Answer:
E = 124.7 N / C
Explanation:
Let's analyze the exercise: the microwave creates an electromagnetic wave of frequency F = 2.45 GHz, this wave is introduced into the microwave cavity and is reflected on the metal walls, which is why one or more standing waves are formed.
The electric field of the standing wave is
I = E²
E =√I
where I is the intensity of the radiation.
What is it
I = P / A
where P is the effective emission power, almost all the power of the microwave and A is the area of the cavity, in the most used microwaves
P = 700 W and the area is A = 25 x 18 cm² = 0.045 m²
I = 700 / 0.045
I = 15555.56 W/m²
let's calculate the electric field
E = √15555.56
E = 124.7 N / C
What effect does the Duck Velocity have on the waves seen by the observer?
Towards the boat:
Away from the boat:
Same as the boat:
Why is it important for equipment for sport to be strong? To protect us
Answer:
To protect us.
Explanation:
For ex. your dunking on a basketball hoop if that wasn't strong you would fall on your back and get injured.
4- What force must be applied to a surface area of 0.0025m , to create a pressure ol
200.000Pa?
A 5.0-m radius playground merry-go-round with a moment of inertia of 1,630 kg m2 is rotating freely with an angular speed of 1.6 rad/s. Two people, each having a mass of 69.5 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on
Answer:
The right solution is "0.511".
Explanation:
Given:
Initial moment of inertia,
= 1630 kg.m²
Radius,
= 5 m
Angular speed,
= 1.6 rad/s
Now,
The moment of inertia after stepping on will be:
= [tex]1630+2\times (69.5\times (5)^2)[/tex]
= [tex]1630+2\times (69.5\times 25)[/tex]
= [tex]5105 \ Kg.m^2[/tex]
hence,
As per the question, the angular speed is conserved, then
⇒ [tex]1630\times 1.6=5105\times \omega'[/tex]
[tex]2608=5105\times \omega'[/tex]
[tex]\omega'=\frac{2608}{5105}[/tex]
[tex]=0.511[/tex]
An electron that has an instantaneous velocity of is moving through the uniform magnetic field (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity
Answer:
Explanation:
From the given information:
The missing value are as follows:
The velocity [tex]v ^{\to} = (2.0 \times 10^6 \ m/s) \hat i + (3.0 \times 10^6 \ m.s) \hat j[/tex]
The uniform magnetic field [tex]B^{\to} = (0.030 T)\hat i -(0.15 T) \hat j[/tex]
The force on electron as a result of the Magnetic field is:
[tex]F^\to = q(v^\to \times B^\to)[/tex]
here;
Change of electron [tex]q = -1.6 \times 10^{-19} C[/tex]
Then,
[tex]v^{\to} \times B^{\to} = \left|\begin{array}{ccc}2\times10^6&3\times 10^6&0\\0.03&-0.15&0\end{array}\right|[/tex]
[tex]= \hat i(0-0) -\hat j ( 0-0) +\hat k (-3\times 10^5 -0.9\times 10^5) \\ \\ =-3.9 \times 10^5 \ k[/tex]
∴
[tex]F^\to = q(v^\to \times B^\to)[/tex]
[tex]F^{\to} = -1.6 \times 10^{-19} \times -3.9 \times 10^5 \\ \\ F^{\to} = 6.24 \times 10^{-14 } \ N[/tex]
For the proton with the same velocity:
[tex]q = 1.6 \times 10^{-19 } \ C \\ \\ F = q (v^\to * B^\to) \ \\ \\ F^\to = q(-3.9 \times 10^5) \\ \\ F = (1.6 \times 10^{-19}) (-3.9 \times 10^5) \\ \\ \mathbf{F ^{\to} = -6.24 \times 10^{-14} \ N}[/tex]
A sound wave with a frequency of 700 Hz and a wavelength of 5 m travels through a liquid. How fast does sound travel through the liquid?
A.
140 m/s
B.
0.007 m/s
C.
3500 mHz
D.
3500 m/s
There are three 20.0 Ω resistors connected in series across a 120 V generator
Answer:
That is equal to R1 + R2. If three or more unequal (or equal) resistors are connected in series then the equivalent resistance is: R1 + R2 + R3 +…, etc. One important point to remember about resistors in series networks to check that your maths is correct.
Can someone help me
An ideal massless spring with a spring constant of 2.00 N/m is attached to an object of 75.0 g. The system has a small amount of damping. If the amplitude of the oscillations decreases from 10.0 mm to 5.00 mm in 15.0 s, what is the magnitude of the damping constant b
Answer: 0.00693
Explanation:
Given
Spring constant [tex]k=2\ N/m[/tex]
Mass of object [tex]m=75\ g[/tex]
The amplitude of the oscillation decreases from 10 mm to 5 mm in 15 s
Equation of amplitude for the ideal spring-mass system is
[tex]\Rightarrow A=A_oe^{-\frac{bt}{2m}}\quad \quad [\text{b=damping constant}]\\\text{Insert the values}\\\\\Rightarrow 5=10e^{\frac{b\times 15}{2\times 0.075}}\\\\\Rightarrow e^{-\frac{b\times 15}{2\times 0.075}}=0.5\\\\\text{Taking natural log both sides}\\\\\Rightarrow \ln \left(e^{-\frac{b\times 15}{2\times 0.075}}\right)=\ln 0.5\\\\\Rightarrow -\dfrac{15b}{0.15}=-0.693\\\\\Rightarrow b=0.00693[/tex]
During forced exhalation, such as when blowing up a balloon, the diaphragm and chest muscles create a pressure of 72.0 mm Hg between the lungs and chest wall. What force in newtons does this pressure create on the 605 cm2 surface area of the diaphragm
Answer:
F = 580.75 N
Explanation:
The force can be calculated through the following formula:
[tex]F = PA[/tex]
where,
F = Force = ?
P = Pressure = (72 mm Hg)(133.322 Pa/1 mm Hg) = 9599.18 Pa
A = Area = (605 cm²)(1 m²/10⁴ cm²) = 0.0605 m²
Therefore,
[tex]F = (9599.18\ Pa)(0.0605\ m^2)[/tex]
F = 580.75 N
How are soil and air similar?
Answer:
The air in the soil is similar in composition to that in the atmosphere with the exception of oxygen, carbon dioxide, and water vapor. In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.036%.
hope this helps
have a good day :)
Explanation:
paano matutugunan o matutulungan ng pamahalaan at ng mga guro yubg mga estudyanteng nakararanas nag stress at anxiety.
Answer: how the government and teachers can address or help students experiencing stress and anxiety.
Explanation:
BRAINLIST A wave travels at a constant speed. How does the wavelength change if the
frequency is reduced by a factor of 3? Assume the speed of the wave remains
unchanged.
A. The wavelength does not change.
B. The wavelength increases by a factor of 3.
C. The wavelength decreases by a factor of 3.
D. The wavelength increases by a factor of 9.
A closed loop conductor that forms a circle with a radius of 2.0 m is located in a uniform but changing magnetic field. If the maximum emf Induced in the loop is 5.0 V what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Answer:
The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.
Explanation:
Given;
radius of the circular loop, r = 2.0 m
maximum induced emf, E = 5.0 V
The emf induced in a magnetic field is given as;
[tex]emf = \frac{d\phi}{dt} \\\\\phi = AB\\\\emf = A\frac{dB}{dt} \\\\\frac{dB}{dt} = \frac{emf}{A} \\\\where;\\A \ is \ the \ area \ circular \ l00p = \pi r^2 = \pi (2)^2 = 4\pi \ m^2\\\\\frac{dB}{dt} = \frac{5}{4\pi} \\\\\frac{dB}{dt} = 0.398 \ T/s[/tex]
Therefore, the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.
