Answer:
a) emf = 0.01804 V
b) I = 0.03 A
Explanation:
a) The emf is calculated by using the following formula:
[tex]|emf|=|\frac{d\Phi_B}{dt}|=|\frac{d(A\cdot B)}{dt}|[/tex] [tex]=A|\frac{dB}{dt}|[/tex]
A: area of the loop = 0.0820m^2
B: magnitude of the magnetic field
dB/dt: change of the magnetic field, in time: 0.220 T/s
Where ФB is the magnetic flux, the surface vector and magnetic vector are perpendicular between them, and the area A is constant.
You replace the values of A and dB/dt in the equation (1):
[tex]|emf|=(0.082m^2)(0.220T/s)=0.01804V[/tex]
b) The current in the loop is:
[tex]I=\frac{emf}{R}[/tex]
R: resistance of the loop = 0.600Ω
[tex]I=\frac{0.01804V}{0.600\Omega}=0.03A=30mA[/tex]
a. The emf induced in this loop is 18.04mV.
b. The current induced in the loop is 30.06mA.
a. We know that,
[tex]flux(\phi)=B*A[/tex]
Where B is magnetic field and A is the area.
[tex]emf=\frac{d\phi}{dt}=A*\frac{dB}{dt}[/tex]
Given that, Area , [tex]A=0.0820m^{2},B=3.80T,\frac{dB}{dt}=0.220T/s[/tex]
Substituting all values in above equation.
[tex]emf=0.0820*0.220=0.01804V=18.04mV[/tex]
b. Resistance, [tex]R=0.600ohm[/tex]
Current induced in the loop is,
[tex]I=\frac{emf}{R}=18.04/0.6=30.06mA[/tex]
Hence, the emf induced in this loop is 18.04mV.
The current induced in the loop is 30.06mA.
Learn more:
https://brainly.com/question/15477464
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track and the train reaches it 76s later. The acceleration is -0.26s^2. Find its speed at the next signal.
Answer:
3.2 m/s
Explanation:
Given:
Δx = 1000 m
v₀ = 23 m/s
a = -0.26 m/s²
t = 76 s
Find: v
This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:
v = at + v₀
v = (-0.26 m/s²) (76 s) + (23 m/s)
v = 3.2 m/s
Or:
Δx = ½ (v + v₀) t
(1000 m) = ½ (v + 23 m/s) (76 s)
v = 3.3 m/s
Or:
v² = v₀² + 2aΔx
v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)
v = 3.0 m/s
Or:
Δx = vt − ½ at²
(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²
v = 3.3 m/s
As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.
uring a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? 26.7 N 16.7 N 13.3 N 107 N 40.0 N
Answer:
107 N, option d
Explanation:
Given that
mass of the ball, m = 0.2 kg
initial velocity of the ball, u = 20 m/s
final velocity of the ball, v = -12 m/s
time taken, Δt = 60 ms
Solving this question makes us remember "Impulse Theorem"
It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"
Mathematically, it is represented as
FΔt = m(v - u), where
F = the average force
Δt = time taken
m = mass of the ball
v = final velocity of the ball
u = initial velocity of the ball
From the question we were given, if we substitute the values in it, we have
F = ?
Δt = 60 ms = 0.06s
m = 0.2 kg
v = -12 m/s
u = 20 m/s
F = 0.2(-12 - 20) / 0.06
F = (0.2 * -32) / 0.06
F = -6.4 / 0.06
F = -106.7 N
Thus, the magnitude is 107 N
N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2 and third one is C/4, forth one is C/8 and so on. Namely, capacitance of a capacitor is one-half of the previous one. What is the equivalent capacitance of this parallel combination when N goes to inifinity?
Answer:
2C
Explanation:
The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.
So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.
Using the formula for the sum of the infinite terms of a geometric series, we have:
Sum = First term / (1 - rate)
Sum = C / (1 - 0.5)
Sum = C / 0.5 = 2C
So the equivalent capacitance of this parallel connection is 2C.
A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale will read:___________.
A) more than 129 lb
B) 129 lb
C) less than 129 lb
D) It is impossible to answer this question without knowing the acceleration of the elevator.
Answer:
C) less than 129 lb.
Explanation:
Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards with magnitude a .
