Answer:
(a) Measured change in mass (Δm) = BVL/Rg
(b) Total measured mass M' = M - BVL/Rg
Explanation:
Current (I) across is coil is given by the formula;
I = V/R ------------------------1
The magnetic force is given by the formula;
Fb = B*I*L -------------------2
Putting equation 1 into equation 2, we have;
Fb = B*V*L/R -------------------3
Change in mass (Δm) is given as:
Δm = Fb/g -----------------------4
Putting equation 3 into equation 4, we have;
Δm = BVL/Rg
Therefore, change in mass (Δm) = BVL/Rg
2. Since B runs from North to South and current running from East to West, then the magnetic force is directed upward.
Therefore,
Total measure mass M' = M - BVL/Rg
The number of tornado deaths in the United States in the 2000s was less than 40 percent the number that occurred in the 1950s, even though there was a significant increase in population. Suggest a likely reason for the decline in the death toll.
ALSO..
A television meteorologist is able to inform viewers about the intensity of an approaching hurricane. However, the meteorologist can report the intensity of a tornado only after it has occurred. Why is this the case?
Answer:
Answer
Explanation:
The followings are the reasons for the reduction in the death toll from Tornadoes occurence in the 2000s compared to 1950s;
(a) Improvement in technology and rapid natural occurence prediction. Through the use of advanced radar systems the early detection of natural occurence e.g tornadoes, providing warning signal to the residents and tracking their possible pathways is made possible.
Fast spread of information through television channels, Radio, Social media etc., help in fast spreading of information, thereby sending early signals to people thus making them prepare for its occurence.
Improvement of protective measures, provision of housing infrastructure for the displaced. With increase in technology, the protective measures are also improving.
Quick Government and medical interventions for casualties. Conducting Tornado drills help people to practice to take cover in a specified location during tornadoes.
TV meteorologist are able to inform viewers of the approaching tornadoes and hurricanes which can be tracked from the time of its origin i.e., from a depression to a storm. Advanced radar data help in delineating the possible pathways of the hurricane.
Tornadoes on the other hand form in a thunderstorm or during a frontal activity. In thunderstorms there are vortices formed which are pre-cursors of tornadoes.. A vortice when touches the ground forms a tornado. Once a tornado is on the ground it becomes easy for radars to measure windpeeds. Hence a TV meteorologist can report the intensity of the Tornado.
Gases are a unique state of matter because the volume of a gas can be compressed or expanded based on its temperature. Jacque Charles studied these changes by observing hot air balloons. For a given mass, as the gas is heated its volume expands and as it cools the volume contracts. Charles found that the volume is directly proportional to temperature when the mass and pressure remain constant. Which equation represents the relationship in Charles' law?
A)(P1V1/T1)=(P2V2/T2)
B) (V1/T1)=(V2/T2)
C) (V1/n1)=(V2/n2)
D) P1V1=P2V2
Answer:
B)
Explanation:
Hi there,
Notice in the prompt it is saying given mass, which means quantity of gas is constant. This eliminates option C, as this is Avogadro's Law (the more gas you have, the more space it takes up). It also states pressure remains constant, so it eliminates D, Boyle's Law (pressure is inversely related with volume), and A, The Combined Gas Law (combines Charles, Boyle's, and Gay-Lussac's Laws).
Charles' Law is telling us that as a gas is heated, the volume increases. This makes sense, as when gases are heated the average kinetic energy (energy of motion and movement) of the system is higher because of heat increase. When the system's kinetic energy is higher, it will have the tendency to occupy more space to balance out the increase in energy. Think of people who become more jittery in an already crowded space, they will push outward to accompany more space to balance.
So, answer is B.
If you liked this solution, hit Thanks or give a Rating!
thanks,
Dentists use special lamps that consist of a concave mirror and a small, bright light source that is fixed on the principal axis of the mirror. When the light source is placed 5.0 cm from the mirror, the reflected light is focused in a bright spot at distance 70.0 cm from the mirror.
(a) Determine the radius of curvature of the mirror?
Answer: 140cm
Explanation:
Radius of curvature r = 2f
Where f is the focal point
f = 70cm, therefore
r =2 x 70 = 140cm
Answer:
The radius of curvature is 9.35 cm.
Explanation:
Given:
u = -5 cm
v = -70 cm
The radius of curvature of the mirror can be obtained from the expression:
[tex]\frac{1}{v} +\frac{1}{u} =\frac{2}{R} \\-\frac{1}{v} -\frac{1}{u} =\frac{2}{R} \\-\frac{1}{70} -\frac{1}{5} =\frac{2}{R} \\-0.014-0.2=\frac{2}{R} \\R=-9.35cm[/tex]
A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.8 cm away, in a time 3.5 10-7 s. What is the magnitude of the electric field
Answer:
3.53 N/C
Explanation:
Electric field = F / q where F is the force in N and q is charge on the electron
F = mass of an electron × a ( acceleration in m/s)
using equation of motion to solve for the acceleration
s ( distance ) = ut + 0.5 at² since the electron is starting from rest then ut = 0
2s / t² = a
F = me × ( 2s / t²)
E electric field = me × ( 2s / t²) / q = me × 2s / ( t² × q)
me, mass of an electron = 9.11 × 10⁻³¹ kg
E = (9.11 × 10⁻³¹ kg × 2 × 0.038 m) / ( (3.5 × 10⁻⁷s)² × 1.6 × 10⁻¹⁹ C) = 0.0353 × 10² N/C = 3.53 N/C
Answer: The magnitude of the electric field is 3.53 N/C
Explanation: Please see the attachments below
9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.
Calculate the work done by the student
Answer:
6370 J
Explanation:
By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m
[tex]W = E_p = mgh [/tex]
where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference
[tex]W = 50*9.8*13 = 6370 J[/tex]
Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen that the 13 th dark fringe is 8.57 cm away from the center of the central maximum. How far is the screen located from the slit
Final answer:
The width of the slit is 1.0 μm.
Explanation:
When light passes through a single slit, it undergoes diffraction, which causes interference patterns. The width of the central peak in the diffraction pattern is related to the width of the slit and the wavelength of the light. In this case, the width of the central peak is given as 5.0 mm and the wavelength is given as 600 nm.
Using the formula for the width of the central peak, we can solve for the width of the slit:
Width of slit = (wavelength * distance to screen) / (number of the peak * distance to the peak)
Substituting the given values into the formula, we find that the width of the slit is 1.0 μm.
An AC generator with an output rms voltage of 42.0 V at a frequency of 60.0 Hz is connected across a 20.0-?F capacitor. Find the following. (a) capacitive reactance ? (b) rms current A (c) maximum current in the circuit A (d) Does the capacitor have its maximum charge when the current takes its maximum value? Yes No
Explanation:
answer and explanation is in the picture
The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts horizontally on an object on the Moon, producing an acceleration equal to am. In the absence of friction and drag, the same force acting on the same object on Earth would produce an acceleration i.e. equal to
a. 6am
b. (1/6)am
c. am
d. 0
e. 9.8 m/s2
Answer:
c.[tex]a_m[/tex]
Explanation:
We are given that
Acceleration due to gravity on the moon=[tex]a_m[/tex]
Acceleration due to gravity on the earth=[tex]a_e[/tex]
[tex]g_m=\frac{1}{6}g_e[/tex]
Net force due to am on an object on moon=[tex]F_{net}=ma_m[/tex]
There is no friction and no drag force and there is no gravity involved
Then, the force acting on an object on earth=[tex]F=ma_e[/tex]
[tex]F=F_{net}[/tex](given)
[tex]ma_m=ma_e[/tex]
[tex]a_e=a_m[/tex]
Hence, option c is true.
A stainless steel tube with an outside diameter of 54 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 190 MPa, determine the maximum load P that the member can support.
Answer:146.26 kN
Explanation:
Given
Outside diameter of tube [tex]d_o=54\ mm[/tex]
thickness of tube [tex]t=5\ mm[/tex]
therefore inner diameter [tex]d_i=54-2\times 5[/tex]
[tex]d_i=44\ mm[/tex]
Cross-section of tube [tex]A=\frac{\pi }{4}(D_o^2-d_i^2)[/tex]
[tex]A=\frac{\pi }{4}\times 980[/tex]
[tex]A=245\pi mm^2[/tex]
Stress developed must be less than the limited value
thus [tex]\frac{P}{A}\leq \sigma [/tex]
[tex]P\leq \sigma A[/tex]
Maximum value of [tex]P=\sigma \times A[/tex]
[tex]P=190\times 245\times \pi [/tex]
[tex]P=146.26\ kN[/tex]
Final answer:
To calculate the maximum load a stainless steel tube can support, one must first find the cross-sectional area based on its outer diameter and wall thickness. Then, using the limit of the normal stress and the area, the maximum axial load can be computed.
Explanation:
To determine the maximum load P that a stainless steel tube can support when it is used as a compression member, we need to use the formula for axial stress in a cylindrical member, which is σ = P/A, where σ is the normal stress and A is the cross-sectional area. Given that the normal stress must not exceed 190 MPa, we need to first calculate the cross-sectional area. For a tube with an outside diameter (d) of 54 mm and a wall thickness (t) of 5 mm, the internal diameter (di) would be d - 2t = 54 mm - (2 x 5 mm) = 44 mm. The cross-sectional area A can be calculated using the formula for the area of a hollow circle, A = π/4 * (d2 - di2).
A = π/4 * ((55)^2 - (44)^2)
A = 855.298 * 10^-6 m2
Once we have the area, we can calculate the maximum load by rearranging the stress formula to P = σ * A. By plugging in the values for σ and A, we can find the value of P that the member can safely support. Remember to convert diameters to meters when calculating the area in square meters for consistency with the stress units of Pascals (Pa).
P = 190 * 855.298 Pa
P = 1.62 MPa
A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 12 m/s. Find the equation of motion.
Answer:
-6sin(2t)
Explanation:
fayemioluwatomisin is correct until the end.
x(0)=0
[tex]x(t)=C_1cos(2t)+C_2sin(2t)\\0=C_1cos(2(0))+C_2sin(2(0))\\C_1=0[/tex]
x'(0)=-12
[tex]x'(t)=2C_2cos(2t)\\-12=2C_2cos(2(0))\\-12=2C_2\\C_2=-6[/tex]
[tex]x(t)=-6sin(2t)[/tex]
The mass-spring system's equation of motion can be determined using Hooke's Law to find the spring constant, and applying the conditions of simple harmonic motion (SHM).
The question relates to the physical concept of a spring-mass system and its simple harmonic motion (SHM). The force exerted on a spring and the resulting stretch can determine the spring constant (k), using Hooke's Law F = kx, where F is the force applied and x is the displacement from the equilibrium position. In this case, we're given that a force of 880 newtons stretches the spring 4 meters, which allows us to calculate the spring constant k = 880 N / 4 m = 220 N/m.
Now, considering that the mass (m) attached to the spring is 55 kilograms and it's released with an upward velocity (v) of 12 m/s from the equilibrium position, the initial conditions for the SHM are established. The equation of motion for a spring-mass system in SHM is x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency (ω = √(k/m)), and φ is the phase constant. The velocity is the first derivative of the displacement function, v(t) = -Aω sin(ωt + φ), and the acceleration is the second derivative of the function.
The maximum velocity and acceleration occur when the displacement is zero and maximum respectively. To fully determine the equation of motion for this particular situation, additional information about the phase constant or initial displacement may be needed, unless it is assumed that the spring is released from the equilibrium point with an upward velocity, in which case the initial displacement would be zero, and the phase constant φ could be determined.
In a ballistics test, a 23 g bullet traveling horizontally at 1100 m/s goes through a 33-cm-thick 400 kg stationary target and emerges with a speed of 950 m/s . The target is free to slide on a smooth horizontal surface.
(a) how long is the bullet in the target?
______S
what average force does it exert on the target?
____N (magnitude only)
(b) What is the targets speed just after the bullet emerges?
_____m/s
Answer:
Explanation:
loss of energy while passing through target by bullet
= 1/2 mu² - 1/2 mv² , m is mass of bullet , u is initial velocity and v is final velocity .
= 1/2 x m ( u² - v² )
= .5 x .023 x ( 1100² - 950² )
= 3536.25 J.
This loss is due to negative work done by friction force
If friction force be F
Work done by friction force = F x .33
F x .33 = loss of kinetic energy
F x .33 = 3536.25
F = 10716 N
impulse of F
F X t , time period during which this force remains active
10716 x t = change in momentum of bullet
= .023 ( 1100 - 950 )
= 3.45
t = 3.45 / 10716
= 3.22 x 10⁻⁴ s.
Average force on the target = friction force created = 10716 N
Impulse by force on target = 10716 x 3.22 x 10⁻⁴
impulse on target = change in momentum of target
= mass of target x its velocity after impact
= 400 v
v = 10716 x 3.22 x 10⁻⁴ / 400
= 86.26 x 10⁻⁴ m /s
When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at the interface separating the two materials. For example, consider a ray of light that travels from air into the water of a lake. As the ray strikes the air-water interface (the surface of the lake), it is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation.
These effects on light propagation occur because light travels at different speeds depending on the medium. The index of refraction of a material, denoted by n , gives an indication of the speed of light in the material. It is defined as the ratio of the speed of light c in vacuum to the speed v in the material, orn=cv
What is the minimum value that the index of refraction can have?
Answer:
First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c). The value of the refractive index of the medium is a measure of its "optical density": Light spreads at maximum speed in a vacuum but slower in others transparent media; therefore in all of them n> 1. Examples of typical values of are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).
The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:
The limit or minimum angle, α lim, is defined as the angle of refraction from which the refracted ray disappears and all the light is reflected. As in the maximum value of angle of refraction, from which everything is reflected, is βmax = 90º, we can know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:
βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1
Explanation:
When a light ray strikes the separation surface between two media different, the incident beam is divided into three: the most intense penetrates the second half forming the refracted ray, another is reflected on the surface and the third is breaks down into numerous weak beams emerging from the point of incidence in all directions, forming a set of stray light beams.
The minimum value of the index of refraction is 1, which occurs in a vacuum where the speed of light is at its maximum and not slowed by any material.
