Which property describes the distance between similar points on a wave?
A. Speed
B. Wavelength
C. Period
D. Frequency
Answer:
B
Explanation:
Answer:
Explanation:
a
In which mode of heat transfer is the convection heat transfer coefficient usually higher, natural convection or forced convection?why?
Which of the following variables was controlled in the design of Study 2?
Answer:
the following variables
3. What is the force of sliding friction
between a block and the ground if the
coefficient of kinetic friction is uk = 0.17
and the block weighs 18 N?
a. 0.52 N
b. 55 N
C. 18 N
d. 3.1 N
Answer:
the force of the friction is A-0.52
QUICK RETURN MOTION IS USED IN a whAt
Answer:
A quick return motion mechanism is used in the shaper and slotter machine in which the circular motion is converted into reciprocating motion so that the slider moves forward and backwards. In the forward direction, the cutting process occurs while in return there is no such cutting.
3. What is the difference between breathing and respiration?
Answer:
Breathing is the biological process of inhaling and exhaling of the gases between the cells and the environment. The mechanism of breathing involves various respiratory structures such as the windpipe, lungs and nose. Respiration, on the other hand, is a chemical process that takes place in the cell.
Explanation:
Health professionals can help those with health issues to __________.
A.
find something other than sports or exercise in which to participate
B.
match exercise programs and sports with their needs and limitations
C.
get completely well before starting any sports or exercise program
D.
find others with the same health issue so they can commiserate on their unfortunate situation
Please select the best answer from the choices provided.
A
B
C
D
Answer:
B
Explanation:
Answer:
B
Explanation:
On a distance-time graph, the faster the object, the _______ the slope of the line. *
A.curvier
B.flatter
C.lower
D.steeper
The following words are used to describe the direction of a moving object EXCEPT... *
A. East
B. North
C.South
D. Today
Answer:
..In a distance-time graph, the slope or gradient of the line is equal to the speed of the object. The steeper the line (and the greater the gradient) the faster the object is moving. Calculate the speed of the object represented by the green line in the graph, from 0 to 3 s.
d) steeper
D)Today
calculate the moment of 3kg and 20cm away from the pivot
Answer:
5.88 Nm
Explanation:
Applying,
Moment(M) = Force(F)×perpendicular distance(d)
M = F×d.................... Equation 1
But,
F = mg.................. Equation 2
Where m = mass, g = acceleration due to gravity.
Substitute equation 2 into equation 1
M = mgd................. Equation 3
From the question,
Given: m = 3 kg, d = 20 cm = 0.2 m
Constant: g = 9.8 m/s²
Substitute these values into equation 3
M = 3×0.2×9.8
M = 5.88 Nm
Can anybody help meh :(
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross-sectional area in 10s is
Answer:
[tex]n=1.56\times 10^{17}[/tex] electrons
Explanation:
Given that,
Total charge = 9 mC = 0.009 C
0.009 C of charge passes through a wire in 3.6 s.
Let q' is the charge that passes through it in 10 s.
So,
[tex]\dfrac{0.009 }{3.6}=\dfrac{q'}{10}\\\\q'=\dfrac{0.009 \times 10}{3.6}\\\\q'=0.025\ C[/tex]
We know that,
q = ne
Where
n is the number of electrons
So,
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{0.025}{1.6\times 10^{-19}}\\\\n=1.56\times 10^{17}[/tex]
So, [tex]1.56\times 10^{17}[/tex]electrons must pass through the cross-sectional area.
Coherent light with wavelength 591 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?
Answer:
The wavelength will be "1.182 μm".
Explanation:
The given values are:
Wavelength
[tex]\lambda=591 \ nm[/tex]
or,
[tex]=591\times 10^-9 \ m[/tex]
Distance,
[tex]d = 3.00 m[/tex]
[tex]n = 1[/tex]
Distance of fringe from center,
[tex]y = 4.84 \ mm[/tex]
We have to find the wavelength of first order dark fringe,
[tex]\lambda = ?[/tex]
As we know,
⇒ [tex]y_{bright} =\frac{1\times \lambda\times L}{d}[/tex]
On putting the given values in the formula, we get
[tex]0.00484=\frac{1\times (591\times 10^{-9})\times 3}{d}[/tex]
On applying the cross multiplication, we get
[tex]\lambda = \frac{0.00484\times 000036632}{0.5\times 3}[/tex]
[tex]=1182\times 10^{-9}[/tex]
or,
[tex]=1.182 \ \mu m[/tex]
7) A meter stick is being balanced on a triangular piece of wood, with two blocks hanging off of
it, a block with a weight of 10. N is hanging 25 cm to the left of the triangle, and there is another
block with an Unknown weight hanging 50. cm to the right of the triangle.