If R be the reaction force
For the elevator is going downwards with acceleration a
mg - R = ma
R = mg - ma
R measures its apparent weight . Spring scale will measure his apparent weight.
So its apparent weight is less than 129 lb .
A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 40.4 Hz, determine the first three overtones. Use 343 m/s as the speed of sound in air.
If the speed of sound is 337 m/s, determine the length of an open tube (open at both ends) that has a fundamental frequency of 233 Hz and a first overtone frequency of 466 Hz.
Answer:
Explanation:
fundamental frequency at closed pipe = 40.4 Hz
overtones are odd harmonics in closed pipe
first three overtones are
3 x 40.4 , 5 x 40.4 , 7 x 40.4 Hz
= 121.2 Hz , 202 Hz , 282.8 Hz .
speed of sound given is 337 , fundamental frequency is 233 Hz
wavelength = velocity of sound / frequency
= 337 / 233
= 1.446 m
for fundamental note in open pipe
wavelength /2 = length of tube
length of tube = 1.446 / 2
= .723 m
= 72.30 cm .
first overtone will be two times the fundamental ie 466. In open pipe all the harmonics are found , ie both odd and even .
someone please help me with this thanks
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R. Assume that the initial height h is great enough so that the car never losses contact with the track.
Required:
a. Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m, g, h, and R.
b. Find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.
Answer:
Explanation:
At height h , potential energy of coaster car having mass m = mgh .
The car will lose potential energy and gain kinetic energy.
height lost by car when it is at the top of loop of radius R
= h - 2R
potential energy lost = mg ( h - 2R )
kinetic energy gained = mg ( h - 2R )
kinetic energy = 0 + mg ( h - 2R )
= mg ( h - 2R )
b )
For the car to remain in contact with the track , if v be the minimum velocity
centripetal force at top = mg
m v² / R = mg
v² = gR
kinetic energy = 1/2 mv²
= 1/2 m x gR
= mgR /
If h be the minimum height that can give this kinetic energy
mg ( h - 2R ) = mgR / 2
h - 2R = R / 2
h = 2.5 R .
Which symbol in a chemical equation separates the reactants from the products?
Answer:
the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔
Explanation:
i hope this will help you
Concerned with citizen complaints of price gouging during past hurricanes, Florida's state government passes a law setting a price ceiling for a bottle of water equal to the market equilibrium price during normal times. After all, it seems unfair that sellers of water gain because of a hurricane.
Answer:idk
Explanation:idk
Answer:
shortage of 50 water bottles
$2
30
Explanation:
A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train then moves at a constant velocity of 42 m/s for 420 s. The train then slows down uniformly at 0.065 m/s^2, until it is brought to a halt. What is the acceleration during the first 5.6 km of travel?
Answer:
0.1575 m/s^2
Explanation:
Solution:-
- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).
- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.
- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).
- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:
[tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]
- We will plug in the given parameters in the equation of motion given above:
[tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]
Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2
g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the dragster at the end of 2.1 s, (b) the final velocity of the dragster at the end of of twice this time, or 4.2 s, (c) the displacement of the dragster at the end of 2.1 s, and (d) the displacement of the dragster at the end of twice this time, or 4.2 s?
The dragster's velocity v at time t with constant acceleration a is
[tex]v=at[/tex]
since it starts at rest.
After 2.1 s, it will attain a velocity of
[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]
or 92.4 m/s.
Doubling the time would double the final velocity,
[tex]v=a(2t)=2at[/tex]
so the velocity would be twice the previous one, 184.8 m/s.
The dragster undergoes a displacement x after time t with acceleration a of
[tex]x=\dfrac12at^2[/tex]
if we take the starting line to be the origin.
After 2.1 s, it will have moved
[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]
or 88 m.
Doubling the time has the effect of quadrupling the displacement, since
[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]
so after 4.2 s it will have moved 352 m.
