Answer:
The frequency does not change, but the wavelength does
Explanation:
Here are the options
A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.
B. The frequency does change, but the wavelength remains unchanged.
C. Both the frequency and wavelength change.
D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.
E. The frequency does not change, but its wavelength does.
When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.
[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]
where [tex]\lambda_o[/tex] indicates wavelength in vacuum
[tex]\lambda_m[/tex] indicates wavelength in medium
n indicates refractive index
v indicates velocity of light wave
c indicates velocity of light
And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.
Therefore the correct option is E
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object
Answer:
Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.
Explanation:
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.630 cm thick and heat conduction occurs through an area of 1.45 ✕ 10−2 m2 at a rate of 500 J/s, what is the temperature difference across it (in °C)? Ceramic has the same thermal conductivity as glass and concrete brick.
Answer:
The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
Explanation:
From the question we are told that
The thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]
The area is [tex]A = 1.45 *10^{-2 } \ m^2[/tex]
The rate is [tex]P = 500 J/s[/tex]
The thermal conductivity is [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]
Generally the rate heat conduction mathematically represented as
[tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]
=> [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]
=> [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]
=> [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth
Answer:
t₀ = 1.55 s
Explanation:
According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in motion relative to the event is not the same as measured by an observer at rest.
It is given as:
t = t₀/[√(1 - v²/c²)]
where,
t = time measured by astronaut in motion = 3.1 s
t₀ = time required according to observer on earth = ?
v = relative velocity = 0.87 c
c = speed of light
3.1 s = t₀/[√(1 - 0.87²c²/c²)]
(3.1 s)(0.5) = t₀
t₀ = 1.55 s
Answer:
The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]Explanation:
Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]
where,
[tex]t_o = 3.1\\\\v = 0.87[/tex]
[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]
For more information visit
A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?
Answer:
-0.24diopters
Explanation:
The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)
So 1/do +1/di = 1/f
1/infinity + 1/-4.2 = 1/f
1/f = 1/-4.2 = -0.24diopters
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 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Complete Question
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Answer:
The force is [tex]F = 0.1602 \ N[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The length of the solenoid is [tex]L = 12 \ cm = 0.12 \ m[/tex]
The current is [tex]I = 36 \ A[/tex]
The diameter is [tex]D = 4.5 \ cm = 0.045 \ m[/tex]
The current carried by the wire is [tex]I = 27 \ A[/tex]
The length of the wire is [tex]l = 2 cm = 0.02 \ m[/tex]
Generally the magnitude of the force on this wire assuming the solenoid's field points due east is mathematically represented as
[tex]F = B * I * l[/tex]
Here B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{L}[/tex]
Here [tex]\mu _o[/tex] is permeability of free space with value [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]
substituting values
[tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]
[tex]B = 0.2225 \ T[/tex]
So
[tex]F = 0.2225 * 36 * 0.02[/tex]
[tex]F = 0.1602 \ N[/tex]
A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction. If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 4.00 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Answer:
B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction
Explanation:
In an electromagnetic wave the electric and magnetic fields are in phase
c = E / B
B = E / c
let's calculate
B = 4.00 / 3 10⁸
B = 1.33 10⁻⁸ T
To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.
If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction
A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 566 nm light? The index of refraction of the coating material is 1.46 and the index of the glass is 1.71.
Answer:
The thickness is [tex]t = 1.415 *10^{-7 } \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]
The index of refraction of glass is [tex]n_g = 1.71[/tex]
The index of refraction of the coating is [tex]n= 1.46[/tex]
Generally the condition for destructive interference is
[tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]
Here m is the order of the interference pattern and given from the question that we are considering minimizing reflection m = 0
t = thickness of the coating
substituting values
[tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]
=> [tex]t = 1.415 *10^{-7 } \ m[/tex]
A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?
Answer:
Pls seeattached file
Explanation:
A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.
What is Resistance?Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.
When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.
According to the question,
R = R₀ (1 + α ΔT)
(1 + 0.0135)R₀ = R₀(1 + α ΔT)
ΔT = (1 + 0.0135) / α
= 0.0135 / 0.0004
= 33.75 °C.
ΔT = [(1 - 0.0135) -1]/0.004
= -33.75 °C
To get more information about Resistance :
https://brainly.com/question/11431009
#SPJ5
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?
Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.
Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m
Explanation: A speed of wave on a string under a tension force can be calculated as:
[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]F_{T}[/tex] is tension force (N)
μ is linear density (kg/m)
Determining velocity:
[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{0.00874 }[/tex]
[tex]|v| =[/tex] 0.0935 m/s
The displacement a pulse traveled in 1.23ms:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]
Δx = 11.5×[tex]10^{-6}[/tex]
With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex] m.