a) Will the weight of the unknown block be more or less than the weight of the 10. N
block?
b) How much torque is the 10. N block creating?
c) How much torque must the Unknown block be creating?
d) What is the weight of the Unknown block?
Explanation:
a) It will weigh less because of its distance from the block.
b) torque = W1×d1 = (10N)(0.25m) = 2.5 N-m
c) torque = W2×d2 = W2×(0.50m)
d) W2×d2 = W1×d1
W2 = (d1/d2)W1
= 5 N
Can someone help me with this please
Answer:
yea im sorry
Explanation:
A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 25.0 N/m. The block rests on a frictionless surface. A 5.70×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.99 m/s and bounced with the same speed of 8.99 m/s in opposite direction. How far does the block compresses the spring?
The total momentum of the block and putty prior to their collision is
(0.454 kg) (0 m/s) + (5.70 × 10⁻² kg) (8.99 m/s) ≈ 0.512 kg•m/s
and the total momentum after the collision is
(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s)
where v is the velocity of the block. Momentum is conserved, so
(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s) = 0.512 kg•m/s
==> v ≈ 2.26 m/s
The total work done on the block by the spring as it gets compressed by a distance x is equal to the change in the block's kinetic energy:
1/2 (25.0 N/m) x ² = 1/2 (0.454 kg) (2.26 m/s) - 0
==> x ≈ 0.202 m ≈ 20.2 cm
What is the density of iron if 5.0 cm^3 has a mass of 39.5 g
Answer:
density =mass
volume
x = 39.5 ÷ 5.0
x =7.9
Answer:
7.9 g/cm3
Explanation:
Density = mass / volume
Density = 39.5 / 5.0
Density = 7.9 g/cm^3
Therefore, density is [tex]7.9 g/cm^{3}[/tex]
A coil with a circular cross section having an area of has 40turns. When the coil is placed inside a 240cmmagnetic field of 2T, the maximum torque is found to be. (a) Calculate the current in the coil. 6210.xNmi(b)What work is required to rotate the coil by , if the initial orientation of the magnetic dipole moment of 180the loop is at with respect to the field B.
Answer:
Explanation:
cross sectional area A = 40 cm² = 40 x 10⁻⁴ m² , n = 40 , Magnetic field B = 2 T . Current = I .
Maximum torque = 2 x 10⁻⁶ Nm
Value of maximum torque = n B A I .
n B A I = 2 x 10⁻⁶ Nm
40 x 2 x 40 x 10⁻⁴ I = 2 x 10⁻⁶ Nm
.32 I = 2 x 10⁻⁶ Nm
I = 2 x 10⁻⁶ / 32 x 10⁻²
= .0625 x 10⁻⁴
= 6.25 x 10⁻⁶
= 6.25 μ A .
b )
Work done in rotating a coil = n B A I ( cos θ₂ - cosθ₁ )
= n B A I ( cos θ₂ - cosθ₁ )
= 2 x 10⁻⁶ ( cos 180 - cos θ )
= 2 x 10⁻⁶ ( -1 - cos θ )
= - 2 x 10⁻⁶ ( 1 + cos θ ) J .
b
Jasmine plays a note of wavelength 22 cm on her clarinet. The speed of
sound in air is 340 m/s. What is the frequency of this note? Give your
answer to 3 significant figures.
Answer:
Frequency = 1,550Hz
Explanation:
To solve this we can use the equation: [tex]f=\frac{v}{\lambda}[/tex]
(frequency = velocity/wavelength).
We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m
Now we can substitute these values into the formula and calculate to solve:
[tex]f=\frac{340}{0.22} \\\\f=1545.454...[/tex]
Simplify to 3 significant figures:
f = 1,550Hz
(Which I believe is just below a G6 if you were interested)
Hope this helped!