An athlete is performing squats in the weight room. The knee is going from anatomical position to 92 degrees and then back to anatomical position each squat. The athlete performs a total of 10 squats. This is done over a time period of 30 seconds. What is the angular acceleration (rad/sec 2) of the knee
Answer:
α = 0.357 ras / s²
Explanation:
This is a rotational kinematics exercise, it tells us that it performs 10 squats in 30 s, for which it performs one squat at t = 3 s, also indicates that the angle of the squat is θ = 92º
θ = θ₀ + w₀ t + ½ α t²
the athlete starts from rest, whereby w₀ = 0 and the initial angle in the vertical position is zero (θ₀=0)
θ = ½ α t²
α = 2 θ /t²
let's reduce the magnitudes to the SI system
θ = 92º (π rad /180º) = 0.511π rad
let's calculate
α = 2 0.5111π /3²
α = 0.1136π rad / s²
α = 0.357 ras / s²
A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What is the longest wavelength that forms an intensity maximum in the fifth order
Answer:
377 nm
Explanation:
Number of lines per meter is, [tex]N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}[/tex]
Grating element is, [tex]d=\frac{1}{N}[/tex]
[tex]=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda[/tex]
[tex]\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\ &=0.377 \times 10^{-6} \mathrm{~m} \\ &=377 \mathrm{~nm}[/tex]
Which of the following is form of energy:
a) Power
b) Light
C) pressure
d) None
Answer:
Explanation:
b) light
A 70 kg stunt pilot begins pulling out of a dive into a vertical circle. If the plane's speed at the lowest point of the circle is 80 m/s, what is the apparent
weight of the pilot at the lowest point of the pullout? The pilot experiences a force of 5g from the centripetal acceleration at the bottom of the dive.
O
3880 N
о
3430 N
0
4116 N
3986 N
Answer:
Apparent weight of pilot due to centripetal acceleration:
m v^2 / R = 5 g m = 5 * 70 * 9.8 = 3430 N
Weight of pilot = 70 * 9.8 = 686 N
Total = 3430 + 686 = 4116 N
Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measured in the laboratory. What is the magnitudeof the velocity of one particle relative to the other
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v[tex]_r[/tex] = ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v[tex]_r[/tex] = ( u + u ) / ( 1 + ( u×u / c²) )
v[tex]_r[/tex] = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v[tex]_r[/tex] = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v[tex]_r[/tex] = 1.904c / ( 1 + 0.906304 )
v[tex]_r[/tex] = 1.904c / 1.906304
v[tex]_r[/tex] = 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
What is the speed of a wave if it has a wavelength of
42 m and a frequency of 7 hertz?
Answer:
♕ [tex]\large{ \red{ \tt{Step - By - Step \: Explanation}}}[/tex]
☃ [tex] \underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} : [/tex]
Frequency ( f ) = 7 HertzWavelength ( λ ) = 42m♨ [tex] \underline {\underline{ \orange{ \large{ \tt{T \: O \: \: F \: I \: N\: D}}}} }: [/tex]
Wave velocity ( v )☄ [tex]\underline{ \underline{ \large{ \pink{ \tt{S\: O \: L \: U \: T\: I \: O \: N}}}}}: [/tex]
✧ [tex] \red{ \boxed{ \large{ \purple{ \sf{Wave \: velocity(v) = Frequency(f) \times Wavelength(λ)}}}}}[/tex]
~Plug the known values and then multiply!
↦ [tex] \large{ \tt{7 \times 42}}[/tex]
↦ [tex] \boxed{ \boxed{ \large{ \bold{ \tt{294 \: m {s}^ {- 1} }}}}}[/tex]
☥ [tex] \large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}} [/tex]
---------------------------------------------------------------
❁ [tex] \underline{ \large{ \red{ \tt{D\: E\: T \: A \: I \: L\: E \: D \: \: I\: N \: F \: O}}}} : [/tex]
Frequency ( f ) : The number of complete waves , set up in a medium in one second is called frequency of the wave. The SI unit of frequency is Hertz ( Hz ). For example : if a sound wave completes 15 compressions and 15 rarefactions in one second , it's frequency is 15 Hz.Wavelength ( λ ) : The distance between two consecutive troughs or crests in a transverse wave or the distance between two consecutive compressions or rarefactions in a longitudinal wave us called wavelength. It is the distance travelled by a wave in a time equal to it's time period. It's SI unit is metre ( m ).Wave velocity ( v ) : The velocity with which a wave propagates in a medium is called wave velocity. It's SI unit is m/s.# KILL : Excuses
KISS : Opportunities
MARRY : Goals
♪ Hope I helped! ♡
☂ Have a wonderful day / night ! ツ
✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✔
▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁
5. Stopping a fast-moving object is harder than stopping a slow-moving
one.
True
False
An irrigation canal has a rectangular cross section. At one point where the canal is 18.2 m wide and the water is 3.55 m deep, the water flows at 2.55 cm/s . At a second point downstream, but on the same level, the canal is 16.3 m wide, but the water flows at 11.6 cm/s . How deep is the water at this point
Answer:
Explanation:
Rate of volume flow at two points will be same at two points .
A₁ V₁ = A₂V₂
A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .
A₁ = 18.2 x 3.55 = 64.61 m²
V₁ = 2 .55 x 10⁻² m/s
A₂ = 16.3 x d = 16.3 d m²
d is depth at second point .
V₂ = 11.6 x 10⁻² m/s
64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d m² x 11.6 x 10⁻² m/s
d = .87 m
so canal is .87 m deep.
An old fashioned string of 80 Christmas lights is wired in series. Each bulb has a resistance of 2 Ohms and the entire string is plugged into a 120V outlet. What is the current passing through each of the bulbs?
The sum of the resistance = 2 ohms x 80 lights = 160 ohms.
Current = Total voltage / total resistance:
Current = 120V / 160 ohms
Current = 0.75 Amps
ADHD is triggered ________ by genetic and neurological factors and ________ by social or environmental ones.
A. less; more
B. somewhat; somewhat
C. little; mostly
D. more; less
Answer:
D More: Less
Explanation:
ADHD is triggered more by genetic and neurological factors and less by social or environmental ones.
What is ADHD?ADHD is one of the most common neurodevelopmental disorders of childhood. It is usually first diagnosed in childhood and often lasts into adulthood. Children with ADHD may have trouble paying attention, controlling impulsive behaviors (may act without thinking about what the result will be), or be overly active.
Moreover, ADHD tends to run in families and, in most cases, it's thought the genes you inherit from your parents are a significant factor in developing the condition. Research shows that parents and siblings of someone with ADHD are more likely to have ADHD themselves.
Therefore, adults with attention deficit hyperactivity disorder (ADHD or ADD) lead anxious lives. The nature of ADHD often makes day-to-day life stressful, creating situations and environments fraught with uncertainty – anxiety's primary fuel.
Learn more about ADHD:
https://brainly.com/question/13719580
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A 1460 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 18.0 m/s. They stick together. In what direction and with what speed do they move after the collision?
Answer in degrees north of east
AND
the speed after the collision in m/s
Answer:
Solution given:
North car
mass[m1]=1460kg
velocity[u1]=27 m/s
mass[m2]=2165kg
velocity [u2]=18m/s
let v be velocity after collision
we have
From the principle of conservation of linear momentum
m1u1+m2u2=(m1+m2)v
1460*27+2165*18=(1460+2165)v
v=[tex] \frac{78390}{3625} [/tex]
v=21.6m/s
the speed after the collision in 21.6 m/s.
For angle.
Tan angle =[tex] \frac{m1u1}{m2u2} [/tex]
Tan angle =[tex] \frac{1460*27}{2165*18} [/tex]
Tan angle=327.74
angle=Tan-¹(327.74)=89.82=90°
in degrees north of east is 90°
The atmospheric features of Neptune are easier to see than those of Uranus because A. Neptune has greater warmth and less haze. B. Neptune has more methane. C. The atmosphere of Uranus rotates differentially. D. Uranus has no significant atmosphere.
Answer:
Option B is the correct answer (Neptune has more methane)Explanation:
From the options given,
The atmospheric features of Neptune are easier to see than those of Uranus because Neptune has more methane
Neptune has small amount of methane and water which gives it blue colour and white patches which distinguish it from uranus
For more information, visit
http://abyss.uoregon.edu/~js/ast121/lectures/lec20.html
In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background signal level is 13 mV, the signal is 91 mV with no attenuator and is 25 mV with an attenuator in place. Calculate pt/pi including the background correction.
Answer:
The answer is "[tex]15.38\%[/tex]"
Explanation:
Background[tex]= 13 \ mv\\\\[/tex]
corrected signal[tex]= 91 \ mv-13\ mv= 78\ mv\\\\[/tex]
with attenuator[tex]=25\ mv-13\ mv= 12\ mv\\\\[/tex]
[tex]\to \frac{p_t}{p_i}=\frac{12}{78}\times 100= 15.38\%[/tex]
At what speed does light travel through water. n=1.33
Answer:
"In water, the speed of light is v01.33=2.26 x 108 m/s."
Explanation:
You have two flashlights that have 0.500-A currents flowing through them. One flashlight has a single 1.50-V battery in its circuit, and the second flashlight has three 1.50-V batteries connected in a chain that provides 4.50 V. How much power is the battery in the first flashlight providing
Answer: 0.75 W
Explanation:
Given
Current flowing [tex]I=0.5\ A[/tex]
The first flashlight has a single 1.5 V battery
The second flashlight has three 1.5 V batteries connected to provide 4.5 V
Power is the product of voltage and current.
For the first flashlight
[tex]\Rightarrow P=1.5\times 0.5\\\Rightarrow P=0.75\ W[/tex]
Thus, the first flashlight providing 0.75 W of power.
Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulley without slipping. The mass of block A is 2.1 kg and the mass of block B is 4.1 kg. Block A is also connected to a horizontally-mounted spring with a spring constant of 358 J/m2. What is the angular frequency (in rad/s) of oscillations of this system
Answer:
[tex]F=1.159[/tex]
Explanation:
From the question we are told that:
Mass of pulley [tex]M=1kg[/tex]
Radius [tex]r=12cm[/tex]
Mass of block A [tex]M_a=2.1kg[/tex]
Mass of block B [tex]m_b=4.1kg[/tex]
Spring constant[tex]\mu= 358 J/m2[/tex]
Generally the equation for Torque is mathematically given by
Since [tex]\sumF=ma[/tex]
At mass A
[tex]T_2-f_3=2.1a[/tex]
At mass B
[tex]4.8-T_1=4.1a[/tex]
At Pulley
[tex]R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}[/tex]
[tex]R(T_1-T_2)=0.55a[/tex]
Therefore the equation for total force F
At mass A+At mass B+At Pulley
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a[/tex]
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a[/tex]
[tex]-f_3+4.1=6.75a[/tex]
[tex]-f_3=6.75a+4.8[/tex]
Since From above equation
[tex]M_{eff}=6.7kg[/tex]
Therefore
[tex]T=2\pi \sqrt{{\frac{M_{eff}}{k}}[/tex]
[tex]T=2\pi \sqrt{{\frac{6.75}{\mu}}[/tex]
[tex]T=0.862s[/tex]
Generally the equation for frequency is mathematically given by
[tex]F=\frac{1}{T} \\F=\frac{1}{0.862}[/tex]
[tex]F=1.159[/tex]