Doubling Tension:
[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{2.0.00874 }[/tex]
[tex]|v| = \sqrt{0.01568}[/tex]
|v| = 0.1252 m/s
Displacement for same time:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]
[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]
With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m
A dentist uses a concave mirror (focal length 2 cm) to examine some teeth. If the distance from the object to the mirror is 1 cm, what is the magnification of the tooth
Answer: 2
Explanation:
1/2=1/1 +1/x
x=-2
magnification= 2/1
magnification=2
3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.
Answer:
Check your router connections then restart your router.
Explanation:
Answer:
Check your router connections then restart your router.
Explanation:
Most internet access comes from routers so the problem is most likely the router.
Kinetic and
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?
11 point)
O raise the board to a higher angle
O press down on the brick in a direction that is perpendicular to the board
O roughen up the texture of the wooden board
o lower the board so it's level with the ground
Answer:
raise the board to a higher angle
Explanation:
Static friction is the force opposite to the applied force.
Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.
So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.
Hence, the correct option is "raise the board to a higher angle".
A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contract- ing when it is cooled from 120.0°C to 10.0°C?
Answer:
42000N
Explanation:
First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.
1) linear thermal expansion coef brass 19e-6 /K
∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm
Second part involves linear elasticity.
for brass, young's modulus is 15e6 psi or 100 GPa
cross-sectional area of rod is π(0.008)² = 0.0002 m²
F = EA∆L/L
F = (100e9)(0.0002)(0.00387) / (1.85)
F = 42000 or 42 kN
If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?
Answer:
[tex]v=1.24\times 10^8\ m/s[/tex]
Explanation:
Given that,
The refractive index of benzene is 2.419
We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,
[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]
So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 10-7 N/A2 .) A. 0.02219 m B. 327 m C. 52 m D. 0.00199 m
Answer:
A. 2.2*10^-2m
Explanation:
Using
Area = length x L/ uo xN²
So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*
3000²)
A = 17.5*10^-3/ 1.13*10^-5
= 15.5*10^-2m²
Area= π r ²
15.5E-2/3.142 = r²
2.2*10^2m
Explanation:
A 26-g rifle bullet traveling 220 m/s embeds itself in a 3.8-kg pendulum hanging on a 2.7-m-long string, which makes the pendulum swing upward in an arc, Determine the vertical and horizontal component of the pendulum's maximum displacement
Answer:
displacements are 0.776m, 0.114m
Explanation:
We were given mass of 26-g rifle bullet , then we can convert to Kg since
Momentum is conserved here.
The initial momentum before impact = (Mi * Vi)
Where Mi= initial given mass
Vi=initial velocity given
= 0.026 * 220 = 5.72 kgm/s
The final momentum after impact is (Mf * Vf )
Mf= final mass
5.72=( 3.82* Vf )
= 5.72/ 3.82
= 1.497 m/s
the speed of the pendulum bob with bullet afterwards= 1.497 m/s
the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.
M = 3.82 kg the mass of the bob containing the bullet
E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J
When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before
E(total) = Mgh + 0 = Mgh = 4.280J
solving for h and substituting,
h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m
Since the height is found,we the angle of the pendulum at the top of the swing can also be determined
A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees
Since A is known, the displacement along the horizontal axis can be calculated as
x = 2.7* sin(A) = 0.776m
therefore, displacement is 0.776m, 0.114m
the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:
[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]
ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:
[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]
[tex]0.8 = e^{-\frac{88}{R} }[/tex]
[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]
[tex]ln(0.8) = -\frac{88}{R}[/tex]
[tex]R = -\frac{88}{ln(0.8)}[/tex]
R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.
Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration
Answer:
The acceleration of the refrigerator together with the objects decreases.
Explanation:
If the mass of the refrigerator is increased by stacking more masses (objects) on it,
and the force applied remains constant, then we know from
F = ma
where
F is the applied force
m is the total mass of the refrigerator and the objects
a is the acceleration of the masses.
If F is constant, and m is increased, the acceleration will decrease
Answer:
The acceleration decreases.
Explanation:
its right
A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface
Answer:
The radius is [tex]r = 3.1905 \ m[/tex]
Explanation:
From the question we are told that
The distance beneath the liquid is [tex]d = 2.70 \ m[/tex]
The refractive index of the liquid is [tex]n_i = 1.31[/tex]
Now the critical value is mathematically represented as
[tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]
substituting values
[tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]
[tex]\theta = 49.76^o[/tex]
Using SOHCAHTOA rule we have that
[tex]tan \theta = \frac{ r}{d}[/tex]
=> [tex]r = d * tan \theta[/tex]
substituting values
[tex]r = 2.7 * tan (49.76)[/tex]
[tex]r = 3.1905 \ m[/tex]
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.
