Answer:
126 mWb
Explanation:
Given that:
length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.
We assume that the magnetic field in the solenoid is constant.
The magnetic flux is given as:
[tex]\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb[/tex]
A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?
Answer:
(a) Fw = 101.01 N
(b) W = 282.82 J
(c) Fg = 382.2 N
(d) N = 368.61 N
(e) Net force = 0 N
Explanation:
(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:
[tex]F_g-F_w=0[/tex] (1)
Fg: gravitational force over the object
Fw: worker's force
However, in an incline you have that the gravitational force on the object, due to its weight, is given by:
[tex]F_g=Wsin\theta=Mg sin\theta[/tex] (2)
M: mass of the ice block = 39 kg
g: gravitational constant = 9.8m/s^2
θ: angle of the incline
You calculate the angle by using the information about the distance of the incline and its height, as follow:
[tex]sin\theta=\frac{0.74m}{2.8m}=0.264\\\\\theta=sin^{-1}(0.264)=15.32\°[/tex]
Finally, you solve the equation (1) for Fw and replace the values of all parameters:
[tex]F_w=F_g=Mgsin\theta\\\\F_w=(39kg)(9.8m/s^2)sin(15.32\°)=101.01N[/tex]
The worker's force is 101.01N
(b) The work done by the worker is given by:
[tex]W=F_wd=(101.01N)(2.8m)=282.82J[/tex]
(c) The gravitational force on the block is, without taking into account the rotated system for the incline, only the weight of the ice block:
[tex]F_g=Mg=(39kg)(9.8m/s^2)=382.2N[/tex]
The gravitational force is 382.2N
(d) The normal force is:
[tex]N=Mgcos\theta=(39kg)(9.8m/s^2)cos(15.32\°)=368.61N[/tex]
(e) The speed of the block when it slides down the incle is constant, then, by the Newton second law you can conclude that the net force is zero.
g You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 78 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
F=49.48 N
Explanation:
Given that
Diameter , d= 30 mm
Holding pressure = 70 % P
P=Atmospherics pressure
We know that
P= 1 atm = 10⁵ N/m²
The force per unit area is known as pressure.
[tex]P=\dfrac{F}{A}[/tex]
[tex]F=P\times A[/tex]
[tex]F=0.7\times 10^5\times \dfrac{\pi}{4}\times 0.03^2\ N[/tex]
Therefore the force will be 49.48 N.
F=49.48 N
A student is investigating the relationship between sunlight and plant growth for her science expieriment. Determine which of the following tables is set up correctly
The question is incomplete as it does not have the options which have been provided in the attachment.
Answer:
Option-D
Explanation:
In the given question, the effect of the sunlight on the growth of the plant has been studied. The values provided in the Option-D can be considered correct as the values are measured in the decimal value up to two decimal value.
The values are measured after the first week, second week, and the initial readings. The difference in the values provided in Option-D does not show much difference as well as are up to two decimal places.
Thus, Option-D is the correct answer.
The shortest path from a starting point to an endpoint, regardless of the path
taken, is called the
A. vector addition
B. sum
C. shortest vector
D. resultant displacement
Answer:
answer is C shortest vector
Answer:the answer is resultant displacement
Explanation:
A body moving with uniform acceleration has two points (5,15) and (20,60) on the (v t) graphs of its motion. calculate the acceleration
Answer:
3m/s²
Explanation:
The slope on a velocity time graph represents the acceleration, so if you simply use the slope formula, you can find the acceleration between those two points.
m=rise/run
m=(60-15)/(20-5)
m=45/15
m=3 m/s ² squared (therefore this is your constant acceleration from those two points).
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 8.1 rad/s. The drawing shows the position of the block when the spring is unstrained. This position is labeled x= 0 m. The drawing also shows a small bottle located 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 m, and is then thrown to the left. In order for the block to knock over the bottle,it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.
Answer:
v₀ = 0.5058 m/s
Explanation:
From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m
Now, the potential energy of the block at x = 0.08 m is ½kx²
where;
k is the spring constant given by; k = ω²m
ω is the angular velocity of the oscillation
m is the mass of the block.
Thus, potential energy of the spring at the bottle(x = 0.08 m) is;
U = ½ω²m(0.08m)²
Also, potential energy of the spring at the bottle(x = 0.05 m) is;
U = ½ω²m(0.05m)²
and the kinetic energy of the block at x = 0.05 m is;
K = ½mv₀²
Thus;
½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²
Inspecting this, ½m will cancel out to give;
ω²(0.08)² = ω²(0.05)² + v₀²
Making v₀ the subject, we have;
v₀ = ω√((0.08)² - (0.05)²)
So,
v₀ = 8.1√((0.08)² - (0.05)²)
v₀ = 0.5058 m/s
A 110.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 8900.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.