A 9.83 -m ladder with a mass of 20.3 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 242 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.64 rad/s2 about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder
Answer:
The net torque will be "1366.33 Nm".
Explanation:
The given values are:
Length of ladder,
[tex]\tau_{242N} = 9.83 \ m[/tex]
then,
[tex]\tau=4.92[/tex]
Mass,
= 20.3 kg
Now,
The net torque will be:
⇒ [tex]\tau_{net}=\tau_{mg}+\tau_{242}[/tex]
On putting the given values, we get
⇒ [tex]=4.92\times (-mg)+9.83\times (+242)[/tex]
⇒ [tex]=4.92\times (-21\times 9.8)+9.83\times 242[/tex]
⇒ [tex]=-1012.53+2378.86[/tex]
⇒ [tex]=1366.33 \ Nm[/tex]
Which is true of the interaction of charges?
A. like charges repel, unlike charges, attract
B. like charges attract, unlike charges, repel
C. like charges repel, unlike charges, repel too
D. like charges attract, unlike charges, are attracted too
Answer:
A. like charges repel, unlike charges, attract
Explanation:
A charge can be positive or negative.
Like charges, example positive and positive or negative and negative, repel each other, while unlike charges, example positive and negative attract each other.
From the given options, the correct option is A.
A. like charges repel, unlike charges, attract
The statement which is true for the interaction of charges is the like charges repel, unlike charges, attract (A).
What is the property of interacting charges?When the two charges or the charged bodies interact each other, the force of attraction or repulsion acts between them. This force of attraction or repulsion can be find out using the coulombs' law.
This force of attraction or repulsion is identified by the sign of the particular charge. According to the property of interacting charges-
Like charges-For the like charges, when two charges interact to each other, then the charges repel each other.Unlike charges-For the unlike charges, when two charges interact to each other, then the charges attract each other.Thus, the statement which is true for the interaction of charges is the like charges repel, unlike charges, attract (A).
Learn more about the property of interacting charges here;
https://brainly.com/question/12455287
For a damped oscillator with a mass of 360 g, a spring constant 130 N/m and a damping coefficient of 86 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 11 cycles
Answer:
A/A₀ = 0.5106
Explanation:
To do this, we need to use several formulas and expressions. First, we need to know the period of time of the oscillator. This can be calculated using the following expression:
ω = 2π/T (1)
But angular frequency (ω) can be also be calculated using:
ω = √(k/m) (2)
Using (1) and (2), we can solve for the period T:
2π/T = √(k/m) (mass in kg)
2π/T = √(130/0.360)
2π/T = √361.11
2π/T = 19.003
T = 2π/19.003 = 0.331 s
Now, the amplitude A at a time t, is:
A = x exp(-bt/2m) (3)
At time 0, A = x. so A₀ = x
The problem states that we have 11 cycles respect to the initial amplitude, so expression (3) can be rewritten as:
A = x exp(-b(17t/2m)) using b as kg/s = 0.086 kg/s
Replacing the data we have:
A = x exp(-0.086(17*0.331)/2*0.36)
A = x exp(-0.086 * 7.815)
A = x exp(-0.6721)
A = 0.5106x (4)
Now, doing the ratio with the innitial we have:
A / A₀ = 0.5106x / x
The ratio is:
A/A₀ = 0.5106Hope this helps
The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner surface is reflective. When light is incident on the uncoated side it passes throughthe lens, reflects off the silver coating, and passes back through the lens. The overall effect is that of amirror with focal length 5.0 cm. What is the index of refraction of the lens material
Answer:
n = 1.4
Explanation:
Given,
R1 = 18 cm, R2 = -18 cm
From lens makers formula
1/f = (n - 1)(1/18 + 1/18) = (n-1)/9
f = 9/(n-1)
Power, P = 1/f ( in m) = (n-1)/0.09
Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens
Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09
Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05
n = 1.4
Convert 0.700 to scientific notation
Answer:
7 × 10^-1
Explanation:
Hope this helps :) mark branliest if you could!
Answer: Therefore, the scientific notation is [tex]7 *10 {}^{ - 1}[/tex]
Step by step explanation:To solve a scientific notation you must count how many zeros they have, this always starts from the first 0, in some cases you always start by putting any number but that is not 0 at the beginning, these exercises are no longer negative scientific notations, if not they are positive.
