Explanation:
Water is wet, in the sense of being a liquid which flows easily, because its viscosity is low, which is because its molecules are rather loosely joined together.
Answer:
Liquid water is not itself wet, but can make other solid materials wet.
Wetness is the ability of a liquid to adhere to the surface of a solid, so when we say that something is wet, we mean that the liquid is sticking to the surface of a material.
Whether an object is wet or dry depends on a balance between cohesive and adhesive forces. Cohesive forces are attractive forces within the liquid that cause the molecules in the liquid to prefer to stick together. Cohesive forces are also responsible for surface tension. If the cohesive forces are very strong, then the liquid molecules really like to stay close together and they won't spread out on the surface of an object very much. On the contrary, adhesive forces are the attractive forces between the liquid and the surface of the material. If the adhesive forces are strong, then the liquid will try and spread out onto the surface as much as possible. So how wet a surface is depends on the balance between these two forces. If the adhesive forces (liquid-solid) are bigger than the cohesive forces (liquid-liquid), we say the material becomes wet, and the liquid tends to spread out to maximize contact with the surface. On the other hand, if the adhesive forces (liquid-solid) are smaller than the cohesive forces (liquid-liquid), we say the material is dry, and the liquid tends to bead-up into a spherical drop and tries to minimize the contact with the surface.
Water actually has pretty high cohesive forces due to hydrogen bonding, and so is not as good at wetting surfaces as some liquids such as acetone or alcohols. However, water does wet certain surfaces like glass for example. Adding detergents can make water better at wetting by lowering the cohesive forces . Water resistant materials such as Gore-tex fabric is made of material that is hydrophobic (water repellent) and so the cohesive forces within the water (liquid-liquid) are much stronger than the adhesive force (liquid-solid) and water tends to bead-up on the outside of the material and you stay dry.
A scientist decreases the wavelength of the light used in a double-slit experiment and keeps every other aspect the same. What will be true
about the new interference pattern seen on the screen compared to the original interference pattern? (point)
O The spacing between the bright fringes will increase.
O The spacing between the dark fringes will remain the sam
same.
O The spacing between the bright fringes will decrease.
O The spacing between the dark fringes will increase.
A scientist decreases the wavelength of the light used in a double-slit experiment and keeps every other aspect the same. The new interference pattern seen on the screen compared to the original interference pattern is the spacing between the bright fringes will increase. Thus, the correct option is A.
What is Interference pattern?Interference is the net effect of the combination of two or more wave forms or trains which are moving on intersecting or coincident paths in a medium. The effect of this pattern is that of the addition of the amplitudes of the individual waves at each point which are affected by more than one wave in the medium.
A scientist when decreases the wavelength of the light used in a double-slit experiment and it keeps every other aspect of the wave the same. The new interference pattern seen on the screen as compared to the original interference pattern is that the spacing between the bright fringes will increase.
Therefore, the correct option is A.
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As water is cooled, its density increases until it reaches about
O A. 25° C
B. -2° C
O c. 4°C
O D.O°C
Answer:
4 degrees C
Explanation:
this is just a 'known'
a boat travels 500 m down a strait river. if starts from the rest accelerates uniformly to a velocity of 5m/s. how long does this take?
PLSSSS HELPPPPP NOWWW!!!!!
Two children are playing tetherball, in which a ball at the end of a cord spins around a pole. After a really good hit, the ball makes three complete revolutions in 2.0 s. What is the angular speed of the ball?
Answer:
3 pi R / s or 9.42 R/s or 540 degrees/sec
Explanation:
Ball covers 2 pi radians each rotation
2 pi * 3 R / 2 s = 3 pi R / s
the angular speed of the ball is 3 pi R / s or 9.42 R/s or 540 degrees/sec
What is angular speed ?Angular speed can be defined as the measures of speed of how fast the central angle of a rotating body changes with respect to time
Angular speed mainly implies how quickly the rotation of an object occur means it is described as the change in the angle of the object per unit of time.