Required:
a. What is the velocity of the truck right after the collision?
b. What is the change in mechanical energy of the car?
Answer:
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of the bodies before collision is equal to the sum of momentum of bodies after collision.
Momentum = Mass*velocity
BEFORE COLLISION
The momentum of a 110.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction = 110*25 = 2750kgm/s
The momentum of a 8900.0 kg truck with a speed of 20.000 m/s in an easterly direction = 8900*20 = 178000kgm/s
Sum of momentum before collision = 2750 + 178000 = 180,750 kgm/s
AFTER COLLISION
The momentum of the car will be 110*18 = 1980kgm/s
The momentum of the truck = 8900v where v is the velocity of the truck after collision.
Sum of momentum after collision = 1980 + 8900v
Applying the conservation law;
180750 = 1980 + 8900v
8900v = 180750-1980
8900v = 178770
v = 178770/8900
v = 20.09m/s
Velocity of the truck after collision is 20.09m/s
Note that the collision is inelastic i.e the body moves with different velocities after collision
b) The mechanical energy experienced by the bodies is kinetic energy.
Kinetic energy = 1/2mv²
Sum of the Kinetic energy before collision = 1/2(110)*25²+1/2(8900)*20²
= 34375 + 1780000
= 1,814,375Joules
Sum of kinetic energy after collision = 1/2*(110)*18²+1/2(8900)*20.09²
= 17820+1796056.045
= 1,813,876.045Joules
Change in mechanical energy = 1,813,876.045Joules - 1,814,375Joules
= -498.955Joules
Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why?
Answer:
The ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil.
Explanation:
Oil is more less dense than water. Thus, the molecules that make up the oil are larger than those that that make up water, so they cannot pack as tightly together as the water molecules will do. Hence, they will take up more space per unit area and are we can say they are less dense.
So, we can conclude that the ball filled with water will reach the bottom of the container first this is because oil is less dense than water and so the glass ball filled with oil will be a lot less denser than the one which is filled with water.
The displacement (in meters) of a particle moving in a straight line is given by the equation of motion:
s = 4/t^2, where t is measured in seconds.
Required:
Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.
Answer:
At [tex]t = 1\; \rm s[/tex], the particle should have a velocity of [tex]-8\; \rm m \cdot s^{-1}[/tex].At [tex]t = 2\; \rm s[/tex], the particle should have a velocity of [tex]-1\; \rm m \cdot s^{-1}[/tex].At [tex]t = 3\; \rm s[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{27}\; \rm m \cdot s^{-1}[/tex].For [tex]a > 0[/tex], at [tex]t = a \; \text{second}[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{a^3}\; \rm m \cdot s^{-1}[/tex].
Explanation:
Differentiate the displacement of an object (with respect to time) to find the object's velocity.
Note that the in this question, the expression for displacement is undefined (and not differentiable) when [tex]t[/tex] is equal to zero. For [tex]t > 0[/tex]:
[tex]\begin{aligned}v &= \frac{\rm d}{{\rm d}t}\, [s] = \frac{\rm d}{{\rm d}t}\, \left[\frac{4}{t^2}\right] \\ &= \frac{\rm d}{{\rm d}t}\, \left[4\, t^{-2}\right] = 4\, \left((-2)\, t^{-3}\right) = -8\, t^{-3} =-\frac{8}{t^3}\end{aligned}[/tex].
This expression can then be evaluated at [tex]t = 1[/tex], [tex]t = 2[/tex], and [tex]t = 3[/tex] to obtain the required results.
An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron's probability of tunneling through the barrier increase
Answer:
factor that the electron's probability of tunneling through the barrier increase 2.02029
Explanation:
given data
kinetic energy = 10.1 eV
height = 18.2 eV
width = 1.00 nm
wavelength = 546 nm
solution
we know that probability of tunneling is express as
probability of tunneling = [tex]e^{-2CL}[/tex] .................1
here C is = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
here h is Planck's constant
c = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-10.1) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c = 2319130863.06
and proton have hf = [tex]\frac{hc}{\lambda } = {1240}{546}[/tex] = 2.27 ev
so electron K.E = 10.1 + 2.27
KE = 12.37 eV
so decay coefficient inside barrier is
c' = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
c' = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-12.37) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c' = 1967510340
so
the factor of incerease in transmisson probability is
probability = [tex]e^{2L(c-c')}[/tex]
probability = [tex]e^{2\times 1\times 10^{-9} \times (351620523.06)}[/tex]
factor probability = 2.02029
You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to be 2.40 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. What frequency is required?
Answer:
f = 1.69*10^5 Hz
Explanation:
In order to calculate the frequency of the sinusoidal voltage, you use the following formula:
[tex]V_L=\omega iL=2\pi f i L[/tex] (1)
V_L: voltage = 12.0V
i: current = 2.40mA = 2.40*10^-3 A
L: inductance = 4.70mH = 4.70*10^-3 H
f: frequency = ?
you solve the equation (1) for f and replace the values of the other parameters:
[tex]f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz[/tex]
The frequency of the sinusoidal voltage is f
A 4.00 kg ball is moving at 4.00 m/s to the EAST and a 6.00 kg ball is moving at 3.00 m/s to the NORTH. The total momentum of the system is:___________.A. 14.2 kg m/s at an angle of 48.4 degrees SOUTH of EAST.B. 48.2 kg m/s at an angle of 24.2 degrees SOUTH of EAST.C. 48.2 kg m/s at an angle of 48.4 degrees NORTH of EAST.D. 24.1 kg m/s at an angle of 24.2 degrees SOUTH of EAST.
E. 24.1 kg m/s at an angles of 48.4 degrees NORTH of EAST.
Answer:
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EASTExplanation:
Momentum = mass*velocity of a body
For a 4.00 kg ball is moving at 4.00 m/s to the EAST, its momentum = 4*4 = 16kgm/s
For a 6.00 kg ball is moving at 3.00 m/s to the NORTH;
its momentum = 6*3 = 18kgm/s
Total momentum = The resultant of both momentum
Total momentum = √16²+18²
Total momentum = √580
total momentum = 24.1kgm/s
For the direction:
[tex]\theta = tan^{-1} \frac{y}{x}\\\theta = tan^{-1} \frac{18}{16}\\ \theta = tan^{-1} 1.125\\\theta = 48.4^{0}[/tex]
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be 70.0 km directly north of his starting point.
a. What was the length of the third leg of his trip?b. What was the direction of the third leg of his trip?
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Suppose I have an infinite plane of charge surrounded by air. What is the maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs
Answer:
[tex]53.1\mu C/m^2[/tex]
Explanation:
We are given that
Electric field,E=[tex]3\times 10^6V/m[/tex]
We have to find the value of maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs.
We know that
[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]3\times 10^6=\frac{\sigma}{2\times 8.85\times 10^{-12}}[/tex]
[tex]\sigma=3\times 10^6\times 2\times 8.85\times 10^{-12}[/tex]
[tex]\sigma=5.31\times 10^{-5}C/m^2[/tex]
[tex]\sigma=53.1\times 10^{-6}C/m^2=53.1\mu C/m^2[/tex]
[tex]1\mu C=10^{-6} C[/tex]
Consider the double slit experiment for light. Complete each statement as it would apply to Young's experiment (for each statement select "Increases", "Decreases", or "Cannot be Predicted"). If a variable is not mentioned, consider it to remain unchanged.Required:a. If the distance to the screen decreases, fringe separation:_______?b. If the frequency of the light used increases, fringe separation:_______?c. If the wavelength of the light used decreases, fringe separation:_______?d. For the fringe separation to remain unchanged, wavelength__________ while the distance to the screen decreases.e. If slit separation decreases, fringe separation :_______?f. If slit separation decreases and the distance to the screen decreases, fringe separation :_______?g. If the distance to the screen triples and slit separation doubles, fringe separation :_______?
Answer:
a) DECREASE , b) Decreases , c) DECREASE , d) the wavelength must increase , e) increasses,
Explanation:
Young's double-slit experience is explained for constructive interference by the expression
d sin θ = m λ
as in this case, the measured angles are very small,
tan θ = y / L
tan θ = sin θ / cos θ = sin θ
sin θ= y L
d y / L = m Lam
we can now examine the statements given
a) if the distance to the screen decreases
y = m λ / d L
if L decreases and decreases.
