Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
[tex]F_{e}[/tex] - W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} = [tex]K_{e}[/tex] + U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
A generator consists of 1000 turns of wire around a coil of area 10cm2 (0.0010m 2 ). Calculate the peak output voltage of this generator when operating at 100 cycles/second in a 0.50 tesla field.
Answer:
The peak output voltage of this generator is 314.2 V.
Explanation:
Given;
the number of turns of the coil, N = 1000 turns
area of the coil, A = 0.001 m²
angular frequency of the coil, f = 100 cycles/seconds
magnitude of the magnetic field, B = 0.5 T
The peak output voltage of this generator is given by;
E = NBAω
Where;
ω is the angular velocity = 2πf
E = NBA(2πf)
E = 1000 x 0.5 x 0.001(2 x π x 100)
E = 314.2 V
Therefore, the peak output voltage of this generator is 314.2 V.
A force of pounds makes an angle of with a second force. The resultant of the two forces makes an angle of to the first force. Find the magnitudes of the second force and of the resultant.
This question is incomplete, the complete question is;
A force of 193 pounds makes an angle of 79°14' with a second force. The resultant of the two forces makes an angle of 27°0' to the first force. Find the magnitudes of the second force and of the resultant.
Answer:
magnitudes of the second force (vector) is 110.84 lb
the resultant force force has a magnitude of 239.85 lb
Explanation:
Given that;
Magnitude of resultant vector R = ?
Direction of resultant vector α = 27°0'
Magnitude of vector p = 193
Magnitude of vector Q = ?
Angle between two vectors ∅ = 79°14'
Using the formula
tan∝ = [ Qsin∅ / P + Qcos∅]
tan27°0' = [ Qsin79°14' / 193 + Qcos79°14' )
we cross multiply
(193tan27°0') + (Qcos79°14'tan27°0' ) = Qsin79°14'
Q = 193tan27°0' / (sin79°14' - cos79°14'tan27°0')
Q = 110.84 lb
Therefore magnitudes of the second force (vector) is 110.84 lb
Now
R = √( p² + Q² + 2PQcos∅ )
R = √( 193² + 110.84² + ( 2 × 193 × 110.84 × cos79°14'))
R = 239.85 lb
Therefore the resultant force force has a magnitude of 239.85 lb
g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before interacting. From this information, find the time interval required for the strong interaction to occur.
Answer:
Time, [tex]t=1.07\times 10^{-23}\ s[/tex]
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]
So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]
Which statement about friction is true? (1 point)
o
Static friction and kinetic friction in a system always act in opposite directions of each other and in the same direction as the
applied force
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the
applied force
Static friction and kinetic friction in a system always act in opposite directions of each other and in the opposite direction of the
applied force
O
Static friction and kinetic friction in a system always act in the same direction as each other and in the same direction as the
applied force.
Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer
Explanation:
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.
What is friction?Friction is the force that prevents one hard material from scooting or rolling over the other.
Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.
We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.
Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.
In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.
Thus, the correct option is B.
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If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y direction, the wave is traveling in what direction? Explain your answer.
Answer:
Explanation:
The direction of propagation of electromagnetic wave
is given by the direction of vector E x B where E is electrical field , B is magnetic field .
Given Electric field = E i because it is along x axis
Magnetic field = Bj because it is along y axis
E x B = Ei x Bj
= EB k .
so direction of E x B is along k direction or z - axis so wave is propagating along z - axis .
The direction of motion of electromagnetic wave will be +z-direction.
Electromagnetic waves are waves that consist of the electric field and magnetic field.
The electric and magnetic fields are perpendicular to each other and the wave propagates in the direction perpendicular to both the fields.
Now, the direction of wave motion can be estimated by taking the cross-product of directional unit vectors of the electric and magnetic fields.
The electric field is in the +y direction and the magnetic field is in the +x-direction.
So, the direction of the wave will be,
[tex]i\times j=k[/tex]
Therefore, the direction of motion of electromagnetic wave will be +z-direction.