To calculate the speed of a rotational motion the angular speed value need to be known, hence The angular speeds formula is used to calculate the distance traveled by an object in terms of rotation and revolutions per unit of time.
The unit of angular speed of an object is radian per second and Both angular speed, angular velocity are determined by using the same formula, angular velocity is vector quantity which describes both magnitude and direction.
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What are the essential components of successfully competing in the field events:
Power, technique, and timing
Power, speed, and strength
Technique, flexibility, and speed
None of the above
Answer:
Power, technique, and timing
Explanation:
I just did the quiz and it was correct!! Hope this helps!
List the chemicals and apparatus
Answer:
Beaker
Graduated Cylinders
Volumetric flask
Test tube
Funnel
What season would Texas be having at point D?
Answer:
Almost winter so fall
Explanation:
Hope this helps! Please let me know if you want me to elaborate more or think my answer is incorrect. Brainliest would be MUCH appreciated. Have a wonderful day!
A 75-W light bulb is turned on. It has an operating voltage of 120 V. (A)How much current flows through the bulb? (B)What is the resistance of the bulb? (C)How much energy is used each second?
Required Answer:
Given:
Power (P) = 75 W Voltage (V) = 120 V(A)
As we know that,
I = P/V↠ Current = 75/120
↠ Current = 0.625 A
(B)
R = V/IHere,
R is resistanceV is Voltage, and I is current↠ Resistance = 120/0.625
↠ Resistance = 192 Ω
(C)
Energy = P × tHere,
P is power T is time↠ Energy = 75 × 1
↠ Energy = 75 J
Two horses pull horizontally on ropes attached to a stump. The two forces Fu and F2 that they apply to the stump are such that the net (resultant) force R has a magnitude equal to that of Fu and makes an angle of 90° with Fu. Let F1 = 1300 N and R = 1300 N also. Find the magnitude of F and its direction (relative to Fi)
For two horses pull horizontally on ropes attached to a stump, the magnitude of F and its direction is mathematically given as
F=1711 N
dF= 135 degrees
What is the magnitude of F and its direction?
Generally, the equation for the components of F1 is mathematically given as
For The x axis
1210 - F2x = 0
F2x = 1210
For the y axis
F2y = R
F2y = 1210
In conclusion
F = sqrt(2)*1210
F=1711 N
Fopt the direction of F2
dF= 90 + 45
dF= 135 degrees
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-What is the potential energy at point A?
-What is the kinetic energy at point A?
-What is the kinetic energy at point B?
-What is the potential energy at point D?
-What is the kinetic energy at point D?
Answer:
1560520156Explanation:
Assuming there is no friction or other force involved, recall that energy is conserved in a system as long as no external force acts on the system.
Using the data from point C, we can find out that the total energy of the system is 156 because [tex]E = K+Pe[/tex].
Since at point A the object doesn't move, it has Kinetic energy of 0, because [tex]K=\frac{1}{2} mv^2[/tex], therefore [tex]0=\frac{1}{2} m0^2[/tex]. However at point A it has maximum Potential energy, because [tex]Pe=mgh[/tex].
At point B, we can find the Kinetic energy by using [tex]E = K+Pe[/tex]. Substitute values:
[tex]156=104+K\\52=K[/tex]
At point D, the object has maximum kinetic energy and no potential energy, therefore it's the opposite of point A.
• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.
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A supertanker uses a windlass (a type of winch) like the one shown in the figure below to hoist its 21,200-kg anchor. Determine the force that must be exerted on the
outside wheel to lift the anchor at constant speed, neglecting friction and assuming the anchor is out of the water.
IN
doorg
-0.45 m
- 1.5m
The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
Force exerted outside the wheelThe force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.
∑τ = 0
Let the distance traveled by the load = 1.5 mLet the radius of the wheel or position of the force = 0.45 m∑τ = R(mg) - r(F)
rF = R(mg)
0.45F = 1.5(21,200 x 9.8)
F = 6.925 x 10⁵ N.
Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
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theshold frequency ???
Answer:
The threshold frequency is defined as the minimum frequency of the incident radiation below which photoelectric emission or emission of electrons is not possible. The threshold frequency refers to the frequency of light that will cause an electron to dislodge emit from the surface of the metal.
a true statement of
kinetic theory
Answer:
real kinetic theory means that kinetic energy
What is refractive index?
[tex] \huge \mathbb \red{HEY \: THERE ♡}[/tex]
Refractive Index:The ratio of the sine of angle of incidence to the sine of angle of refraction in case of lens for light of a given colour and given pair of media is constant. Note** It is also called Snell's Law of Refraction. [tex] \mathsf \orange{\frac{sine \: i}{sine \: r} = constant}[/tex][tex] \huge \mathbb\pink{HOPE \: IT \: HELPS}[/tex]
Spring
The pulling force acting on an object is 350.0 N. It also is being affected by a 50.0 N
kinetic friction force. As a result, its acceleration is 3.0 m/s?, what is its mass?
A. 67 kg
B. 83 kg
C. 100 kg
D. 600 kg
E. 0.0 kg
If the gravitational force of the Farth nulling you"down" in th
Answer:
100 kg
Explanation:
Net force = 350 - 50 = 300 N
F= ma
300 N = m * 3 m/s^2
100 kg = mass
Why is there two π in the formula of a mathematical pendulum (T=2π √l/g)?
Answer:
The π is from the initial formula of period T=2π/omega
15C of charge flow through the filament of a light bulb in 22 seconds. What is the strength of the current in the filament?
Answer:
Current(I)= charge(q)/time
I=15/22
I=0.680.68A
Explanation:
Current is flow of charge per unit time. Always make sure the time is in seconds. istthe
A wave is sent back and forth along a rope 4 m long with a mass of 0.6 kg by exerting a force a force of 30 N. Calculate the linear mass density of the rope (in kg/m).
The linear mass density of the rope in the given motion of the wave is 0.15 kg/m.
Linear mass density of the ropeThe linear mass density of the rope in the given motion of the wave is determined by dividing the mass of the rope with the length of the entire rope.
The linear mass density of the rope is calculated as follows;
μ = m/L
μ = 0.6 kg / 4 m
μ = 0.15 kg/m
Thus, the linear mass density of the rope in the given motion of the wave is 0.15 kg/m.
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1. A drag racer accelerates from rest at 18ft/sec^2. How long does it take to acquire a speed of 60mph? What is required?
2. A contestant ran a 100-m dash in 10.6sec. What was his speed a) In feet per second and b) in miles per hr?
(1) The time taken for the drag racer to accelerate is 4.89 s
(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.
Formula:
t = (v-u)/a........... Equation 1Where:
t = timev = Final velocityu = initial velocitya = acceleration.From the question,
Given:
a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²v = 60 mph = (60×0.44704) = 26.82 m/su = 0 m/s ( from rest)Substitute these values into equation 1
t = (26.82-0)/5.4864t = 4.89 seconds(2) To calculate the speed of the contestant, we use the formula below
Formula:
s = d/t............ Equation 1Where:
s = speed of the contestantd = distancet = time.From the question,
Given:
d = 100 mt = 10.6 sSubstitute these values into equation 1
s = 100/10.6s = 9.43 m/s(a) In feet = (9.43/0.3048) = 30.94 ft/s
(b) in miles per hr = (9.43×2.24) = 21.12 miles per hr
Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
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A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is 5.0 × 10 seconds, what is the value of impulse
A.
0.26 newton seconds
B.
0.58 newton seconds
C.
1.82 newton seconds
D.
2.20 newton seconds
A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is [tex] \sf{5.0 \times 10^{-4}} [/tex] seconds, what is the value of impulse ?
So, the value of impulse is (A). 0.26 newton seconds.
IntroductionHi ! Here, I will help you about the impulse. Impulse is described as a quantity which expresses the integral of the amount force respect to time. The amount of impulse will be proportional to the amount of force or time (the greater value of the force or the value of time, the value of impuls is greater too). The relationship between impulse, force, and time is expressed in this equation :
[tex] \boxed{sf{\bold{I = F \times \Delta t}}} [/tex]
With the following condition:
I = impulse that occurs (N.s)F = force that given (N)[tex] \sf{\Delta t} [/tex] = interval of the time (s) Problem SolvingWe know that :F = force that given = 520 N[tex] \sf{\Delta t} [/tex] = interval of the time = [tex] \sf{5.0 \times 10^{-4} \: s}[/tex]What was asked :
I = impulse that occurs = ... N.sStep by step :
[tex] \sf{I = F \times \Delta t} [/tex]
[tex] \sf{I = 520 \times (5.0 \times 10^{-4}} [/tex]
[tex] \sf{I = 2,600 \times 10^{-4}} [/tex]
[tex] \sf{I = 2.60 \times 10^3 \times 10^{-4}} [/tex]
[tex] \sf{I = 2.60 \times 10^{3 + (-4)}} [/tex]
[tex] \sf{I = 2.60 \times 10^{-1} = \boxed{0.26 \: N.s}} [/tex]
Conclusion :So, the value of impulse is (A). 0.26 newton seconds
Difine Ripple Factor
-,- ty!!
Answer:
The answer is in the picture above.
Calculate the frequency of a sound wave produced when a tennis racquet string is plucked. The tension of the string is 274 N, the mass of the string is 28 kg and the length of the string is 0.74 m.
Answer:
Explanation:
The velocity of sound from the plucked string in the tennis racquet is:
[tex]v=\sqrt{\frac{FL}{m}}=\sqrt{\frac{274(0.74)}{28}}\approx 2.69 ms^{-1}[/tex]
Then the frequency will be:
[tex]f=\frac{v}{2L}=\frac{2.69}{2(0.74)}\approx 1.81 Hz[/tex]
A positive charge +Q is moving to the right and experiences a vertical (upward) magnetic force. In which direction is the magnetic field?
O into the screen
O upward
O out of the screen
O to the left
O to the right
Answer:
out of the screen
Explanation:
because charge +Q is already moving upward
A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 km above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850 km/h, the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00g directed at an angle of 30.0° above the horizontal. For reasons of safety, the rocket should be at least 1.00 km in front of the airliner when it climbs through the airliner’s altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an x-t graph showing the motions of both the rocket and the airliner; and (iii) a y-t graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner.
Answer:
Explanation:
A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 km above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850 km/h, the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00g directed at an angle of 30.0° above the horizontal. For reasons of safety, the rocket should be at least 1.00 km in front of the airliner when it climbs through the airliner’s altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an x-t graph showing the motions of both the rocket and the airliner; and (iii) a y-t graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner.
The surface of a lake has an area of 15.5 km2. What is the area of the lake in m2?
Answer:
1.55 x 10⁷ m²
Explanation:
Unit conversion
km² → m² 1 km² = 10⁶ m²Solving
15.5 km² = 10⁶ x 15.5 m²155 x 10⁵ m²1.55 x 10⁷ m²Mrs. dela Vega had 27 students in her class last year and this year she had 30 students in her math class. What is the percent of difference?
Burns are a particular safety hazard when handling
Answer:
flammable liquids
Explanation:when it catches on fire your going to get burned.hope this helps.
A 0.5 kg turntable with a radius of 15 cm is rotating 78 times per minute.
A. What is the angular momentum of the turntable?
B. A 100 gram bar which is as long as the diameter of the turntable is dropped on the
turntable such that the bar is rotating about its center. What is the new angular velocity of the turntable-bar system?
C. In order to stop the turntable and bar in just one rotation, how much force must be applied to the edge of the turntable (assuming the force is applied perpendicular to the radius)?