An example:[tex]8840 = 8.84 \times 10 {}^{3} [/tex]
Always to write a positive exercise you should always take the following:
The coefficient "C" must be determined by traversing the decimal point. You must also determine the exponent "n" always counting how many places the comma went, if it went to the left "n" it will be a positive exponent, but if "n" goes to the right it will then be negative.And finally, the leading and trailing zeros of the coefficient are eliminated.
So, Therefore, The result of the scientific notation of the problem is [tex]7 *10 {}^{ - 1}[/tex]
You can see more about scientific notation in: https://brainly.com/question/17947470
While diving in cancun Mexico where the seawater has a density of 1,015 kg/m3 Nana observed that her pressure meter device reading was 3.75 atm. The reading at sea level is standard 1.0 atm. At what depth is she diving when the meter read 3.75 atm g
Answer:
The depth of the diver is 28.01 m
Explanation:
Given;
density of the seawater, ρ = 1,015 kg/m³
standard sea level pressure, P₀ = 1.0 atm = 101,325 Pa
the final reading of her pressure, P₁ = 3.75 atm = 379968.75 Pa
acceleration due to gravity, g = 9.8 m/s²
Let the depth she was diving at the final pressure = h
This depth is calculated as;
P₁ = P₀ + ρgh
P₁ - P₀ = ρgh
[tex]h = \frac{ P_1 \ - \ P_o}{\rho g} = \frac{379968.75 \ - \ 101325}{1015 \ \times \ 9.8} = 28.01 \ m[/tex]
Therefore, the depth of the diver is 28.01 m
How are satellites used in GPS?
to keep the International Space Station in orbit
to communicate with the International Space Station
to communicate with possible extraterrestrial life
to pinpoint specific locations on Earth
Answer:
To pinpoint specific locations on Earth
Explanation:
The satellite has a full specific view of a specific area just like other satellites do. Satellites communicate together which also communicates with your GPS to give you an accurate location where you are or where you want to be. That is how satellites are used in GPS.
Hope this helps you out :)
Among all the electromagnetic waves (EM), which has the highest frequency? *
A. INFRARED RADIATION
B. RADIO WAVE
C. ULTRAVIOLET
D. GAMMA RAYS
How do transverse waves move? *
A. The particles move up and down to the direction in which the waves travel.
B. The particles move back and forth to the direction in which the waves travel.
C. The particles move in circular paths to the direction in which the waves travel.
D. The particles move back and forth and side to side to the direction in which the waves travel.
20. How do we know that waves carry energy? *
A. Waves causes disturbances
B. Waves can set other objects into motion
C. Waves must travel in a medium
D. Waves move back and forth
21. What happens when waves pass by? *
A. The particles vibrate
B. The particles stop moving
C. The particles move forward
D. The particles remain stationary
22. Which of the following correctly relates wavelength and frequency? *
A. The higher the wavelength, the higher the frequency.
B. The lower the wavelength, the higher the frequency.
C. The lower the wavelength, the lower the frequency.
D. The wavelength is directly related to frequency.
23. Which of the following is the fastest? *
A. A wave wit h 10 m wavelength and a frequency of 30 Hz.
B. A wave with 30 m wavelength and a frequency of 2 Hz.
C. A wave with 20 m wavelength and a frequency of 2 Hz
D. A wave with 5 m wavelength and a frequency of 60 Hz.)
24. How can we generate mechanical waves? *
A. By producing light energy
B. By producing horizontal motion
C. By vibrating a medium
All of the above
25. A high energy electromagnetic wave has ____ *
A. Low frequency
B. High Frequency
C. Zero wave speed
D. Highest wave speed
first is gamma Ray's, last is d
At what speed is the kinetic energy of a particle twice its Newtonian value?
While scuba diving, you have injured yourself and must signal the boat on the surface for help. As you always do when diving, you have five laser pointers with you, each emitting light of a different color: red, orange, yellow, green, and blue. When you shine the yellow light up toward the boat, the beam makes such a small angle with the surface of the water that all the light is reflected back into the water.
Required:
a. If you are not able to move closer to the boat, which color or colors of light might you try?
b. Are these other lights likely to be successful in penetrating the surface of the water?