Hope this helps!
Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
A plano-convex glass lens of radius of curvature 1.4 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.6. Determine the radii of the first and second bright fringes in the reflected light.
Given that,
Radius of curvature = 1.4 m
Wavelength = 520 nm
Refraction indexes = 1.6
We know tha,
The condition for constructive interference as,
[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]
Where, [tex]\lambda=wavelength[/tex]
We need to calculate the radius of first bright fringes
Using formula of radius
[tex]r_{1}=\sqrt{2tR}[/tex]
Put the value of t
[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=0.603\ mm[/tex]
We need to calculate the radius of second bright fringes
Using formula of radius
[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=1.04\ mm[/tex]
Hence, The radius of first bright fringe is 0.603 mm
The radius of second bright fringe is 1.04 mm.
An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
Current, I = 1.23 A
Explanation:
Given that,
Inductance, L = 35 mH
Resistance, R = 12 ohms
Potential difference, V = 18 V
We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :
[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)
Here,
[tex]I_o[/tex] is final current
[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]
[tex]\tau[/tex] is time constant
[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]
So, equation (1) becomes :
[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]
So, after 5 ms the current in the circuit is 1.23 A.
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?
Answer:
the child is 1.581 m far from the fence
Explanation:
The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.
From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:
[tex]x - x_o = u_xt[/tex]
[tex]\mathtt{x = u_xt \ \ \ since (x_o = 0)}[/tex] ---- (1)
the equation of the motion y is :
[tex]\mathtt{y - y_o =u_yt - 0.5 gt^2}[/tex]
[tex]\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}[/tex]
[tex]\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }[/tex]
[tex]\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}[/tex]
[tex]\mathtt{1 = 5.14t - 4.9t^2}[/tex]
[tex]\mathtt{4.9t^2 - 5.14t +1 = 0}[/tex]
By using the quadratic formula, we have;
[tex]\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }[/tex]
where;
a = 4.9, b = -5.14 c = 1
[tex]= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }[/tex]
[tex]= \mathbf{ 0.791 \ \ OR \ \ 0.258} }[/tex]
In as much as the ball is traveling upward, then we consider t= 0.258sec.
From equation (1)
[tex]\mathtt{x = u_x(0.258)}[/tex]
[tex]\mathtt{x = ucos 40^0 (0.258)}[/tex]
[tex]\mathtt{x = 8 \ cos 40^0 (0.258)}[/tex]
[tex]\mathbf{x = 1.581 \ m}[/tex]
Thus, the child is 1.581 m far from the fence
A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current
Answer:
Explanation:
From the question we are told that
The radius is [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]
The current is [tex]I = 4.5 \ A[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{J}{\sigma }[/tex]
Where [tex]\sigma[/tex] is the conductivity of aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]
J is the current density which mathematically represented as
[tex]J = \frac{I}{A}[/tex]
Here A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (1.4*10^{-3})^2[/tex]
[tex]A = 6.158*10^{-6} \ m^2[/tex]
So
[tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]
[tex]J = 730757 A/m^2[/tex]
So
[tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]
[tex]E = 0.021 \ N/C[/tex]
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
Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3
Answer:
v₂ = 9 v
Explanation:
For this exercise in fluid mechanics, let's use the continuity equation
v₁ A₁ = v₂ A₂
where v is the velocity of the fluid, A the area of the pipe and the subscripts correspond to two places of interest.
The area of a circle is
A = π R²
let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint
In this case v₁ = v and the area is
A₁ = π R²
in the second point
A₂= π (R / 3)²
we substitute in the continuity equation
v π R² = v₂ π R² / 9
v = v₂ / 9
v₂ = 9 v
Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?
Answer:
10.75 A
The current is in opposite direction since it causes a repulsion force between the wires
Explanation:
Force per unit length on the wires = 4.30×10^−5 N/m
distance between wires = 2.6 cm = 0.026 m
current through one wire = 0.52 A
current on the other wire = ?
Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as
[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]
where [tex]F/l[/tex] is the force per unit length on the wires
[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A
[tex]I_{1}[/tex] = current on the first wire = 0.520 A
[tex]I_{2}[/tex] = current on the other wire = ?
r = the distance between the two wire = 0.026 m
substituting the value into the equation, we have
4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] = [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]
4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]
[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A
The current is in opposite direction since it causes a repulsion force between the wires.