The answer is DECREASE
b) if the frequency increases
the wave speed is
c = λ f
λ = c / f
we substitute
y = (m / d l) c / f
in this case if if the frequency is increased the separation decreases
Decreases
c) If the wavelength decreases
separation decreases
DECREASE
d) if it is desired that the separation does not change while the separation to the Panamanian decreases the wavelength must increase
y = (m / d) lam / L
e) if the parcionero between the slits (d) decreases the separation increases
INCREASES
f) t he gap separation decreases and the distance to the screen decreases so well.
Pattern separation remains constant
If you slide down a rope, it's possible to create enough thermal energy to burn your hands or your legs where they grip the rope. Suppose a 30 kg child slides down a rope at a playground, descending 2.5 m at a constant speed.
How much thermal energy is created as she slides down the rope?
Answer:
Q = 735 J
Explanation:
In this exercise we must assume that all the mechanical energy of the system transforms into cemite energy.
Initial energy
Em₀ = U = m g h
final energy
[tex]Em_{f}[/tex] = Q
Em₀ = Em_{f}
m g h = Q
let's calculate
Q = 30 9.8 2.5
Q = 735 J
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and position. A motorcyclist travels around a curved path that has a radius of 250 ft . While traveling around the curved path, the motorcyclist increases speed by 1.05 ft/s2 . Part A - Finding the time interval for the motorcyclist to reach a given acceleration If the motorcyclist starts from rest, determine the time needed to reach an acceleration of 4.25ft/s2. Express your answer to three significant figures and include the appropriate units.
Answer:
a^2 = ar^2 + al^2 where ar is the radial acceleration and al is the
linear acceleration - since vectors ar and al are at right angles
ar^2 = a^2 - al^2 = 4.25^2 - 1.05^2
ar = 4.12 ft/s^2
ar = V^2 / R where ar is the radial acceleration
So V^2 = ar * R = 4.12 * 250 = 1030 ft^2/s^2
V = 32.1 m/s the linear speed of the cycle
Also, V = al t or t = V / al = 32.1 / 1.05 = 30.6 sec
A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s after this, the ball reaches the bottom of the incline. What was the magnitude of the ball's acceleration, assume it to be constant
Answer: The ball's acceleration is 2.35 m/s2
Explanation: Please see the attachment below
Answer:
The acceleration is [tex]a= 2.4 \ m/s^2[/tex]
Explanation:
From the question we are told that
The distance covered is [tex]d = 87 \ m[/tex]
The time taken is [tex]t = 8.6 \ s[/tex]
Time taken reach the bottom is [tex]t_b = 1 \ s[/tex]
According to the equation of motion
[tex]S = ut + \frac{1}{2} at^2[/tex]
since the ball started at rest u = 0 m/s
substituting values
[tex]87 = 0 + \frac{1}{2} * a * (8.6)^2[/tex]
=> [tex]a = \frac{2 * 87}{8.6^2}[/tex]
=> [tex]a= 2.4 \ m/s^2[/tex]
The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spring with a transverse wave showing the position of the mass on the spring will make a transverse wave shape if bouncing up and down and moving horizontally. The shortest spring has the mass at the top of each crest and are labeled A and E respectively. The position where the mass in in the trough of the transverse wave would be labeled C and stretches the spring the farthest. The middle length springs has masses where the equilibrium of the material of the medium containing the transverse wave would be at the equilibrium of the material would be and are labeled B and D. Which describes the system at point D? The velocity has the maximum upward value because the acceleration upward is at a maximum.
Answer:
D. "The net force is zero, so the acceleration is zero"
Explanation:
edge 2020
The system at point D is defined by "The net force is zero, so the acceleration is zero"
What is Net force?When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.
Given the image which shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spring with a transverse wave showing the position of the mass on the spring will make a transverse wave shape if bouncing up and down and moving horizontally.
The shortest spring has the mass at the top of each crest and are labeled A and E respectively. The position where the mass in in the trough of the transverse wave would be labeled C and stretches the spring the farthest. The middle length springs has masses where the equilibrium of the material of the medium containing the transverse wave would be at the equilibrium of the material would be and are labeled B and D.
At D, the net force becomes zero. This makes the system to be in equilibrium or moving with constant velocity. The acceleration becomes zero.
Thus, the system at point D is defined as "The net force is zero, so the acceleration is zero"
Learn more about net force.
https://brainly.com/question/18031889
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Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.
Answer:
Q1_new = 515.68 µC
Q2_new = 246.82 µC
Explanation:
Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.
Then it is only necessary to calculate the charge on each capacitor:
Q1 = 5.85 µF * 250 V = 1462.5 µC
Q2 = 2.8 µF * 250 V = 700 µC
Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.