The minimum value that the index of refraction can have is for a vacuum, where light travels at its fastest and the speed of light is not slowed down by any medium. According to the definition of the index of refraction, n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material. Since the speed of light in a vacuum is the highest possible speed for light and cannot be exceeded, the index of refraction n has a minimum value of 1 in a vacuum. That is because when we plug in the values into the equation, with both c and v being equal (since v is the speed of light in a vacuum), the ratio n = c/v equals 1. Therefore, no material can have an index of refraction less than 1.
the coefficient of static friction between the bed of a truck and a box resting on it is 0.67. The truck is traveling along a straight horizontal road at 30 m/s. what is the least distance in which the truck can stop if the box is not to slide/
Answer:
68.46 m.
Explanation:
Given,
coefficient of friction,μ = 0.67
speed of truck, v = 30 m/s
distance travel by the truck to stop = ?
Now,
Calculation of acceleration
we know,
f = m a
and also
f = μ N = μ mg
Equating both the forces equation
m a = μ mg
a = μ g
a = 0.67 x 9.81
a = 6.57 m/s²
Now, using equation of kinematics
v² = u² + 2 a s
0 = 30² - 2 x 6.57 x s
s = 68.46 m
Hence, the minimum distance travel by the truck is equal to 68.46 m.
The truck would need to stop in a minimum distance of approximately 68.5 meters to prevent the box from sliding off, based on the given parameters and considering the law of inertia and the role of static friction.
To determine the minimum distance the truck can stop without the box sliding off, we need to consider the law of inertia and the role of static friction.
The force must meet or exceed the force exerted by the truck's deceleration to keep the box from sliding.
Given that the initial velocity of the truck is 30 m/s and the box has a mass of 500 kg, we can use the formula for static friction, which is F = μN, where F is the force of friction, μ is the coefficient of static friction, and N is the normal force.
In this case, N equates to the gravitational force on the box, so N = mg = 500 kg * 9.8 m/s² = 4900 N. Substituting these values in the formula for static friction, we get F = 0.67 * 4900 N = 3283 N.
The force of friction, in terms of motion, is also equivalent to mass times acceleration (F = ma), so we can set this equal to the static friction we calculated and solve for acceleration: 3283 N = 500 kg * a. Solving for a gives an acceleration of approximately 6.57 m/s².
Using motion equations, specifically v² = u² + 2as where v is the final velocity (0 m/s), u is the initial velocity (30 m/s), a is the acceleration (-6.57 m/s² because it's deceleration), and s is the distance, we can compute for the distance s which gives a value of approximately 68.5 meters.
Learn more about Physics of Static Friction here:
https://brainly.com/question/28151605
#SPJ6
The probable question is :-
A truck is moving along a straight horizontal road at 30 m/s. A box is placed on the bed of the truck, and the coefficient of static friction between the bed and the box is 0.67. What is the least distance in which the truck can come to a stop without causing the box to slide? Consider the deceleration of the truck due to braking.
At some point, not close to its ends, within a solenoid of arbitrary length, calculate the approximate magnetic field if the solenoid carries a current 10.0 A and has 110.0 turns per meter of the solenoid's length.
Answer: The approximate magnetic field is 1.38 × 10^-3T
Explanation: Please see the attachment below
The approximate magnetic field at a point within the solenoid is 4.4 × 10^-4 T.
Explanation:To find the magnetic field strength at a point within a solenoid, we can use the equation B = µ0nI, where B is the magnetic field strength, µ0 is the permeability of free space, n is the number of turns per unit length, and I is the current. Given that the solenoid carries a current of 10.0 A and has 110.0 turns per meter of its length, we can substitute these values into the equation to calculate the magnetic field strength:
B = (4π × 10-7 T·m/A)(110.0 turns/m)(10.0 A) = 4.4 × 10-4 T
Therefore, the approximate magnetic field at a point within the solenoid is 4.4 × 10-4 T.
Learn more about Magnetic field within a solenoid here:https://brainly.com/question/32192685
#SPJ11
g In a lab experiment, you are given a spring with a spring constant of 15 N/m. What mass would you suspend on the spring to have an oscillation period of 0.97 s when in SHM
Answer:
0.358 kg
Explanation:
From simple harmonic motion,
T = 2π√(m/k)................ Equation 1
Where T = period of the spring, k = spring constant of the spring, m = mass suspended, π = pie
make m the subject of the equation
m = kT²/4π².................. Equation 2
Given: k = 15 N/m, T = 0.97 s, π = 3.14
Substitute into equation 2
m = 15(0.97²)/(4×3.14²)
m = 14.1135/39.4384
m = 0.358 kg.
Hence mass suspended = 0.358 kg
0.356kg
Explanation:The period, T, of a mass on a spring of spring constant, k, in simple harmonic motion can be calculated as follows;
T = 2π [tex]\sqrt{\frac{m}{k} }[/tex] --------------------(i)
Where;
m = mass
From the question;
T = 0.97s
k = 15N/m
Taking π = 3.142 and substituting the values of T and k into equation (i) as follows;
0.97 = 2 x 3.142 x [tex]\sqrt{\frac{m}{15} }[/tex]
0.97 = 6.284 x [tex]\sqrt{\frac{m}{15} }[/tex]
[tex]\frac{0.97}{6.284}[/tex] = [tex]\sqrt{\frac{m}{15} }[/tex]
0.154 = [tex]\sqrt{\frac{m}{15} }[/tex]
Square both sides
0.154² = [tex]\frac{m}{15}[/tex]
0.0237 = [tex]\frac{m}{15}[/tex]
m = 0.356
Therefore, the mass that could be suspended on the spring to have an oscillation period of 0.97s when in SHM is 0.356kg
Two identical steel balls, each of mass 67.8 g, are moving in opposite directions at 4.80 m/s.They collide head-on and bounce apart elastically. By squeezing one of the balls in a vise while precise measurements are made of the resulting amount of compression, you find that Hooke's law is a good model of the ball's elastic behavior. A force of 15.9 kN exerted by each jaw of the vise reduces the diameter by 0.130 mm. Model the motion of each ball, while the balls are in contact, as one-half of a cycle of simple harmonic motion. Compute the time interval for which the balls are in contact.
Answer:
Explanation:
To detrmine the time interval at which the balls are in contact.
Given information
The mass of each steal ball 67.8g. The speed of ball towards each other is 4.80 m/s. The exerted force by each jaw 15.9 kN and the forece reduce the diameter by 0.130 mm.
Expression for the effective spring constant ball is shown below.
K = |F|/|x|
Here,
k is a spring constant
F is the force exerted on the ball
x is dispalcement due force
substitute 15.9 kN for F and 0.130 mm in above equation
K = (15.9 kN)(1X10³N) / (0.130 mm)(1x10⁻³m/1mm)122 x 10⁶ N/mThe spring constant is 122 x 10⁶ N/m
An object with a mass of 2.3 kg has a force of 6.2 newtons applied to it. What is the resulting acceleration of the object?
Answer:2.7m/s^2
Explanation:
mass=2.3kg
Force=6.2Newton
Acceleration=force ➗ mass
Acceleration=6.2 ➗ 2.3
Acceleration=2.7m/s^2
The resulting acceleration of the object is 2.70 m/s^2.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given that:
Mass of the object: m = 2.3 kg.
Force applied on it: F = 6.2 Newtons.
Now from Newton's 2nd law of motion: it can be stated that:
force = mass × acceleration
acceleration = force/mass
= 6.2 Newton/2.3 kg
= 2.70 m/s^2.
Hence, the acceleration of the object is 2.70 m/s^2.
Learn more about acceleration here:
brainly.com/question/12550364
#SPJ2
A thin metal cylinder of length L and radius R1is coaxial with a thin metal cylinder of length L and a larger radius R2. The space between the two coaxial cylinders is filled with a material that has resistivity rho. The two cylinders are connected to the terminals of a battery with potential difference ΔV, causing current I to flow radially from the inner cylinder to the outer cylinder. Part A Find an expression for the resistance of this device in terms of its dimensions and the resistivity. Express your answer in terms of some or all of the variables R1, R2, L, and rho. R = nothing
Answer:
The expression for resistance is [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
Explanation:
Generally flow of charge at that point is mathematically given as
[tex]J = \frac{I}{2 \pi r L}[/tex]
Where L is length of the cylinder as given the question
The potential difference that is between the cylinders is
[tex]\delta V = -E dr[/tex]
Where is the radius
Where E is the electric field that would be experienced at that point which is mathematically represented as
[tex]E = \rho J[/tex]
Where is the [tex]\rho[/tex] is the resistivity as given the question
considering the formula for potential difference we have
[tex]\delta V = -[\frac{\rho I}{2 \pi r L} ]dr[/tex]
To get V we integrate both sides
[tex]\int\limits^V_0 {\delta V} \, = \int\limits^{R_2}_{R_1} {\frac{\rho I}{2 \pi L r} } \, dr[/tex]
[tex]V = \frac{\rho I}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
According to Ohm law
[tex]V= IR[/tex]
Now making R the subject we have
[tex]R = \frac{V}{I}[/tex]
Substituting for V
[tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
The resistance of a coaxial cylindrical configuration with radii R1 and R2, length L, and resistivity rho, is given by the formula R = (rho / (2 * pi * ln(R2/R1))).
