Answer: 55 kmph
Explanation: Divide 110 by 2
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8,32 meters per second take the speed of sound as 340 meters per second calculate frequency
Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, = 39 kHz
velocity of the bat, [tex]v_b[/tex] = 8.32 m/s
speed of sound in air, v = 340 m/s
The apparent frequency of sound striking the wall is calculated as;
[tex]f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz[/tex]
The frequency reflected by the stationary wall to the bat is calculated as;
[tex]f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz[/tex]
[tex]f_s\approx 41 \ kHz[/tex]
a) Find the current in the 1 Ω resistor.
b ) Find the current in the 8 Ω resistor.
c ) Find the current in the 5 Ω resistor.
PLEASE HELP I NEED THIS TODAY
Answer:
a)I=V/R
39.5 amp
Explanation:
because the voltage in serious with 1ohm resistor
Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius of curvature of 15.0 cm. The refractive index of the glass is 1.50. What is the distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod? Is the image in the air or in the glass?
Answer:
the distance of image from the vertex is 45 cm and the image formed is in the glass.
Explanation:
distance of object, u = - infinity
radius of curvature, R = - 15 cm
refractive index, n = 1.5
Let the distance of image is v.
Use the formula
[tex]-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm[/tex]
The image is in the glass.
Please helppppppp!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
I think the Bulb No. 2 will stop emitting light if the bulb No. 1 burns out.
DUE IN 1 MINUTE PLS ANSWER QUICK ...A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the
ocean floor and is received 15s later. What is the distance between the submarine and the ocean floor?
Answer:
s = 11250 m = 11.25 km
Explanation:
The distance covered by the sound wave while traveling from submarine to ocean floor and then back to submarine can be given as follows:
[tex]s = vt[/tex]
but, the distance between the floor and the submarine will be half of this value:
[tex]s = \frac{1}{2}vt[/tex]
where,
s = distance between submarine and ocean floor = ?
v = velocity of sound = 1500 m/s
t = time taken for the round trip = 15 s
Therefore,
[tex]s = \frac{1}{2}(1500\ m/s)(15\ s)[/tex]
s = 11250 m = 11.25 km
A pair of glasses uses a nonreflective coating of index of refraction 1.4 to minimize reflection of light with wavelength 500nm. If the index of refraction of the glass is 1.5, what is the minimum non-zero thickness of the coating
Answer:
d = 178.57 10⁻⁹ m
Explanation:
For this exercise we must find the thickness to minimize the reflection, so the interference for the reflection must be destructive.
To find the expression we must take into account, two things:
* When the light goes from an index mordant medium to one with a higher refractive incoe, it undergoes a phase change of 180 (pi radians)
* within the film the wavelength of light is modulated by the index of refraction
λₙ = λ₀/ n
In this case the light passes from the air to the reflective layer and undergoes a phase change of ∫π rad, then it is reflected in the film-glass layer where it undergoes another phase change of π rad, therefore the total change of phase is 2π radians, this change is the or changes its value
period of the trigonometric functions, therefore its value does not change
the expression for destructive interference is
d sin θ = (me + ½) λₙ
d sin θ = (m + ½) λ₀ / n
the minimum thickness occurs for m = 0 and if we take perpendicular incidence the sine = 1
d = λ₀ /2 n
l
et's calculate
d = 500 10⁻⁹ /( 2 1.4)
d = 178.57 10⁻⁹ m
Fill in the blanks. Power station produce electricity at __________.
Answer:
heat
Explanation:
heat is the answer hdhhdhdhdhdhdhdhdhdhd
A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 2.50 s after the bird lets go, the fish lands in the ocean. (a) Just before reaching the ocean, what is the horizontal component of the fish's velocity in m/s
Answer:
Horizontal Component of Fish's Velocity = 2.6 m/s
Explanation:
In this scenario, we will neglect the effects of the air resistance on the small fish. Since there is no resisting force available in the horizontal direction. Therefore, the horizontal component of the velocity of the fish will remain equal to the horizontal component of the velocity of the seagull and it will remain the same throughout the whole motion.
Horizontal Component of Fish's Velocity = Constant Horizontal Speed of Seagull
Horizontal Component of Fish's Velocity = 2.6 m/s
A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B particles with a combined energy of 720 units are added to the container, the system is allowed to come to thermal equilibrium.Part A) At equilibrium, how many energy units does each A particle have?Part B) At equilibrium, how many energy units does each B particle have?
Answer:
"8 units" is the appropriate answer.
Explanation:
According to the question,
Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.
Now,
The total energy will be:
= [tex]480+720[/tex]
= [tex]1200 \ units[/tex]
The total number of particles will be:
= [tex]50+100[/tex]
= [tex]150[/tex]
hence,
Energy of each A particle or each B particle will be:
= [tex]\frac{1200}{150}[/tex]
= [tex]8 \ units[/tex]
Who is a socio-economically disadvantaged child? Explain any four ways forhelping such a child
Explanation:
A socio-economically disadvantaged child is the one who is disadvantaged in terms of social position and economic position. Such children have limited resources in terms of education, money and future options. Four ways of helping such children are as follows:
1) Help them in education - You can help such children by giving them free tuition. If you belong to a well off family, you can get them admitted in schools as well. Provide them with books and uniform.
2) Encourage them to do well in school and pursue their passions.
3) Provide them with meals, if they do not have access to regular meals.
4) encourage them to go school regularly.
Answer:
sorry i dont know the answer bit mark me as BRAINLISTPOR FA ME PODRIAN AYUDAR. URGENTE!!!!
¿Cuál es la energía cinética de un 0,01kg bala viajando a una velocidad de 700Sra?
Answer:
K.E = 2450 Joules
Explanation:
Given the following data;
Mass = 0.01 kg
Velocity = 700 m/s
To find the kinetic energy;
La energía cinética (K.E) se puede definir como una energía que posee un objeto o cuerpo debido a su movimiento.
Matemáticamente, la energía cinética viene dada por la fórmula;
[tex] K.E = \frac{1}{2}mv^{2}[/tex]
Sustituyendo en la fórmula, tenemos;
K.E = ½ * 0.01 * 700²
K.E = 0.005 * 490000
K.E = 2450 Joules
Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of planets from earliest to latest.
a. Small clumps of matter stick together via the process of accrection to form plantesimals a few hundred kilometers in diameter
b. Dust keeps matter inside the disk cool long enough for planet formation to start
c. Planetisimals begin to accrete, forming protoplanets
d. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter
e. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
Answer: See explanation
Explanation:
The evolutionary stages for the formation of planets from earliest to latest will be:
1. Dust keeps matter inside the disk cool enough for planet formation to start
2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.
3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.
4. Planetesimals begin to accrete, forming protoplanets.
5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on expansion, what is its temperature when the initial volume has tripled
Answer:
Explanation:
It is a case of adiabatic expansion .
[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]
T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.
Given ,
V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .
[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]
[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]
1.552 = 292 / T₂
T₂ = 188 K .
Instead of changing the the frequency you can change the tension to produce the next higher harmonic. What percentage of the tension would produce the next higher harmonic?
Answer:
[tex]\frac{T}{T_o} = ( 1 + \frac{1}{n} )^2[/tex]
Explanation:
This is a string resonance exercise, the wavelengths in a string held at the ends is
λ = 2L₀ / n
where n is an integer
the speed of the wave is
v = λ f
f = v /λ
the speed of the wave is given by the characteristics of the medium (string)
v = [tex]\sqrt{\frac{T}{\mu } }[/tex]
we substitute
f = [tex]\frac{n}{2L_o} \ \sqrt{\frac{T}{\mu } }[/tex]
to obtain the following harmonic we change n → n + 1
f’ = [tex]\frac{n+1}{2L_o} \ \sqrt{\frac{T_o}{\mu } }[/tex]
In this case, it tells us to change the tension to obtain the same frequency.
f ’= \frac{n}{2L_o} \ \sqrt{\frac{T}{\mu } }
how the two frequencies are equal
[tex]\frac{n+1}{2L_o} \sqrt{\frac{T_o}{ \mu } } = \frac{n}{2L_o} \sqrt{\frac{T}{\mu } }[/tex]
(n + 1) [tex]\sqrt{T_o}[/tex] = n [tex]\sqrt{T}[/tex]
[tex]\frac{T}{T_o} = ( \frac{n+1}{n} )^2[/tex]
[tex]\frac{T}{T_o} = ( 1 + \frac{1}{n} )^2[/tex]
this is the relationship of the voltages to obtain the following harmonic,
A 9.2 resistor and a 5.2 resistor are connected in series with a battery. The potential difference across the 5.2 resistor is measured as 12 V. Find the potential difference across the battery.
Answer:
V = 33.12 V
Explanation:
Given that,
A 9.2 resistor and a 5.2 resistor are connected in series with a battery.
The potential difference across the 5.2 resistors is measured as 12 V.
In series combination of resistors, the current is same throughout the circuit. Let I is the current. Using Ohm's law,
[tex]I=\dfrac{V}{R}[/tex]
Put all the values,
[tex]I=\dfrac{12}{5.2}\\\\I=2.30\ A[/tex]
The equivalent resistance is :
R = 9.2 + 5.2
= 14.4 ohms
Let V os the potential difference across the battery. So,
[tex]V=2.3\times 14.4\\\\V=33.12\ V[/tex]
So, the potential difference across the battery is equal to 33.12 V.
