Answer:
2 to 3 weeks
Explanation:
A seated musician plays a C4 note at 262 Hz . How much time Δ does it take for 346 air pressure maxima to pass a stationary listener?
Answer:
t = 1.32 s
Explanation:
We are given;. Frequency of C4 note; F_c = 262 Hz
In conversions, we know that 1 Hz = 1 cycle/s
Thus, F_c = 262 cycles/s
Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.
346 air pressure maxima denotes that the air pressure maxima is 346 cycles.
Thus, time will be;
t = 346 cycles/262 cycles/s
t = 1.32 s
The time taken for the musical note to pass the stationary listener is 1.32 s.
The given parameters:
frequency of the C4 note, f = 262 Hzair pressure maximum, n = 346The frequency of a sound wave is defined as the number of cycles completed per second by the wave.
[tex]F = \frac{n}{t}[/tex]
where;
t is the time to compete the maximum cycleThe time taken for the musical note to pass the stationary listener is calculated as follows;
[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]
Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.
Learn more here:https://brainly.com/question/15613196
A radio signal has a frequency of 1.023 x 108 HZ. If the speed of the signal in air is 2.997 x 108m/s, what is the wavelength of the signals? а 7.15 m b 5.23 m C 2.93 m d 0.93 m
Answer:
2.93 m (which agrees with answer "C" on the list)
Explanation:
Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:
Wavelength = 2.997 10^8 / 1.023 10^8 = 2.93 m
A car traveling initially at a speed of 20 m/s accelerates to a speed of 31 m/s over a distance of 45 meters.
What is the magnitude of the car's acceleration?
Answer:abc defg hijk lmnop qrs tuv wx y and z
Explanation: now i know my abc's
If a car has a centripetal acceleration of 7m/s2 over a radius of 7m. How fast is it going
a. 7 m/s
b 1 m/s
с o m/s
d 49 m/s
Answer:
7 m/s (agrees with answer a in your list)
Explanation:
Recall that the centripetal acceleration is defined by the square of the tangential velocity divided by the radius of the rotational motion:
[tex]a_c=\frac{v_t^2}{R}[/tex]
then the tangential velocity is extracted from here as:
[tex]a_c=\frac{v_t^2}{R} \\v_t^2=a_c * R\\v_t=\sqrt{a_c * R}[/tex]
in our case, this becomes:
[tex]v_t=\sqrt{7*7} = 7 \,\,m/s[/tex]
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
15 points.
An object of mass 100 kg is observed to accelerate at a rate of 15
m/s/s. Calculate the force required to produce this acceleration.
Answer:
its 0.5 for all i beleive
Explanation:
what are the laws of newton
Answer:
Explanation:
These are the laws of Newton
Answer:
the first law, an object will not change its motion unless a force acts upon it. the 2nd one, the force of an object is equal to its mass times it acceleration. the 3rd one is when 2 objects interact, they apply forces to each other of equal magnitude and opposite direction.
Use the drop-down menus to complete each sentence.
A plant grows toward a sunny window. This response is an example of
✔ phototropism
.
Sometimes a plant grows around a tree for support. This response is an example of
✔ thigmotropism
.
The roots of a plant grow toward a water source. This response is an example of
✔ hydrotropism
.
The roots of a plant grow down into the soil. This response is an example of
✔ gravitropism
.
the answers are already there
Answer:
The correct answer is - phototropism, thigmotropism, hydrotropism, and gravitropism in order ( already match correctly).
Explanation:
phototropism is a phenomenon in which plants grow towards the light or sun which is accomplished by the hormone auxin in the cells far from the light.
Thigotropism is a type of plant growth that occurs around the tree to support itself which is a touch stimuli response.
The movement of the plant in the direction of the source of the water is known as hydrotropism. In which stimuli is humidity or the water concentration.
The movement of the plant or roots of the plants towards the soil or earth is known as gravitropism here gravity is the stimuli.
Answer:
Use the drop-down menus to complete each sentence.
A plant grows toward a sunny window. This response is an example of
phototropism
.
Sometimes a plant grows around a tree for support. This response is an example of
thigmotropism
.
The roots of a plant grow toward a water source. This response is an example of
hydrotropism
.
The roots of a plant grow down into the soil. This response is an example of
gravitropism
.
Explanation:
A long wire carries a current 5 A from west to east. A magnetic compass pointing North is placed underneath the wire at a distance of 2 mm. What is the deflection of the compass when it is placed under the wire?
Answer:
no deflection
Explanation:
current is flowing from west to east. As the magnetic field of a long wire carrying current is circular, its direction will be north below the wire and south above the wire (according to the right hand rule). So, when the compass is placed underneath the wire, it will still point towards the north direction.
The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force
Answer:
5N
Explanation:
Given parameters:
Original length = 22cm
Spring constant, K = 50N/m
New length = 32cm
Unknown
Force applied = ?