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A particle with charge q is placed at the center of the cavity. The net charge on the inner surface of the conducting shell is
Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
A factory worker pushes a 30.0 kg crate a distance of 3.7 m along a level floor at constant velocity by pushing downward at an angle of 30∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
a) [tex]F = 210.803\,N[/tex], b) [tex]W_{F} = 779.971\,J[/tex], c) [tex]W_{f} = 235.683\,J[/tex], d) [tex]W_{N} = 0\,J[/tex]; [tex]W_{g} = 544.289\,J[/tex], e) [tex]W_{net} = 0\,J[/tex]
Explanation:
a) The net force exerted on the crate is:
[tex]\Sigma F = F - m\cdot g \cdot \sin \theta - \mu_{k}\cdot m\cdot g \cdot \cos \theta = 0[/tex]
The magnitud of the force that the work must apply to the crate is:
[tex]F = m\cdot g \cdot \sin \theta + \mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]
[tex]F = (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 30^{\circ} + 0.25 \cdot (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]
[tex]F = 210.803\,N[/tex]
b) The work done on the crate due to the external force is:
[tex]W_{F} = (210.803\,N)\cdot (3.7\,m)[/tex]
[tex]W_{F} = 779.971\,J[/tex]
c) The work done on the crate due to the external force is:
[tex]W_{f} = (63.698\,N)\cdot (3.7\,m)[/tex]
[tex]W_{f} = 235.683\,J[/tex]
d) The work done on the crate due the normal force is zero, since such force is perpendicular to the motion direction.
[tex]W_{N} = 0\,J[/tex]
And, the work done by gravity is:
[tex]W_{g} = (147.105\,N)\cdot (3.7\,m)[/tex]
[tex]W_{g} = 544.289\,J[/tex]
e) Lastly, the total work done is:
[tex]W_{net} = W_{F} - W_{f} - W_{g} - W_{N}[/tex]
[tex]W_{net} = 779.971\,J - 235.683\,J - 0\,N - 544.289\,J[/tex]
[tex]W_{net} = 0\,J[/tex]
EXPLANATION ⛔
A 20 gram mass is suspended from meter rod at 20cm. The meter rod is balanced on 40cm mark. Weight of meter rod is
A. 0.4N
B. 0.6N
C. 6N
D. 60N
Answer:b
Explanation:I’m just trynna get more money dude
pls what is the difference between Ac power and dc power
Answer:
The difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction, or "forward." In AC, electrons keep switching directions, sometimes going "forward" and then going "backward."
What caused the disappearance of land bridges?
A. Volcanic outgassing
B. Shrinking of the polar ice caps
C. Beginning of an ice age
D. A mass extinction
Answer: B
Explanation:
I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges
Answer:B :)
Explanation:
A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)
Answer:
The angular momentum is [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The mass of the woman is [tex]m = 50 \ kg[/tex]
The angular speed of the rim is [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]
The mass of the disk is [tex]m_d = 110 \ kg[/tex]
The radius of the disk is [tex]r_d = 4.0 \ m[/tex]
The moment of inertia of the disk is mathematically represented as
[tex]I_D = \frac{1}{2} m_d r^2_d[/tex]
substituting values
[tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]
[tex]I_D = 880 \ kg \cdot m^2[/tex]
The moment of inertia of the woman is
[tex]I_w = m * r_d^2[/tex]
substituting values
[tex]I_w = 50 * 4^2[/tex]
[tex]I_w =800\ kg[/tex]
The moment of inertia of the system (the woman + the large disk ) is
[tex]I_t = I_w + I_D[/tex]
substituting values
[tex]I_t = 880 +800[/tex]
[tex]I_t =1680 \ kg \cdot m^2[/tex]
The angular momentum of the system is
[tex]L = I_t w[/tex]
substituting values
[tex]L = 1680 * 5.024[/tex]
[tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the
runner's kinetic energy will become:
(a) four times as great as it was.
(b) half the magnitude it was.
(c) energy is conserved.
(d) twice as great as it was.
(e) four times less than it was.
Answer:
It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). If a positron is accelerated by a constant electric field of magnitude 286 N/C, find the following.
(a) Find the acceleration of the positron. m/s2
(b) Find the positron's speed after 8.70 × 10-9 s. Assume that the positron started from rest. m/s
Answer:
a) a = 5.03x10¹³ m/s²
b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]
Explanation:
a) The acceleration of the positron can be found as follows:
[tex] F = q*E [/tex] (1)
Also,
[tex] F = ma [/tex] (2)
By entering equation (1) into (2), we have:
[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]
Where:
F: is the electric force
m: is the particle's mass = 9.1x10⁻³¹ kg
q: is the charge of the positron = 1.6x10⁻¹⁹ C
E: is the electric field = 286 N/C
[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]
b) The positron's speed can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed =?
[tex]V_{0}[/tex]: is the initial speed =0
t: is the time = 8.70x10⁻⁹ s
[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]
I hope it helps you!
How many significant figures does 0.09164500561 have?
Answer:
10 Sig Figs
Explanation:
Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.
The Great Lakes are all part of what? The Mississippi River The St. Lawrence Seaway A large body of salt lakes The Missouri River
Answer:
St Lawrence Sea way
Explanation:
The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.
You drive in straight line at 20 m/s for 10 miles, then at 30m/s for an other 10 miles what is your average speed
Answer:
25 m/s
Explanation:
Data provided in the question
20 m/s for 10 minutes
And, the 30 m/s for another 10 minutes
Based on the above information, the average speed is
As we know that
[tex]Average\ speed = \frac{Total\ distance}{Total\ time}[/tex]
[tex]= \frac{20\times10\times60 + 30\times10\times60 }{20\times60}[/tex]
= 25 m/s
1 hour = 60 minutes
1 minute = 60 seconds
Hence, the average speed is 25 m/s
In the question, there are miles is given but instead of this we use the minutes as we have to find out the average speed and time should not be in miles it should be in minutes, hour or seconds
Therefore we considered the same
Although these quantities vary from one type of cell to another, a cell can be 2.2 micrometers in diameter with a cell wall 40 nm thick. If the density (mass divided by volume) of the wall material is the same as that of pure water, what is the mass (in mg) of the cell wall, assuming the cell to be spherical and the wall to be a very thin spherical shell?
Answer:
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
Explanation:
First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:
A = 4πr²
where,
A = Surface Area = ?
r = Radius of Cell = Diameter/2 = 2.2 μm/2 = 1.1 μm = 1.1 x 10⁻⁶ m
Therefore,
A = 4π(1.1 x 10⁻⁶ m)²
A = 15.2 x 10⁻¹² m²
Now, we find the volume of the cell wall. For that purpose, we use formula:
V = At
where,
V = Volume of the Cell Wall = ?
t = Thickness of Wall = 40 nm = 4 x 10⁻⁸ m
Therefore,
V = (15.2 x 10⁻¹² m²)(4 x 10⁻⁸ m)
V = 60.82 x 10⁻²⁰ m³
Now, to find mass of cell wall, we use formula:
ρ = m/V
m = ρV
where,
ρ = density of water = 1000 kg/m³
m = Mass of Wall = ?
Therefore,
m = (1000 kg/m³)(60.82 x 10⁻²⁰ m³)
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
The mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
Since we assume the cell to be spherical and the wall to be a thin spherical shell, the volume of the cell wall V = At where
A = surface area of cell = 4πR² where R = radius of cell = 2.2 μm/2 = 1.1 × 10⁻⁶ m and t = thickness of cell wall = 40 nm = 40 × 10⁻⁹ m.Volume of cell wallSo, V = 4πR²t
Substituting the values of the variables into the equation, we have
V = 4πR²t
V = 4π(1.1 × 10⁻⁶ m)² × 40 × 10⁻⁹ m.
V = 4π(1.21 × 10⁻¹² m²) × 40 × 10⁻⁹ m.
V = 193.6π × 10⁻²¹ m³
V = 608.21 × 10⁻²¹ m³
V = 6.0821 × 10⁻¹⁹ m³
V ≅ 6.082 × 10⁻¹⁹ m³
Mass of the cell wallWe know that density of cell wall, ρ = m/v where m = mass of cell wall and V = volume of cell wall.
Making m subject of the formula, we have
m = ρV
Since we assume the density of the cell wall to be equal to that of pure water, ρ = 1000 kg/m³
So, m = ρV
m = 1000 kg/m³ × 6.082 × 10⁻¹⁹ m³
m = 6.082 × 10⁻¹⁶ kg
Converting to mg, we have
m = 6.082 × 10⁻¹⁶ kg × 10⁶ mg/kg
m = 6.082 × 10⁻¹⁰ mg
So, the mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
Learn more about mass of cell wall here:
https://brainly.com/question/13173768
When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³
so here P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex] [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]
solve it we get
P2 = 87.9 lb/ft³
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temperature of 322 K. Over the course of one hour, the engine absorbs 1.37 x 105 J from the hot reservoir and exhausts 7.4 x 104 J into the cold reservoir.
1) What is the power output of this engine?
2) What is the maximum (Carnot) efficiency of a heat engine running between these two reservoirs?
3) What is the actual efficiency of this engine?
Answer:
The power output of this engine is [tex]P = 17.5 W[/tex]
The the maximum (Carnot) efficiency is [tex]\eta_c = 0.7424[/tex]
The actual efficiency of this engine is [tex]\eta _a = 0.46[/tex]
Explanation:
From the question we are told that
The temperature of the hot reservoir is [tex]T_h = 1250 \ K[/tex]
The temperature of the cold reservoir is [tex]T_c = 322 \ K[/tex]
The energy absorbed from the hot reservoir is [tex]E_h = 1.37 *10^{5} \ J[/tex]
The energy exhausts into cold reservoir is [tex]E_c = 7.4 *10^{4} J[/tex]
The power output is mathematically represented as
[tex]P = \frac{W}{t}[/tex]
Where t is the time taken which we will assume to be 1 hour = 3600 s
W is the workdone which is mathematically represented as
[tex]W = E_h -E_c[/tex]
substituting values
[tex]W = 63000 J[/tex]
So
[tex]P = \frac{63000}{3600}[/tex]
[tex]P = 17.5 W[/tex]
The Carnot efficiency is mathematically represented as
[tex]\eta_c = 1 - \frac{T_c}{T_h}[/tex]
[tex]\eta_c = 1 - \frac{322}{1250}[/tex]
[tex]\eta_c = 0.7424[/tex]
The actual efficiency is mathematically represented as
[tex]\eta _a = \frac{W}{E_h}[/tex]
substituting values
[tex]\eta _a = \frac{63000}{1.37*10^{5}}[/tex]
[tex]\eta _a = 0.46[/tex]
When an electromagnetic wave falls on a white, perfectly reflectingsurface, it exerts a force F on that surface. If the surfaceis now painted a perfectly absorbing black, the force that the samewave would exert on the surface is:___________.
A) F
B) F/2
C) F/4
D) 2F
E) 4F
Answer:
B. F/2
Explanation:
The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave
i.e Prad = u
For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.
i.e Prad = 2u.
Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2
A skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stopA skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stop
Before the bus starts moving, the bus and the skater are both standing still.
When the bus starts moving and pulls away from the bus-stop, the skater stays right where she is.
The people outside on the sidewalk see her standing still, and they see the bus moving out from under her.
The other passengers on the bus see her rolling backwards down the aisle, toward the back of the bus.
A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but Uk = 0. Suddenly, a wind from the northeast exerts a force of 3.70 N on the skater.a) Use work and energy to find the skater's speed after gliding 100 m in this wind.b) What is the minimum value of Ug that allows her to continue moving straight north?
Answer:
a. 2.668 m/s
b. 0.00494
Explanation:
The computation is shown below:
a. As we know that
[tex]W = F\times d[/tex]
[tex]KE = 0.5\times m\times v^2[/tex]
As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.
F = 3.70 cos 45 = 2.62 N
[tex]W = F \times d = 2.62 N \times 100 m[/tex]
[tex]W = 261.6 N\times m[/tex]
We know that
KE1 = Initial kinetic energy
KE2 = kinetic energy following 100 m
The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.
So, the equation is
KE2 = KE1 - W
[tex]0.5 m\times v2^2 = 0.5 m\ v1^2 - W[/tex]
Now solve for v2
[tex]v2 = \sqrt{v1^2 - {\frac{2W}{M}}}[/tex]
[tex]= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}[/tex]
= 2.668 m/s
b. Now the minimum value of Ug is
As we know that
Ff = force of friction
Us = coefficient of static friction
N = Normal force = weight of skater
So,
[tex]Ff = Us\times N[/tex]
Now solve for Us
[tex]= \frac{Ff}{N}[/tex]
[tex]= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}[/tex]
= 0.00494
EASY! WILL GIVE BRAINLIEST!
Find the conductivity of a conduit with a cross-sectional area of 0.50 cm2 and a length of 15 cm if its conductance G is 0.050 ohm-1.
σ = _____ ohm-1cm-l
3
75
1.5
0.0017
the answer is 1.5 hope this helps
Answer:
1.5
Explanation:
0.5=σ/(15/0.5)
σ=3/2 or 1.5