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
for lunch. If the lily pads are spaced 2.4 m apart, and Ferdinand jumps with a
speed of 5.4m/s, taking 0.60 s to go from lily pad to lily pad, at what angle
must Ferdinand make each of his jumps?
Answer:
θ = 33°
Explanation:
Here, we can use the formula for the total time of flight of a projectile to calculate the launch angle of frog:
[tex]T = \frac{2\ u\ Sin\theta}{g} \\\\Sin\theta = \frac{Tg}{2u}[/tex]
where,
θ = launch angle = ?
T = Total time of flight = 0.6 s
g = acceleration due to gravity = 9.81 m/s²
u = launch speed = 5.4 m/s
Therefore,
[tex]Sin\theta = \frac{(0.6\ s)(9.81\ m/s^2)}{(2)(5.4\ m/s)}\\\\\theta = Sin^{-1}(0.545)[/tex]
θ = 33°
Your tutor says atomic nuclei are converted to nuclear energy in a nuclear reaction. why should you seek a new tutor?
Answer:
The answer is "Momentum and net Energy(that contains mass)".
Explanation:
No matter which type of nuclear reaction, the momentum always is maintained. According to the formula, E = mc^2 of Einstein, energy conservation on the atomic level requires the change in mass of its core which may have been shared throughout the reaction as mass and energy, that's why we can say that the nuclei aren't fully converted into electrical energy rather rearranged and the mass could be changed and dispensed to the nucleon as energy.
4) The SI unit of time is second. why?
Answer:
Second, fundamental unit of time, now defined in terms of the radiation frequency at which atoms of the element cesium change from one state to another. The second was formerly defined as 1/86,400 of the mean solar day—i.e., the average period of rotation of the Earth on its axis relative to the Sun.
source britannica
Explanation:
A 1.2 kg mass is suspended from the ceiling by a string. A second horizontal string holds the mass at rest next to the wall. The angle between the string and the ceiling is 65o. What is the tension force in the horizontal rope
Answer:
The tension in the horizontal string is 25.2 N.
Explanation:
mass, m = 1.2 kg
Angle, A = 65 degree
The tension in the string is T.
So, T cos A = m g
T x cos 65 = 1.2 x 9.8
T x 0.4226 = 11.76
T = 27.83 N
Tension in the horizontal string, T' = T sin A = 27.83 x sin 65 = 25.22 N
Answer:
Explanation:
See the figure attached . T₁ and T₂ are tension in the inclined and horizontal string .
The vertical component of T₁ will balance weight and horizontal component will balance the tension T₂.
T₁ sin 65 = mg and T₁ cos 65 = T₂
Dividing ,
Tan 65 = mg / T₂
T₂ = mg / tan 65
= 1.2 x 9.8 / 2.1445
= 5.5 N
Two people that have identical weight are holding onto a massless pole while standing on horizontal frictionless ice. 1)If the guy on the left starts to pull on the pole, where do they meet
Answer:
Explanation:
From the missing image attached below, it is obvious that there no external force. This implies that they cannot change their position by merely just pulling the ropes. As a result, there will be no movement and no net force will exist.
So, if there is no external force;
The center of mass of the two people is:
[tex]X_{cm}= \dfrac{m_1x_1+m_2x_2}{m_1+m_2} \\ \\ X_{cm}= \dfrac{m(-3m)+m(+3m)} {m+m}\\ \\ X_{cm}= \dfrac{0}{2m} \\ \\ X_{cm} =0[/tex]
Thus, In the system, no movement occurs and all forces remain the same.
Engineers are working on a design for a cylindrical space habitation with a diameter of 7.50 km and length of 29.0 km. The habitation will simulate gravity by rotating along its axis. With what speed (in rad/s) should the habitation rotate so that the acceleration on its inner curved walls equals 8 times Earth's gravity
Answer:
The speed will be "0.144 rad/s".
Explanation:
Given that,
Diameter,
d = 7.50 km
Radius,
R = [tex]\frac{7.5}{2} \ Km[/tex]
Acceleration on inner curve,
= 8 times
Now,
As we know,
⇒ [tex]\omega^2R=8g[/tex]
or,
⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]
⇒ [tex]=\sqrt{\frac{78.4}{3750} }[/tex]
⇒ [tex]=\sqrt{0.0209}[/tex]
⇒ [tex]=0.144 \ rad/s[/tex]
The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N
Answer:
8760 N
Explanation:
think this is the right answer :)
A ballistic pendulum is a device for measuring the speed of a projectile. The projectile is launched horizontally and embeds in a stationary block on the end of a string. The block-projectile system swings upward after the collision, reaching a maximum height. Which of the following statements is correct about the collision between the projectile-block system?
a. Kinetic energy of the system is conserved.
b. Linear momentum of the system is conserved.
c. Linear momentum of the system is not conserved.
d. The total mechanical energy of the system is conserved
A FBD of a rocket launching into space should include:
Answer:
Explanation:
the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.
1./ The upward net force on the space shuttle at launch is 10,000,000 N. What is the least amount of charge you could move from its nose to the launch pad, 60 m below, and thereby prevent it from lifting off
Answer:
[tex]Q=2C[/tex]
Explanation:
From the question we are told that:
Force [tex]F=10000000N[/tex]
Distance [tex]d=60m[/tex]
Where
[tex]Q_1=Q_2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{KQ^2}{r^2}[/tex]
Where
[tex]K=9*10^9[/tex]
Therefore
[tex]Q^2=\frac{Fr^2}{K}[/tex]
[tex]Q^2=\frac{10000000*60^2}{98*10^9}[/tex]
[tex]Q=\sqrt{\frac{10000000*60^2}{9*10^9}}[/tex]
[tex]Q=2C[/tex]
Select the statement that is true:
A. Food chains and food webs have nothing to do with each other.
B. A food web and a food chain are different names for the same thing.
C. A food chain is a part of a food web.
D. A food web is part of a food chain.
Answer:
C
Explanation:
A food chain outlines who eats whom. A food web is all of the food chains in an ecosystem.
Answer:
C
Hope this helped :D
John is going to use a rope to pull his sister Laura across the ground in a sled through the snow. The rope makes an angle of 25 with the ground He is pulling horizontally with a constant force of 400 N. John and manages to get the sled going from 0 to 4 m/s in 5 s. The force due to friction on the sled is 310 N. What is the mass of Laura and the sled combined
which one of the following is not a simple machine
1 ladder
2 wheel Barrow
3 pulley
4 electric pole
A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that d is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.
Required:
a. What is the tension in the cable?
b. Compute the energy stored in the electric field after the top plate was raised.
Answer:
A) F = V²E_o•πr²/2d²
B) U = E_o•Aπr²V²/2d
Explanation:
A) Since we have two circular plates, the formula for the electric field is expressed as;
E = V/d
Where;
V is voltage
d is distance
However, the net electric field produced is given by;
E' = V/2d
The tension in the cable can then be expressed as;
F = qE'
Where q is charge
Thus;
F = qV/2d - - - (eq 1)
We also know that;
C = q/V = E_o•A/d
A is area = πr²
Thus;
q/V = E_o•πr²/d
q = VE_o•πr²/d
Let's put VE_o•πr²/d for q in eq 1 to get;
F = V²E_o•πr²/2d²
B) formula for the energy stored in the electric field is;
U = ½CV²
From earlier, we saw that; C = E_o•A/d
Thus;
U = ½E_o•AV²/d
A = πr²
Thus;
U = E_o•Aπr²V²/2d
A single-story retail store wishes to supply all its lighting requirement with batteries charged by photovoltaic cells. The PV cells will be mounted on the horizontal rooftop. The time-averaged lighting requirement is 10 W/m2 , the annual average solar irradiance is 150 W/m2 , the PV efficiency is 10%, and the battery charging/discharging efficiency is 80%. What percentage of the roof area will the PV cells occupy
Answer:
83.33% of the roof area will be occupied by the PV cells
Explanation:
Given the data in the question;
time-averaged lighting requirement [tex]P_{lighting[/tex] = 10 W/m²
the annual average solar irradiance [tex]q_{solar[/tex] = 150 W/m²
the PV efficiency η[tex]_{pv[/tex] = 10% = 0.1
battery charging/discharging efficiency η[tex]_{battery[/tex] = 80% = 0.8
we know that; Annual average power to the light = [tex]P_{lighting[/tex] × A[tex]_{roof[/tex]
Now, the electrical power delivered by the solar cell battery system will be;
⇒ [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
[tex]P_{lighting[/tex]A[tex]_{roof[/tex] = [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
Such that;
A[tex]_{pv[/tex] = [tex]P_{lighting[/tex]A[tex]_{roof[/tex] / [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = [tex]P_{lighting[/tex] / [tex]q_{solar[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
so we substitute
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / [ 150 W/m² × 0.1 × 0.8 ]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / 12 W/m²
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 0.8333
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = (0.8333 × 100)%
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 83.33%
Therefore, 83.33% of the roof area will be occupied by the PV cells.