Solution:
The force applied on a spring can be derived using the expression below;
Force = KE
k is the spring constant
E is the extension
extension = new length - original length
extension = 32cm - 22cm = 10cm
convert the extension from cm to m;
100cm = 1m;
10cm will give 0.1m
So;
Force = 50N/m x 0.1m = 5N
If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (3a1)
Question 10 options:
There cannot be any forces applied to the ball.
There must be exactly one force applied to the ball.
The net force applied to the ball is zero.
The net force applied to the ball is directed to the right.
Answer:
C. The net force applied to the ball is zero.
Explanation:
From Newton's second law of motion;
F = ma
Where F is the force on an object, m is its mass and a is its acceleration.
Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.
Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.
So that;
F = m x 0
= 0
No force is applied on the object.
Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.
A 37.0-kg child swings in a swing supported by two chains, each 3.06 m long. The tension in each chain at the lowest point is 410 N. (a) Find the child's speed at the lowest point.______m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)_______ N(upword)
Answer:
1. 6.15m/s
2. 820N
Explanation:
The total upward force
= 410x2
= 820
g = 9.81
a = v²/r
= 2xT - msg = m x v²/r
= 820-37*9.81 = 37v²/3.06
= 820-362.97 = 37v²/3.06
= 457.03 = 12.09v²
To get v²
V² = 457.03/12.09
V² = 37.8
V = √37.8
V = 6.15m/s
B. We already have the answer to this question
The force exerted is simply gotten by this calculation
2x410
= 820N
Sona wants to install room heater in her living room. She had only two options, either to install heater at the top of the window or near the ground level. . Which method of installing room heater would be the effective way . Why or Why not ?
In 1993, Wayne Brian threw a spear at a record distance of 201.24 m. (This is not an official sports record because a special device was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with respect to the horizontal. What was the initial speed of the spear? 2. Find the maximum height and time of flight of the spear in problem #1.
I really don't know how to do any of this please help me :(
Answer:
V₀ = 45.81 m/s
H = 70.45 m
T = 5.36 s
Explanation:
The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 201.24 m
V₀ = Initial Speed = ?
θ = Launch Angle = 35°
g = 9.8 m/s²
Therefore,
201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²
V₀ = √[(201.24 m)/(0.095 m/s²)
V₀ = 45.81 m/s
Now, for maximum height:
H = V₀² Sin² θ/g
H = (45.81 m/s)² Sin² 35°/9.8 m/s²
H = 70.45 m
For the total time of flight:
T = 2 V₀ Sin θ/g
T = 2(45.81 m/s) Sin 35°/9.8 m/s²
T = 5.36 s
Calculate the distance covered by a bus moving at a rate of 11.5km/h with a time of 2060seconds
Answer:
6.56km
Explanation:
Given parameters:
Speed = 11.5km/hr
Time = 2060s
Unknown:
Distance covered = ?
Solution:
Speed is distance divided by the time taken.
Speed = [tex]\frac{distance}{time}[/tex]
Distance = Speed x time
Let us convert the seconds to hours;
3600s = 1hr
2060s = [tex]\frac{2060}{3600}[/tex] = 0.57hr
Now
Distance = 11.5km/hr x 0.57hr = 6.56km
A car is moving at an average speed of 20 meters per second. This is equivalent to
Answer:
44.73 MP/H or 71.98 KM/H
Explanation:
I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?
a. 250 kg
c. 10 Kg
b. .002 Kg
d. 45 Kg
Answer:
a. 250kg I think it's the right answer. hope it helps:)
Answer:
C.10
Explanation:
because when you divide 50 divided by 5 = 10
2. An ambulance traveling at 20 m/s emits a sound at 500 Hz. What frequency does a person standing on the corner of a street detect?
What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren
Answer:
77.6 Hz
Explanation:
What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren.
The given parameters are:
Observer speed = 115 km/h
Source frequency = 753 Hz
Speed of sound = 342 m/s
First convert km/h to m/s
Observer speed = (115 × 1000) / 3600
Observer speed = 31.94 m/s
The frequency received by the person will be:
F = fv / ( V - v )
Where
F = frequency received by the person
f = siren frequency
V = speed of sound
v = speed of the ambulance
Substitute all the parameters into the formula
F = (753 × 31.94) / ( 342 - 31.94 )
F = 24050.82 / 310.06
F = 77.568 Hz
Therefore, the frequency received by a person is approximately 77.6Hz
if a toy car has a centripetal acceleration of 50 m/s2 and was making the turn at 10 m/s. what was his radius
a 2m
ь 500m
C 5m
d.25m
Answer:
Explanation:
The centripetal acceleration is expressed as;
a = v²/r
a is the acceleration = 50m/s²
v is the velocity = 10m/s
r is the radius
To get the radius
r = v²/a
r = 10²/50
r = 100/50
r = 2m
Hence its radius is 2m
A crate is pulled due south with a force of 350. N. What other force must be applied if the
net force on the crate is 425 N due north? Enter the magnitude (with units) and direction
(north, south, east, west).
Answer:
775 N due North.
Explanation:
If the crate is pulled South with 350 N force, and the net force on the crate results into 425 N due North, then the other force (F) acting must be larger than the 350 N, and pointing North:
F - 350 N = 425 N
F = 425 N + 350 N = 775 N due North.