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An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.440 mm.
1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
2. What would be the angular position of the second-order, two-slit, interference maxima in this case?
3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?
4. What is the intensity at the angular position of θ2?
Answer:
1) θ = 0.00118 rad, 2) θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4) I / Io = 0.216
Explanation:
In the double-slit interference phenomenon it is explained for constructive interference by the equation
d sin θ = m λ
1) the first order maximum occurs for m = 1
sin θ = λ / d
θ = sin⁻¹ λ / d
let's reduce the magnitudes to the SI system
λ = 520 nm = 520 10⁻⁹ θ = 0.00118 radm
d = 0.440 mm = 0.440 10⁻³ m ³
let's calculate
θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)
θ = sin⁻¹ (1.18 10⁻³)
θ = 0.00118 rad
2) the second order maximum occurs for m = 2
θ = sin⁻¹ (m λ / d)
θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)
θ = 0.00236 rad
3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains
I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )
where the function sinc = sin x / x
and b is the width of the slits
we caption the values
x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)
x = 2.21
I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²
remember angles are in radians
I / I₀ = cos² (3.0945) [0.363] 2
I / I₀ = 0.9978 0.1318
I / I₀ = 0.1738
4) the maximum second intensity is
I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)
x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)
x = 4.41
I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²
I / Io = cos² 6.273 0.216
I / Io = 0.216
.
a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?
Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .
Explanation:
The average speed of the car is 31,680 meters per hour.
To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.
Given:
Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours
Distance traveled (d) = 5,280 meters
Average Speed (v) = Distance (d) / Time (t)
Average Speed (v) = 5280 meters / (1/6) hours
To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:
Average Speed (v) = 5280 meters × (6/1) hours
Average Speed (v) = 31,680 meters per hour
Hence, the average speed of the car is 31,680 meters per hour.
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The magnetic field at the center of a 1 cm diameter loop is 2.5 mT. If a long straight wire carries the same current as the loop of wire, at what distance from the wire is the magnetic field 2.5 mT? A. 0.10 m B. 1.6x10-3 m C. 0.01 m D. 20 m
Answer:
B. 1.6 x 10⁻³ m
Explanation:
The magnetic field at the center of the loop is given by;
[tex]B = \frac{\mu_o I }{2R}[/tex]
Where;
μ₀ is the permeability of free space
I is the current in the loop
R is the radius of the circular loop
B is the magnetic field
Given;
diameter of the loop = 1cm
radius of the loop, r = 0.5 cm = 0.005 m
magnetic field, B = 2.5mT = 2.5 x 10⁻³ T
The current in the loop is calculated as;
[tex]I = \frac{2BR}{\mu_o} \\\\I = \frac{2*2.5*10^{-3}*0.005}{4\pi*10^{-7}} \\\\I = 19.89 \ A[/tex]
The magnetic at a distance from the long straight wire is calculated as;
[tex]B = \frac{\mu_o I}{2\pi d}[/tex]
where;
d is the distance from the wire;
[tex]d = \frac{\mu_o I}{2\pi B} \\\\d = \frac{4\pi *10^{-7} * 19.89}{2\pi *2.5*10^{-3}} \\\\d = 1.6 *10^{-3} \ m[/tex]
Therefore, the distance from the wire where the magnetic field is 2.5 mT is 1.6 x 10⁻³ m.
B. 1.6 x 10⁻³ m
This question involves the concepts of the magnetic field due to a loop and a current-carrying wire and current.
A long straight wire carrying the same current as the loop of wire has a magnetic field of 2.5 mT at a distance of b "B. 1.5 x 10⁻³ m".
The magnetic field at the center of a loop of wire is given by the following formula:
[tex]B=\frac{\mu_o I}{2r}[/tex]
where,
B = Magnetic Field = 2.5 mT = 2.5 x 10⁻³ T
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = ?
r = radius = diameter/2 = 1 cm/2 = 0.5 cm = 0.005 m
Therefore,
[tex]I = \frac{(2.5\ x\ 10^{-3}\ T)(2)(0.005\ m)}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]
I = 19.9 A
Now, the magnetic field at a distance from the straight wire is given by the following formula:
[tex]B=\frac{\mu_o I}{2\pi R}[/tex]
where,
R = distance from wire = ?
Therefore,
[tex]R = \frac{(4\pi \ x \ 10^{-7}\ N/A^2)(19.9\ A)}{2\pi(2.5\ x\ 10^{-3}\ T)}[/tex]
R = 1.6 x 10⁻³ m
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A diffraction grating with 200 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At what angle from the beam axis will the first order peak occur if the tube emits light with wavelength of 617.3 nm
Answer
123.5 x 10 ^-3 radian
Explanation:
Given the Width of slit a = 1 x 10⁻³ / 200
a = 5x 10⁻⁶ m .
angle at which first order peak is formed
= λ / a (where λ is wavelength and a is width of slit)
given λ = 617.3 x 10⁻⁹ m
a = 5 x 10⁻⁶
θ = 617.3 x 10⁻⁹ / 5 x 10⁻⁶
= 123.5x 10⁻³ radian .
first order peak is formed at an angle of 123.5 x 10⁻³ radian .
Explanation:
If an electromagnetic wave has components Ey = E0 sin(kx - ωt) and Bz = B0 sin(kx - ωt), in what direction is it traveling?
Answer:
Its traveling in the +x direction
Explanation:
The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.
The electromagnetic wave is travelling in the +x direction.
Electromagnetic waves are waves formed as a result of vibrations between an electric field and a magnetic field.
Given that:
Ey = E0 sin(kx - ωt)
Hence the electric field is moving in the +y direction.
Also, Bz = B0 sin(kx - ωt)
Hence the magnetic field is moving in the +z direction
Electric fields and magnetic fields (E and B) in an EM wave are perpendicular to each other and are also perpendicular to the direction of propagation of the wave.
Therefore the direction of the wave is travelling in the +x direction.
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A defibrillator is a device used to shock the heart back to normal beat patterns. To do this, it discharges a 15 μF capacitor through paddles placed on the skin, causing charge to flow through the heart. Assume that the capacitor is originally charged with 5.0 kV .Part AWhat is the charge initially stored on the capacitor?3×10−9 C7.5×104 C7.5×10−2 C7.5×10−5 CPart BWhat is the energy stored on the capacitor?What is the energy stored on the capacitor?1.9×108 J380 J190 J1.9×10−4 JPart CIf the resistance between the two paddles is 100 Ω when the paddles are placed on the skin of the patient, how much current ideally flows through the patient when the capacitor starts to discharge?5×105 A50 A2×10−2 A5×10−2 APart DIf a defibrillator passes 17 A of current through a person in 90 μs . During this time, how much charge moves through the patient?If a defibrillator passes 17 {\rm A} of current through a person in 90 {\rm \mu s} . During this time, how much charge moves through the patient?190 mC1.5 C1.5 mC17 C
Answer:
a) q = 7.5 10⁻² C , b) 190 J , c) I₀ = 50 A , d) 1.5 mC
Explanation:
The expression for capacitance is
C = q / DV
q = C DV
let's reduce the magnitudes to the SI system
ΔV = 5 kV = 5000 V
C = 15 μF = 15 10⁻⁶ F
t = 90 μs = 90 10⁻⁶ s
q = 15 10⁻⁶ 5000
q = 7.5 10⁻² C
b) the energy in a capacitor is
U = ½ C ΔV²
U = ½ 15 10⁻⁶ 5000²
U = 1,875 10² J
answer 190 J
c) At the moment the discharge begins, all the current is available and it decreases with time,
whereby
V = I R
in the first instant I = Io
I₀ = V / R
I₀ = 5000/100
I₀ = 50 A
but this is for a very short time
answer 50 A
d) The definition of current is
i = dq / dt
in this case they give us the total current and the total time, so we can find the total charge
i = q / t
q = i t
q = 17 90 10⁻⁶
q = 1.53 10⁻³ C
answer is 1.5 mC
PLEASE HELP FAST WILL GIVE BRAINLIEST The sentence, "The popcorn kernels popped twice as fast as the last batch," is a(n) _____. 1.experiment 2.hypothesis 3.observation 4.control
The answer is 3. Observation
Explanation:
The sentence "The popcorn kernels popped twice as fast as the last batch" is the result of observing or measuring the time popcorn kernels require to pop. In this context, the sentence best matches the word "observation" which the term used in the Scientific method to refer to statements that are the result of studying a phenomenon, either through the senses such as sight or through precise instruments that allow scientists to understand numerically variables such as time, speed, temperature, etc.
Just wondering if I did this right
Yeah
All they are all correct
?a wire is stretched 30% what is the percentage change in resistance
Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]
So that;
[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])
= (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])
= [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
[tex]R_{o}[/tex] = 1.69R
Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.
The change in resistance of the wire = [tex]R_{o}[/tex] - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
Consult Interactive Solution 27.18 to review a model for solving this problem. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 653 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference
Answer:
Explanation:
In the given case for destructive interference , the condition is,
path difference = (2n+1)λ /2 where n is an integer and λ is wavelength
2 μ d = (2n+1)λ /2
Putting λ = 653 nm
for minimum thickness n = 0
2 μ d = 653 / 2 nm
= 326.5 nm
For constructive interference the condition is
2 μ d = n λ₁
326.5 nm = n λ₁
λ₁ = 326.5 / n
For n = 1
λ₁ = 326.5 nm ,
or , 326.5nm .
Longest wavelength possible is 326.5
What is an understood decimal
If the solenoid is 45.0 cm long and each winding has a radius of 8.0 cm , how many windings are in the solenoid
Answer:
The number of windings is 1.
Explanation:
The radius of the solenoid = 8.0 cm = 0.08 m
Length of the solenoid = 45.0 cm = 0.45 m
number of turn = ?
circumference of each winding = 2πr = 2 x 3.142 x 0.08 = 0.503 m
The number of windings = (Length of the solenoid)/(circumference of each winding)
==> 0.45/0.503 = 0.89 ≅ 1
How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m
Answer:
Explanation:
volume of water being lifted
= π r² h , where r is radius of cylinder and h is height of cylinder
= 3.14 x5² x 10
= 785 m³
mass of water = 785 x 10³ kg
mass of this much of water is lifted so that its centre of mass is lifted by height
10 / 2 = 5m .
So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity
= 785 x 10³ x 9.8 x 5
= 38.465 x 10⁶ J
(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
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A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?Please help i will give brainliest!!!! lower the board so it's level with the ground ____ roughen up the texture of the wooden board ___ raise the board to a higher angle ____ press down on the brick in a direction that is perpendicular to the board _____
Answer:
to overcome the out of friction we must increase the angle of the plane
Explanation:
To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.
X axis
fr - Wₓ = m a (1)
Y axis
N- [tex]W_{y}[/tex] = 0
N = W_{y}
let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
the friction force has the formula
fr = μ N
fr = μ Wy
fr = μ mg cos θ
from equation 1
at the point where the force equals the maximum friction force
in this case the block is still still so a = 0
F = fr
F = (μ mg) cos θ
We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.
This is the force that balances the friction force, any force slightly greater than F initiates the movement.
Consequently, to overcome the out of friction we must increase the angle of the plane
the correct answer is to increase the angle of the plane
A planar electromagnetic wave is propagating in the x direction. At a certain point P and at a given instant, the magnitude of the electric field of the wave is 0.082 V/m . What is the magnetic vector of the wave at point P at that instant?
A) (0.27 nT)k
B) (-0.27 nT)k
C) (0.27 nTİ
D) (6.8 nT)k
E) (-6.8 nT))
Answer:
b
Explanation:
A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?
Answer:
A. 0.0374C
B. 0.012F
C. 18 ohms
Explanation:
See attached file
Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form _______. A. pure elements B. molecules C. metals D. the periodic table
Answer:
Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form pure elements
The answer is option A
Answer:
its molecues
Explanation:
during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the image
Answer:
lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.
Ocean waves with a wavelength of 120 m are coming in at a rate of 8 per minute. What is their speed?
Explanation:
We know that,
[tex]v(wave \: speed) = f(frequency) \times \alpha (wavelength)[/tex]
frequency (f) = 1 / t (sec) = 8/60 = 0.13 Hz
V ( wave speed) = 0.13 * 120 = 16 m/sec
The speed of the given wave is equal to 15.96 m/s.
What are frequency and wavelength?The frequency of the wave can be defined as the number of oscillations of a wave in one second. The frequency has S.I. units which can be expressed as per second or hertz (Hz).
The wavelength can be described as the distance between the two adjacent points in phase. Two crests or two troughs of a wave are separated by a distance is called wavelength.
The relationship between wavelength (λ), frequency (ν), and wave speed (V):
V = νλ
Given, the frequency of the wave, ν = 8 min⁻¹ = 0.133 s⁻¹
The wavelength of the wave, λ = 120 m
The speed of the waves can calculate from the above-mentioned relationship:
V = νλ = 120 × 0.133 = 15.96 m/s
Therefore, the speed of the wave is equal to 15.96 m/s.
Learn more about wavelength and frequency, here:
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A bicycle has wheels that are 60 cm in diameter. What is the angular speed of these wheels when it is moving at 4.0 m/s
Answer:
13.33 rad/s
Explanation:
Applying,
v = ωr......................... Equation 1
Where v = linear speed, ω = angular speed and r = radius.
Note that,
r = d/2................. Equation 2
Where d = diameter of the wheel.
Substitute equation 2 into equation 1
v = ωd/2............... Equation 3
make ω the subject of the equation
ω = 2v/d................ Equation 4
Given: v = 4 m/s, d = 60 cm = 0.6 m
Substitute these values into equation 4
ω = 2(4)/0.6
ω = 13.33 rad/s
Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion
A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less than the actual distance, the same as the actual distance, or more than the actual distance? Explain.
Answer:
p = -q
he distance is equal to the current distance, so the distance does not change
Explanation:
For this exercise we can solve it using the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies
1 / p = - 1 / q
p = -q
Therefore the distance is equal to the current distance, so the distance does not change
Please help!
Much appreciated!
Answer:
your question answer is 22°
Two spheres A and B of negligible dimensions and masses 1 kg and √3 kg respectively, are supported on the smooth circular surface, fixed to the ground with a centre O and radius of 0.1m. The spheres are joined by the cord shown in length π/20 m; determine the angles α and β corresponding to the position of equilibrium of the spheres with respect to the vertical passing through O.
Answer:
α = π/3
β = π/6
Explanation:
Use arc length equation to find the sum of the angles.
s = rθ
π/20 m = (0.1 m) (α + β)
π/2 = α + β
Draw a free body diagram for each sphere. Both spheres have three forces acting on them:
Weight force mg pulling down,
Normal force N pushing perpendicular to the surface,
and tension force T pulling tangential to the surface.
Sum of forces on A in the tangential direction:
∑F = ma
T − m₁g sin α = 0
T = m₁g sin α
Sum of forces on B in the tangential direction:
∑F = ma
T − m₂g sin β = 0
T = m₂g sin β
Substituting:
m₁g sin α = m₂g sin β
m₁ sin α = m₂ sin β
(1 kg) sin α = (√3 kg) sin (π/2 − α)
1 sin α = √3 cos α
tan α = √3
α = π/3
β = π/6