Answer:
c
Explanation:
A uniformly charged insulating rod is bent into the shape of a semicircle of radius R = 5 cm. If the rod has a total charge of Q = 3.10-9C, find the magnitude and direction of the electric field at O, the center of the circle.
Hi there!
We can begin by using Coulomb's Law:
[tex]E = \frac{kq}{r^2}[/tex]
k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)
E = Electric field strength (N/C)
r = distance from point (m)
q = charge (C)
Since this is a continuous charge, we must use calculus.
We can express this as the following:
[tex]q = \lambda L[/tex]
λ = Linear charge density (C/m)
L = Length of rod (m)
Now, since this is an arc, L = s (arc length). Additionally, we must find the differential elements of each:
[tex]dq = \lambda ds\\\\dq = \lambda rd\theta[/tex]
Our new equation is:
[tex]dE = \frac{kdq}{r^2}\\\\dE = \frac{k\lambda rd\theta}{r^2}[/tex]
However, we will only take the cosine component of the electric field since the vertical components will cancel out. (Electric fields are a vector). Therefore:
[tex]dE = \frac{k\lambda rd\theta}{r^2}cos\theta\\\\dE = \frac{k\lambda}{r}cos\theta d\theta[/tex]
Integrate. For a semicircle, the bounds will be from -π/2 to π/2.
[tex]E = \frac{k\lambda}{r}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}} {cos\theta} \, d\theta\\\\E = \frac{k\lambda}{r}sin\theta\left \|{\frac{\pi}{2}} \atop {-\frac{\pi}{2}}} \right. \\\\E = \frac{k\lambda}{r}(1 - (-1)) = \frac{2k\lambda}{r}[/tex]
We need to solve for λ, which is Q/ L:
[tex]\lambda = \frac{3.10 \times 10^{-9} C}{\pi (0.05)} = 1.9735 \times 10^{-8} \frac{C}{m}[/tex]
Now, plug and solve for the electric field strength:
[tex]E = \frac{2(8.99\times 10^9)(1.9735\times 10^{-8})}{0.05} = \boxed{7096.783 \frac{N}{C}}[/tex]
**A diagram was not provided, but if the hemisphere's focus was to the right, the electric field would be to the right, and etcetera.
A 2 kg solid disk with a radius of 0.22 m has a tangential force of 300N applied to it.
a) What is the torque acting on the disk?
b) What is the moment of inertia of the disk?
c) What angular acceleration is produced by the torque?
d) If the disk starts from rest and the acceleration is constant for 3.0s, what is the angular velocity of the disk at the end of 3.0s?
e) Through what angle in radians has the disk rotated during this time?
(a) The torque acting on the disk is 66 Nm.
(b) The moment of inertia of the disk is 0.05 kgm².
(c) The angular acceleration is produced by the torque is 1,320 rad/s².
(d) The final angular velocity of the disk is 3,960 rad/s.
(e) The angle of rotation of the disk is 5,940 rad.
Torque acting on the diskThe torque acting on the disk is calculated as follows;
τ = Fr
τ = 300 x 0.22
τ = 66 Nm
Moment of inertiaThe moment of inertia of a solid disk is calculated as follows;
I = ¹/₂MR²
I = ¹/₂ x 2 x (0.22)²
I = 0.05 kgm²
Angular acceleration of the diskThe angular acceleration of the disk is calculated as follows;
τ = Iα
[tex]\alpha = \frac{\tau }{I} \\\\\alpha = \frac{66}{0.05} \\\\\alpha = 1,320 \ rad/s^2[/tex]
Angular velocity of the disk after 3 sωf = ωi + αt
ωf = 0 + (1320 x 3)
ωf = 3,960 rad/s
Angle of rotation of the diskωf² = ωi²+ 2αθ
(3,960)² = 0 + 2(1320)θ
θ = (3,960²) / (2 x 1320)
θ = 5,940 rad
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