The question asks for an expression for the resistance of a coaxial cylindrical configuration, where the space between two metal cylinders of length L and radii R1 and R2 is filled with a material of resistivity rho. To find this, we utilize the formula for the resistance R of a material, which is R = rho * (L/A), where A is the cross-sectional area. However, since the current flows radially through the material between the cylinders, we need to consider the formula for resistance in terms of the radii and length of the cylinders. The resistance can be expressed as R = (rho * L) / (2 * pi * L * ln(R2/R1)), simplifying to R = (rho / (2 * pi * ln(R2/R1))).
A solenoidal inductor for a printed circuit board is being redesigned. To save weight, the number of turns is reduced by one-fifth, with the geometric dimensions kept the same. By how much must the current change if the energy stored in the inductor is to remain the same? Select one: a. It must be two times larger. b. It should be one-half as large. c. It should be left the same. d. It must be four times larger. e. No change in the current can compensate for the reduction in the number of turns.
The "it must be five times larger" current change if the energy stored in the inductor is to remain the same.
Explanation:
A current produced by a modifying magnetic field in a conductor is proportional to the magnetic field change rate named INDUCTANCE (L). The expression for the Energy Stored, that equation is given by:
[tex]U= \frac{1}{2} LI^2[/tex]
Here L is the inductance and I is the current.
Here, energy stored (U) is proportional to the number of turns (N) and the current (I).
[tex]L = \frac{\mu_0 N^2 *A}{l}[/tex]
mu not - permeability of core material
A -area of cross section
l - length
N - no. of turns in solenoid inductor
Now,given that the proportion always remains same:
[tex]\frac{N_2}{N_1} = \frac{I_1}{I_2}[/tex]
In this way the expression
[tex]\frac{1}{5} = \frac{I_1}{I_2}[/tex]
[tex]I_2 = I_1 \times 5[/tex]
Thus, it suggest that "it must be five times larger" current change if the energy stored in the inductor is to remain the same.
Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.
Answer:
Velocity between points A and B will be 0.2344 m/s
Complete Question:
A tennis racket is thrown vertically into the air. The center of gravity G has a velocity of 5 m/s upwards. Angular velocity about the x - direction of 1 rad/s and angular velocity about the y - direction of 20 rad/s. Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.
Answer:
a) [tex]v_{A} = 10 m/s[/tex]
b)
[tex]v_{B} = 3i + 0.15 j m/s\\v_{B} = \sqrt{3^{2} + 0.15^{2} }\\v_{B} = 3.004 m/s[/tex]
Explanation:
a) Velocity of point A
The velocity of point A is a combination of the translational and the rotational velocity of the racket.
[tex]v_{A} = v_{G} + v_{r}[/tex]
[tex]v_{G} = 5 m/s[/tex]
[tex]v_{r} = wr[/tex]
r = 25 cm = 0.25 m
w = 20 rad/s
[tex]v_{r} = 20 * 0.25\\v_{r} = 5 m/s[/tex]
[tex]v_{A} = 5 + 5\\v_{A} = 10 m/s[/tex]
b) Velocity of point B
At point B, the linear velocity is in the +ve z-direction while the rotational velocity is in the -ve z-direction:
[tex]v_{G} = 5 m/s[/tex]
[tex]v_{r} = -r w\\v_{r} = - 0.25 * 20\\v_{r} = - 5 m/s[/tex]
[tex]v_{Bz} = v_{G} + v_{r} \\v_{Bz} = 5 -5\\v_{Bz} = 0 m/s[/tex]
In the y - direction, r = 30/2 = 15 cm = 0.15 m
r = 0.15 m
[tex]w_{x} = 1 rad/s[/tex]
[tex]v_{By} = rw_{x} \\v_{By} = 0.15 * 1\\v_{By} = 0.15 rad/s[/tex]
In the x - direction, r = 0.15 m, [tex]w_{y} = 20 rad/s[/tex]
[tex]v_{Bx} = rw_{y} \\v_{Bx} = 0.15 * 20\\v_{Bx} = 3.0 rad/s[/tex]
[tex]v_{B} = 3 i +0.15 j\\[/tex] m/s
The first American to win the Nobel prize in 1907 was:
Fizeau
Roemer
Michelson
Answer:
Michelson
Explanation:
The catcher catches a .5 kg baseball traveling at 40 m/s and the glove is pushed back 10 cm . What average force did the catcher put on the ball?
Answer:
4000 N
Explanation:
First we calculate the acceleration
2as = vf^2 - vi^2
2xax0.1 = 0^2-40^2
0.2 x a = -1600
a = - 8000 m/s^2
F= ma
F= 0.5 x 8000
F 4000N
A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in contact with the floor for 1.5 ms (1.5multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor?
Answer:
1400 N
Explanation:
Change in momentum equals impulse which is a product of force and time
Change in momentum is given by m(v-u)
Equating this to impulse formula then
m(v-u)=Ft
Making F the subject of the formula then
[tex]F=\frac {m(v-u)}{t}[/tex]
Take upward direction as positive then downwards is negative
Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then
[tex]F=\frac {0.3(2--5)}{0.0015}=1400N[/tex]
The average force exerted on the ball by the floor is 1400 N.
What is force?Force is the product of mass and acceleration. The S.I unit of force is Newton (N)
To calculate the average magnitude of the force exerted on the ball by the floor, we use the formula below.
Formula:
F = m(v-u)/t............... Equation 1Where:
F = Force exerted on the ball by the floor.m = mass of the floorv = final velocity of the ballu = initial velocity of the ballt = timeFrom the question,
Given:
m = 0.3 kgv = 5 m/su = -2 m/s (rebounds)t = 1.5×10⁻³ sSubstitute these values into equation 1
F = 0.3[(5)-(-2)]/(1.5×10⁻³)F = 0.3(5+2)/(1.5×10⁻³)F = 2.1/(1.5×10⁻³)F = 1.4×10³F = 1400 NHence, the average force exerted on the ball by the floor is 1400 N.
Learn more about force here: https://brainly.com/question/13164598
A barge floating in fresh water (rho = 1000 kg/m^3) is shaped like a hollow rectangular prism with base area A = 550 m^2 and height H = 2.0 m. When empty the bottom of the barge is located H0 = 0.55 m below the surface of the water. When fully loaded with coal the bottom of the barge is located H1 = 1.35 m below the surface.
Randomized Variables
A = 750 m²
H₀ = 0.55 m
Hᵢ = -1.1 m
(A) Write an equation for the buoyant force on the empty barge in terms of the known data.
(B) Determine the mass of the barge in kilograms.
Answer:
Explanation:
A )
When empty , H₀ length of barge is inside water .
volume of barge inside water = A x H₀
Weight of displaced water = AH₀ x ρ x g
Buoyant force = weight of displaced water = AH₀ ρg
B)
It should balance the weight of barge
Weight = buoyant force
Weight = AH₀ ρg
mass of barge = weight / g
weight / g = AH₀ ρ
= 550 x .55 x 1000
= 302500 kg
An image is located at 4f, where f is the focal length, after it exits the lens. What is the object distance in terms of the focal length?
Answer:
p = 4/3 f
Explanation:
The constructor station describes the position of the image object in relation to the focal length
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distances to the object and the image respectively
the objent distance is
1 / p = 1 / f - 1 / q
they indicate that the image is located at q = 4f this distance is positive because the image is after the lens
1 / p = 1 / f - 1 / 4f
1 / p = 3 / 4f
p = 4f / 3
the object is positive because it is to the left of the lens, according to the optical sign convention
The sign convention is that the distance to the objects is positive if it is on the left side of the lens and the image is positive if it is on the right side of the lens.
While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from a sodium street lamp, radio waves from an AM radio station, radio waves from an FM radio station, and microwaves from an antenna of a communications system. Rank these type of waves in terms of increa
Answer:
AM radio, FM radio, microwaves, sodium light.
Explanation:
Electromagnetic radiation are waves from electromagnetic field which spread through the space or any other material medium and carries radiating energy. Examples includes X rays, radio waves, Gamma rays, etc. Exposure to high level of electromagnetic radiation could be harmful to human body, on the other hand, science as not been able to prove that exposing humans to low level electromagnetic radiation is harmful to our health.
An automobile traveling 71.0 km/h has tires of 60.0 cm diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in 40.0 complete turns of the tires, what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking? (Note: automobile moves without sliding)
Explanation:
We have,
Initial speed of an automobile, u = 71 km/h = 19.72 m/s
Diameter of the tie, d = 60 cm
Radius, r = 30 cm
(a) The angular speed of the tires about their axles is given by :
[tex]\omega=\dfrac{v}{r}\\\\\omega=\dfrac{19.72}{0.3}\\\\\omega=65.73\ rad/s[/tex]
(b) Final angular velocity of the wheel is equal to 0 as its stops. The angular acceleration of the wheel is given by :
[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]
[tex]\theta=40\ rev\\\\\theta=251.32\ rad[/tex]
[tex]0-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_i^2}{2\theta}\\\\\alpha =\dfrac{(65.73)^2}{2\times 251.32}\\\\\alpha =-8.59\ rad/s^2[/tex]
(c) Let the car move a distance d during the braking. So,
[tex]d=\theta r\\\\d=251.32\times 0.3\\\\d=75.39\ m[/tex]
Therefore, the above is the required explanation.
a. The angular speed is 65.73 rad/s.
b. The magnitude of the angular acceleration is -8.59 rad/s².
c. The distance should be 75.39m.
Calculation of the angular speed, magnitude, and the distance:Since
Initial speed of an automobile, u = 71 km/h = 19.72 m/s
Diameter of the tie, d = 60 cm
Radius, r = 30 cm
a. Now the angular speed should be
= v/r
= 19.72/0.3
= 65.73 rad/s
b. Now the magnitude is
= 65.73^2/2*251.32
= -8.59 rad/s^2
c. The distance should be
= 251.32*0.3
= 75.39 m
Learn more about speed here: https://brainly.com/question/18212021
A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each with a radius of 5.00 mm. If water flows through the single pipe at 1.45 m/s, calculate the speed (in m/s) of the water in the narrower pipes.
Answer:
1.24 m/s
Explanation:
Metric unit conversion:
9.25 mm = 0.00925 m
5 mm = 0.005 m
The volume rate that flow through the single pipe is
[tex]\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s[/tex]
This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of
[tex]\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s[/tex]
So the flow speed of each of the narrower pipe is:
[tex]v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}[/tex]
[tex]v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s[/tex]
Near the earth the intensity of radiation from the sun is 1.35 kW/m2. What volume of space in this region contains 1.0 J of electromagnetic energy? (c = 3.0 x 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
The volume is 2.22x10⁵m³
Explanation:
the solution is in the attached Word file
To calculate the volume of space containing 1.0 J of electromagnetic energy near the Earth, use the formula Volume = Energy / Intensity after converting the intensity from kW/m² to W/m².
The volume of space that contains 1.0 J of electromagnetic energy can be found using the formula:
Volume = Energy / Intensity
Given that the intensity is 1.35 kW/m², you need to convert it to W/m² (1 kW = 1000 W).
After converting the intensity, you can then calculate the volume using the given energy of 1.0 J.
The volume of space near Earth containing 1.0 J of electromagnetic energy can be found by first calculating the energy density using the given intensity and speed of light, and then dividing the amount of energy by energy density, resulting in a volume of 2.22 imes 10^5 m^3.
Finding the Volume of Space Containing 1.0 J of Electromagnetic Energy
To find the volume of space near Earth that contains 1.0 J of electromagnetic energy given the intensity (I) from the sun as 1.35 kW/m2, we can use the formula for energy density (u), which is given by the equation u = I/c, where c is the speed of light. Now, to find the volume (V) that contains energy (E), we use the equation V = E/u.
First, let's convert the intensity to watts per square meter: 1.35 kW/m2 = 1350 W/m2. Next, calculate the energy density (u):
u = 1350 W/m2 / (3.0 * 108 m/s) = 4.5 * 10-6 J/m3.
With the energy density known, we can calculate the volume (V) that contains 1.0 J of energy:
V = 1.0 J / (4.5 * 10-6 J/m3) = 2.22 * 105 m3.
Therefore, a volume of 2.22 * 105 m3 near Earth contains 1.0 J of electromagnetic energy.
A 2.40-kilogram mud ball drops from rest at a height of 18.0 m. If the impact between the ball and the ground lasts 0.50 s, what is the magnitude of the average force exerted by the ball on the ground?
Answer:
The magnitude of the force exerted by the ball on the ground during the 0.5 s of contact = 90.16 N
Explanation:
Given,
Mass of mud ball = m = 2.40 kg
Height the ball is released from = y = 18 m
Total contact time of ball and the ground = t = 0.5 s
The Newton's second law of motion explains that the magnitude of the change of momentum is equal to the magnitude of a body's impulse.
Change in momentum = Magnitude of Impulse
Change in momentum = (final momentum) - (initial momentum)
Since the ball is dropped from rest, initial momentum = 0 kgm/s
But to calculate its final momentum, we need the ball's final velocity before hitting the ground.
Using the equations of motion,
u = initial velocity of the ball = 0 m/s (ball was dropped from rest)
v = final velocity of the ball = ?
g = acceleration due to gravity = 9.8 m/s²
y = vertical distance covered by the ball = 18 m
v² = u² + 2gy
v² = 0² + (2)(9.8)(18)
v² = 352.8
v = 18.78 m/s
Final momentum of the ball = (m)(v)
= (2.4) × (18.78) = 45.08 kgm/s
Change in momentum = 45.08 - 0 = 45.08 kgm/s
Impulse = Ft
Change in momentum = Magnitude of Impulse
45.08 = F × (0.5)
F = (45.08/0.5) = 90.16 N
Hope this Helps!!!
Answer:
90.14 N
Explanation:
according to the impulse momentum theorem,
Impulse = change in momentum
Where impulse = force × time and change in momentum = m ( v - u).
The object was initially at rest, hence it initial velocity is zero.
To get the final velocity, we use the formula below
v² = u² + 2gh
Where h = height of the cliff = 18.0m
v² = 2 × 9.8 × 18
v² = 352.8
v = √333.2
v = 18.78 m/s
At t = 0.50s and v = 18.78 m/s, we can get the average force of impact
F×0.50 = 2.4 (18.78 - 0)
F × 0.50 = 2.4 (18.78)
F × 0.50 = 45.072
F = 45.072 /0.50
= 90.14 N