When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:
1462.5 µC - 700 µC = 762.5 µC
Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.
This 762.5 µC will be divided proportionately between the two capacitors.
Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC
Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC
A 50-cm-long spring is suspended from the ceiling. A 410 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 16 cm before coming to rest at its lowest point. It then continues to oscillate vertically. Part A What is the spring constant
Answer:
25.125 N/m
Explanation:
extension on the spring e = 16 cm 0.16 m
mass of hung mass m = 410 g = 0.41 kg
equation for the relationship between force and extension is given by
F = ke
where k is the spring constant
F = force = mg
where m is the hung mass,
and g is acceleration due to gravity = 9.81 m/s^2
imputing value, we have
0.41 x 9.81 = k x 0.16 = 0.16k
4.02 = 0.16k
spring constant k = 4.02/0.16 = 25.125 N/m
A spool is on a horizontal surface (with friction) and pulled to the right with a thread attached to the center of the spool, so that the spool rolls without slipping. The spool has a mass of M, moment of inertia I, and a radius of R.
Requird:
a. What is the relationship between the magnitude of the force of tension and the force of friction?
b. What expressions relates the magnitudes of the translational acceleration of the spool's center of mass a and the spool's rotational acceleration about the center of mass a (recall the spool rolls without slipping on the surface).
Answer:
The correct answer is -
A) [tex]F_{f} =(\frac{r}{R} ) F_{T}[/tex]
B) [tex]\alpha =\frac{a}{R}[/tex]
Explanation:
As it is mention that the spool has mass M, radius R and moment of inertia I. In the first part of the question as the spool is not moving and r < R which means there is net torque = 0
so, [tex]F_{f}{R} = F_{T}{r}[/tex]
and [tex]F_{f} =(\frac{r}{R} ) F_{T}[/tex]
In the second part of the question from the given information, we can express the angular acceleration
= [tex]\alpha =\frac{a}{R}[/tex] ( alpha = angular acceleration and a = translational acceleration)
A) The relationship between the magnitude of the force of tension and the force of friction Ff = (r/R)Ft
B) α = a/R
What is Rotational Acceleration?When As it is mentioned that the spool has mass M, Then radius R, and also a moment of inertia I. Then In the foremost part of the query as the spool is not moving and r < R which means there is net torque = 0
so, Ff R = Ftr
and then Ff = (r/R)Ft
In the second part of the query from the given notification, we can express the angular acceleration
Therefore, = α = a/R ( alpha = angular acceleration and a = translational acceleration)
Find more information about Rotational Acceleration here:
https://brainly.com/question/14001220
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad/s2. Assume no slippage. m/s2 (b) How many revolutions do the tires make in 2.50 s if they start from rest
Answer:
a) The linear acceleration of the car is [tex]4.65\,\frac{m}{s^{2}}[/tex], b) The tires did 7.46 revolutions in 2.50 seconds from rest.
Explanation:
a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:
[tex]\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}[/tex]
Where:
[tex]a_{r}[/tex] - Magnitude of the radial acceleration, measured in meters per square second.
[tex]a_{t}[/tex] - Magnitude of the tangent acceleration, measured in meters per square second.
Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:
[tex]\| \vec a \| = a_{t}[/tex]
[tex]\| \vec a \| = r \cdot \alpha[/tex]
Where:
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]r[/tex] - Radius of rotation (Radius of a tire), measured in meters.
Given that [tex]\alpha = 15\,\frac{rad}{s^{2}}[/tex] and [tex]r = 0.31\,m[/tex]. The linear acceleration experimented by the car is:
[tex]\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)[/tex]
[tex]\| \vec a \| = 4.65\,\frac{m}{s^{2}}[/tex]
The linear acceleration of the car is [tex]4.65\,\frac{m}{s^{2}}[/tex].
b) Assuming that angular acceleration is constant, the following kinematic equation is used:
[tex]\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}[/tex]
Where:
[tex]\theta[/tex] - Final angular position, measured in radians.
[tex]\theta_{o}[/tex] - Initial angular position, measured in radians.
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]t[/tex] - Time, measured in seconds.
If [tex]\theta_{o} = 0\,rad[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\alpha = 15\,\frac{rad}{s^{2}}[/tex], the final angular position is:
[tex]\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}[/tex]
[tex]\theta = 46.875\,rad[/tex]
Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)
[tex]\theta = 7.46\,rev[/tex]
The tires did 7.46 revolutions in 2.50 seconds from rest.
A man claims that he can hold onto a 16.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man in a collision in which he is in one of two identical cars each traveling toward the other at 59.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.05 s.
Find the magnitude of the average force needed to hold onto the child.
N
Answer:
F = -8440.12 N
the magnitude of the average force needed to hold onto the child is 8440.12 N
Explanation:
Given;
Mass of child m = 16 kg
Speed of each car v = 59.0 mi/h = 26.37536 m/s
Time t = 0.05s
Applying the impulse momentum equation;
Impulse = change in momentum
Ft = ∆(mv)
F = ∆(mv)/t
F = m(∆v)/t
Where;
F = force
t = time
m = mass
v = velocity
Since the final speed of the car is zero(at rest) then;
∆v = 0 - v = -26.37536 m/s
Substituting the given values;
F = 16×-26.37536/0.05
F = -8440.1152 N
F = -8440.12 N
the magnitude of the average force needed to hold onto the child is 8440.12 N
A ball is thrown straight upward and falls back to Earth. Suppose a y-coordinate axis points upward, and the release point is the origin. Instantaneously at the top its flight, which of these quantities are zero
a. Displacment
b. Speed
c. Velocity
d. Accerlation
Explanation:
A ball is thrown straight upward and falls back to Earth. It means that it is coming to the initial position. Displacement is given by the difference of final position and initial position. The displacement of the ball will be 0. As a result velocity will be 0.
Acceleration is equal to the rate of change of velocity. So, its acceleration is also equal to 0.
Hence, displacement, velocity and acceleration are zero.
What is the length of the x-component of the vector shown below? A. 65.8 B. 90.6 C. 112 D. 33.2
Answer:
The correct answer is - option b. 90.6
Explanation:
The scalar x-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle
If you shine a light straight down onto that vector, then the length of its shadow on the x-axis is -
x-component = 112· cosine(36°)
x-component = 112 · (0.8090)
x-component = 90.60
Thus, The correct answer is - option b. 90.6
Which statement describes an essential characteristic of data in an experiment?
Answer:
the data must be reliable
Explanation:
what are the strengths and weaknesses of the four methods of waste management?
Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;
* It creates employment
* It keeps the environment clean
* The practice is highly lucrative
* It saves the earth and conserves energy
The weaknesses of the methods of waste management includes;
* The sites are often dangerous
* The process is mostly
* There is a need for global buy-in
* The resultant product had a short life
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107 N/m2). By what volume has 1.0 m3 of water from the surface of the lake been compressed if it is forced down to this depth? The bulk modulus of water is 2.3 × 109 Pa.
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
[tex]K = - V\cdot \frac{dP}{dV}[/tex]
Where:
[tex]K[/tex] - Bulk module, measured in pascals.
[tex]V[/tex] - Sample volume, measured in cubic meters.
[tex]P[/tex] - Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
[tex]-\frac{K \,dV}{V} = dP[/tex]
This resultant expression is solved by definite integration and algebraic handling:
[tex]-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP[/tex]
[tex]-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}[/tex]
[tex]\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}[/tex]
[tex]\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }[/tex]
The final volume is predicted by:
[tex]V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }[/tex]
If [tex]V_{o} = 1\,m^{3}[/tex], [tex]P_{o} - P_{f} = -10132500\,Pa[/tex] and [tex]K = 2.3\times 10^{9}\,Pa[/tex], then:
[tex]V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }[/tex]
[tex]V_{f} \approx 0.996\,m^{3}[/tex]
Change in volume due to increasure on pressure is:
[tex]\Delta V = V_{o} - V_{f}[/tex]
[tex]\Delta V = 1\,m^{3} - 0.996\,m^{3}[/tex]
[tex]\Delta V = 0.004\,m^{3}[/tex]
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Identify the following as combination, decomposition, replacement, or ion exchange reactions NaBr(aq) + Cl2(g) → 2 NaCl(aq) + Br2(g)
Answer:
Replacement
Explanation:
in replacements, like ions replace like. in this equation, we can see that Bromine replaced Chlorine. so, the answer is replacement.
Answer:
Single-replacement or replacement
Explanation:
The single-replacement reaction is a + bc -> ac + b, compare them.
NaBr + Cl2 -> 2 NACl + Br.
AB + C -> AC + B
As you can see they are the same ( even though the b is with the a and not with the c. The formula can be switched around a little with the order of b and c ) ((also like ions replace like ions in replacements, which they are in this))