An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 and the temperature remains constant?
Answer:
2.22 kPaExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
From the question we have
[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]
We have the final answer as
2.22 kPaHope this helps you
Use the information below for the next five questions:
An open organ pipe emits B (494 Hz) when the temperature is 14°C. The speed of sound in air is v≈(331 + 0.60T)m/s, where T is the temperature in °C
Determine the length of the pipe.
What is the wavelength of the fundamental standing wave in the pipe?
What is frequency of the fundamental standing wave in the pipe?
What is the frequency in the traveling sound wave produced in the outside air?
What is the wavelength in the traveling sound wave produced in the outside air?
How far from the mouthpiece of the flute should the hole be that must be uncovered to play D above middle C at 294 Hz? The speed of sound in air is 343 m/s.
Answer: Please see answer in explanation column.
Explanation:
Given that
v≈(331 + 0.60T)m/s
where Temperature, T = 14°C
v≈(331 + 0.60 x 14)m/s
v =331+ 8.4 = 339.4m/s
In our solvings, note that
f= frequency
λ=wavelength
L = length
v= speed of sound
a) Length of the pipe is calculated using the fundamental frequency formulae that
f=v/2L
Length = v/ 2f
= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m
b) wavelength of the fundamental standing wave in the pipe
L = nλ/2,
λ = 2L/ n
λ( wavelength )= 2 x 0.3435/ 1
= 0.687m
c) frequency of the fundamental standing wave in the pipe
F = v/ λ
= 339.4m/s/0.687m=
494.03s^-1 = 494 Hz
d) the frequency in the traveling sound wave produced in the outside air.
This is the same as the frequency in the open organ pipe = 494Hz
e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m
f) To play D above middle c . the distance is given by
L =v/ 2 f
= 343/ 2 x 294
=0.583m
A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is launched to a maximum height 50.2 cm. How much should the spring be compressed to send the ball twice as high?
We know, by conservation of energy :
[tex]\dfrac{kx^2}{2}=mgh[/tex]
Therefore,
[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}[/tex]
Putting given values, we get :
[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm[/tex]
Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.
Hence, this is the required solution.
A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself.
Answer:
m = 876.71 kg
Explanation:
This is an exercise of Archimedes' principle, which states that the thrust on a body is equal to the weight of the dislodged liquid
B = ρ g V
therefore the load that the balloon can lift is
B - W_structure - w_load = 0
w_load = B - W_structure
The volume of the balloon is
v = 4/3 π r³
let's substitute
w_carga = rho g 4/3 π r³ - m_structure g
the air density at T = 25ºc is ρ = 1.18 kg / m³
let's calculate
w_load = 1.18 9.8 4/3 π 7.15³ - 930 9.8
w_load = 17705,77 - 9114
w_ load = 8591.77 N
this corresponds to a mass of
w_load = m g
m = w_load / g
m = 8591.77 / 9.8
m = 876.71 kg
If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the internal energy of the gas?
Answer:
The change in the internal energy of the gas 1,595 J
Explanation:
The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)
This is summarized in the expression:
ΔU= Q - W
where the heat absorbed and the work done by the system on the environment are considered positive.
Taking these considerations into account, in this case:
Q= 500 cal= 2,092 J (being 1 cal=4.184 J) W=500 JReplacing:
ΔU= 2,092 J - 500 J
ΔU= 1,592 J whose closest answer is 1,595 J
The change in the internal energy of the gas 1,595 J
What is Energy in physics?
Answer:
Energy, in physics, the capacity for doing work. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another.
Explanation:
Hope this helps!
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will ________ because the toy car ____
Answer:
If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration.
Explanation:
I hope this helped
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car gains more potential energy at the higher starting point.
When a toy car moves along a sloped racetrack, it converts its potential energy (due to its height above the ground) into kinetic energy (energy of motion). The higher the starting height of the racetrack, the more potential energy the toy car possesses initially.
As the toy car moves down the sloped track, it will accelerate due to the force of gravity. The potential energy is converted into kinetic energy, and the car's speed increases. According to the law of conservation of energy, the total mechanical energy (sum of potential and kinetic energy) remains constant as long as no external forces, such as friction, act on the car.
Therefore, if the starting height of the racetrack is increased, the toy car will have more potential energy to start with. As it moves down the track, it will convert this increased potential energy into kinetic energy, resulting in a higher speed compared to when it starts from a lower height.
To know more about speed here
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explain the relationship among visible light, the electromagnetic spectrum, and sight.
Explanation:
The electromagnetic spectrum is the name given to the full range of frequencies and/or wavelengths that electromagnetic phenomena may have.
Human eyes respond to a small range of wavelengths in that spectrum. That response is called sight. Because humans can see that electromagnetic energy, it is called visible light.
In the video your blood is compared to a __________________ that delivers oxygen to your body and picks up CO2 to be released out when you breath.
Answer:
Delivery truck
Explanation: