased speed.
At a height of 0 meters,
Potential energy = mgh
Potential energy = 2,500×9.8×30
Potential energy = 735,000 J
Kinetic energy = 1/2mv²
Kinetic energy = 1/2×2,500×0
Kinetic energy = 0.
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(a) What is the minimum force of friction required to hold the system of Figure in equilibrium. Let W₁=100N and W₂=59N
(b) What coefficient of static friction between the 100N block and the table ensures equilibrium?
The minimum force of friction required to hold the system of Figure in equilibrium is 50 Newton.
What is frictional force?
Frictional force is the force produced when two surfaces slide against and touch each other. The surface texture and amount of force needing them together have the biggest an impact on these forces. The amount of frictional force is influenced by the object's position and angle.
The forces of adhesion between the sites of contact areas of the surfaces, which are always minutely uneven, are the primary cause of friction between objects. The action of the imperfections of the harder surface tilling across the softer surface shears these "fused" connections, causing friction.
If an object is placed against an object, then the frictional force will be the same as the weight of the object. If an object is pushed against the surface, then the frictional force will be increased and become extra than the weight of the object.
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the braided metal shield is very good at blocking electromagnetic signals from entering the cable and producing noise.
a. true
b. false
This assertion is accurate; electromagnetic signals are effectively blocked from entering the cable and causing noise by the braided metal shield.
What is electromagnetic ?The study of charge, its forces, and the fields that surround it is known as electromagnetic. Two components of the electromagnetic electric field are electricity and magnetism. In science, electromagnetism is just an interaction that takes place between charged particles. It is also the dominant force in interactions between atoms and molecules and is also the second-strongest of a four basic interaction, after the strong force.
How does electromagnetic work and what was its purpose?Electromagnets are wire coils with electricity flowing through them. An electromagnet's wire coils act like magnets when an electric current flows through them because moving charges produce magnetic fields.
Consequently, electromagnetic governs much more than only electricity and magnetism. It is the force that holds oppositely charged electrons to oppositely charged atomic nuclei, allowing for the formation of stable atoms and the occurrence of chemistry, including the chemistry that gives rise to life.
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A bat flies toward a wall, emitting a steady sound of frequency 1.70 kHz. This bat hears its own sound plus the sound reflected by the wall. How fast should the bat fly in order to hear a beat frequency of 8.00 Hz?
A bat is flying at a wall and making a constant 1,70 kHz sound. The bat must fly at a speed of 0.8 m/s in order to detect an 8.00 Hz beat frequency.
Who defines frequency?The frequency of a repeated event is its number of instances per unit time. It differs from angular frequency and is sometimes referred to as sampling rate for clarification. The unit of frequency is hertz (Hz), or one occurrence per second.
Do people own a frequency?The spring constant of a standing human body is approximately 7.5 Hz, and the rate of a sitting position in a cab is typically 4-6 Hz, according to the research that has already been done. Table 1 displays the natural frequencies of the major body parts.
Briefing:Frequency = 1.70 kHz
New frequency = 8.00 Hz
V = ?
8.00 / 1700 = 2 * v / 340m/s
v = 0.8 m/s
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Wax is a nonpolar substance. In which type of substance is it the most soluble?
Answer: C. A hot non-polar substance
Explanation: I am guessing you have options for this, because I've already had this question at school. So C.
The uniform bars, each of mass m, are connected as shown with their rollers of negligible mass confined to move in the vertical and horizontal guides. Determine the equilibrium angle e resulting from the application of the couple M.
The equilibrium angle results from the application of the couple M, if The uniform bars, each of mass m, are connected as shown with their rollers of negligible mass confined to move in the vertical and horizontal guides is [tex]2cos^{-M/{2mgb}}[/tex].
What is couple force?In mechanics, a couple is a pair of equal parallel forces that are acting in opposing directions. A couple only has the ability to cause or prevent the turning of the body. The sum of either force's magnitude and the angle between its action lines is used to calculate the turning effect, or moment, of a pair.
Given:
The bars are uniform and each has a mass of [tex]m[/tex].
Use the equilibrium of the system;
∑ M = 0 about the common point,
[tex]M = 2\times m \times g \times b \times cos(\theta/2)[/tex]
[tex]cos(\theta/2) = M / (2mgb)[/tex]
[tex]\theta / 2 =[/tex][tex]cos^{-M/{2mgb}}[/tex]
[tex]\theta =[/tex] [tex]2cos^{-M/{2mgb}}[/tex]
[tex]2cos^{-M/{2mgb}}[/tex]
[tex]2cos^{-M/{2mgb}}[/tex]
[tex]2cos^{-M/{2mgb}}[/tex]
Therefore, the equilibrium angle results from the application of the couple M, if The uniform bars, each of mass m, are connected as shown with their rollers of negligible mass confined to move in the vertical and horizontal guides is [tex]2cos^{-M/{2mgb}}[/tex].
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Consider the following two-car accident: Two cars of equal mass m collide at an intersection. Driver E was traveling eastward, and driver N, northward. After the collision, the two cars remain joined together and slide, with locked wheels, before coming to rest. Police on the scene measure the length d of the skid marks to be 9 meters. The coefficient of friction μ between the locked wheels and the road is equal to 0.9. (Figure 1) Each driver claims that his speed was less than 14 meters per second (about 31 mph). A third driver, who was traveling closely behind driver E prior to the collision, supports driver E's claim by asserting that driver E's speed could not have been greater than 12 meters per second. Take the following steps to decide whether driver N's statement is consistent with the third driver's contention. Part A
Let the speeds of drivers E and N prior to the collision be denoted by ve and vn, respectively. Find v2, the square of the speed of the two-car system the instant after the collision.
Express your answer terms of ve and vn.
The square of speed of two car system is 11m/s.
Measurements of motion characteristics, such as velocity, can be made using the conservation of momentum principle. Imagine that while on a physics expedition you happen to pass by a frozen lake where a hockey game is being played. Just as he collides, rather brutally for a pick-up game, with another player who is initially at rest, you measure one player's speed as 11.0 m/s. Think about a scenario where car A hits an immovable, static wall. Car A starts off moving at a certain speed (v), and after hitting the wall, it slows to a stop at a speed of zero. Newton's second law of motion, which applies the formula force equals mass times acceleration, uses this equation to describe the force in this circumstance. The acceleration in this scenario is (v - 0)/t, where t is the amount of time required for car A to stop.According to Newton's third law of motion, the wall, which is static and unbreakable, applies an equal force back to the car in response to the force the car is applying in its direction.To study about Newton's third law of motion-
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A student is trying to determine the muzzle
velocity (initial) of a bullet that is too fast to
measure directly. The student fires the 0.05kg
bullet into a 4kg block of wood initially at rest.
After the collision, the block, with the bullet
embedded inside moves to the right with a
speed of 1.87m/s right.
before collision:
Pbullet
Pblock
after collision:
Pcombined
kg m/s
kg m/s
kg m/s
How fast was the bullet fired?
m/s
The muzzle velocity of the bullet is 89.4 m/s.
What is muzzle?Muzzle is the front end of a firearm, where the projectile exits the barrel. It is an important safety feature, which helps to ensure that the firearm is pointed in a safe direction. Muzzles also help to contain the explosive force of the propellant gas, reducing the recoil of the firearm, and helping to protect the shooter from harmful gases and debris. Some firearms have additional features, such as flash suppressors, which help to reduce the muzzle flash created when the firearm is fired, making it easier to aim in low light conditions.
The muzzle velocity of the bullet can be determined by conservation of momentum. Momentum is defined as the mass of an object multiplied by its velocity. Therefore, the momentum of the bullet (Pbullet) before the collision is equal to the combined momentum of the bullet and the block (Pcombined) after the collision. We can calculate the muzzle velocity of the bullet by rearranging the equation as follows:
Pbullet = Pcombined
mv = (m + M)V
v = (m + M)V/m
where m is the mass of the bullet (0.05 kg), M is the mass of the block (4 kg), V is the velocity of the block after the collision (1.87 m/s), and v is the muzzle velocity of the bullet.
Plugging in the values, we get:|
v = (0.05 + 4) * 1.87/0.05 = 89.4 m/s
Therefore, the muzzle velocity of the bullet is 89.4 m/s.
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Given that : A = 4i+6j+12k and B = 3i+12j-7k.
(a) find the magnitude of A and B
(b) find magnitude of A-B
(c) find the sum of A and B
(d)Find the magnitude of A+B
(a) The magnitude of A is 14 and magnitude of B is 14.2.
(b) The magnitude of A - B is 19.95.
(c) The sum of A and B is 7i + 18j + 5k
(d) The magnitude of A + B is 19.95.
What is the magnitude of A and B?The magnitude of A and B is calculated as follows;
| A | = √ ( 4² + 6² + 12² )
| A | = 14
| B | = √ ( 3² + 12² + 7² )
| B | = 14.2
The magnitude of A - B is calculated as follows;
A - B = ( 4i + 6j + 12k ) - (3i + 12j - 7k)
A - B = ( i - 6j + 19 k )
|A-B| = √ (1² + 6² + 19²) = 19.95
The sum of A and B is calculated as follows;
A + B = ( 4i + 6j + 12k ) + (3i + 12j - 7k)
A + B = ( 7i 18j + 5k )
The magnitude of A + B is calculated as follows;
|A+B| = √ (7² + 18² + 5²) = 19.95
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3. If the static friction between wood and concrete is 0.62,
determine the force required to make a wooden block of mass
2 kg start to slide
Refer to the photo taken