Answer:
a. Red and Orange
b. Yes, they are
Explanation:
a. This is because, since the yellow light is reflected back into the water, it undergoes total internal reflection and its wavelength in water is not long enough to allow it penetrate the water surface. Since the wavelength decreases from left to right, both the green and blue light have shorter wavelength than the yellow light, so they get reflected back. The red and orange lights are more likely to penetrate since they have longer wavelengths than the yellow light.
b. This is because the red and orange light have longer wavelengths and are thus more likely to penetrate the water surface and not get reflected back.
In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube which can lead to standing waves. Suppose 0.7 MHz ultrasound is directed through a layer of tissue at a bone 0.55 cm below the surface. Remember, sound waves in the body travel at 1540 m/s. Explain.
Answer:
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
Explanation:
In this experiment the system can be approximated as a tube with one end open and the other closed.
The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m
a node is formed at the closed end and a belly at the open end, so the resonance has the form
λ = 4L 1st harmonic
λ = 4/3 L third harmonic
λ = 4/5 L fifth harmonic
λ = 4L / (2n + 1) n = 0, 1, 2, (2n + 1)
This wave is a standing wave therefore energy density remains in place
P = 1/2 ρ v (w S₀)²
angular velocity is related to frequency
w = 2π f
we substitute
E = P = 2π² ρ v f² S₀²
if this energy per unit area is transformed into heat
E = m c_e DT
let's use the concept of density
ρ = m / V
m = ρ V
if there are no losses in the system
½ π² ρ v f² S₀² = ρ V c_e ΔT
ΔT = [tex]\frac{\pi ^2 \ v \f^2 S_o^2}{2V \ c_e}[/tex]
When analyzing this expression the temperature increase is
* quadratic at the frequency and maximum amplitude of the wave
* proportional to the speed of the wave in the tissue
* inversely proportional to tissue volume
we can approximate the volume of the tissue to the volume of a cylinder tube
V = π r² L
ΔT = [tex]\frac{\pi \ v \ f^2 S_o^2 }{r^2 \ L \ c_e}[/tex]
we calculate
ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
1. Determine the potential energy of a 2kg rock at the top of a hill that is 20m high.
2. Determine the kinetic energy of a 2000 kg roller coaster car that is moving with a speed of 25 m/s.
3. A 80kg freezer is located in an office on the 59 floor of an office building 200 meters above the ground. What is the potential energy of the freezer?
Answer:
1) 392 joules
2) 625,000 joules
3) 156,800 joules
Explanation:
Gravitational potential energy = mgh. m = mass in kg, g = acceleration due to gravity and h = height in meters.
Kinetic energy = 1/2(m)(v)^2, m = mass in kg, v = velocity in meters per second
1) P.E = 2 x 9.8 x 20 = 392 joules
2) K.E = 1/2 x 2000 x (25)^2 = 625,000 joules
3) P.E = 80 x 9.8 x 200 = 156,800 joules
Answer:
[tex]1)\:392\:\text{J}\:(400\:\text{J with one significant figure)},\\2)\:625,000\:\text{J}\:(600,000\:\text{J with one significant figure)},\\3)\:156,800\:\text{J}\:(200,000\:\text{J with one significant figure)},[/tex]
Explanation:
1. The potential energy of an object is given by [tex]PE=mgh[/tex]. Substituting given values, we have:
[tex]PE=2\cdot 9.8\cdot 20=\boxed{392\:\text{J}}[/tex]
2. The kinetic energy of an object is given by [tex]KE=\frac{1}{2}mv^2[/tex]. Substituting given values, we have:
[tex]KE=\frac{1}{2}\cdot 2000\cdot 25^2 =\boxed{625,000\: \text{J}}[/tex]
3. 1. The potential energy of an object is given by [tex]PE=mgh[/tex]. Substituting given values, we have:
[tex]PE=80\cdot 9.8\cdot 200=\boxed{156,800\:\text{J}}[/tex]
why do blades come in different lengths in a jig saw sander?
Answer:
To determine the minimum blade length, add 1" to the workpiece thickness. One type of material, and some materials can be cut by more than one type of blade. No matter the material, there's likely a jigsaw blade designed specifically for. Armed with the right blade, follow these pointers to make your work go (and cut) .
Explanation: