Answer:
It is because there are a odd number of wavelengths fitting onto the string
Explanation:
The general formula for the harmonics on a string loose at one end is f = nv/2L where n = odd integer, v = speed of wave in string and L = length of string.
For a string loose at one end, there are a odd number of wavelengths which can fit onto the string. Since this is the case, we only have harmonics of odd numbers and thus we have even number harmonics missing.
The correct answer is 40 or - 40??
Answer:
40 cm
hope it's help you but you can choose if your ans. is -40 cm
Which players are usually the tallest on their team, and stay close to the basket so they can shoot and rebound the ball?
Mid-fielder
Center
Forward
Guard
Answer:
Center
Explanation:
The center is the tallest player on each team, playing near the basket. On offense, the center tries to score on close shots and rebound. But on defense, the center tries to block opponents' shots and rebound their misses.
Answer:
C
Explanation:
Help me
Shown is a picture of two toy trains getting ready to collide. If the collision is non-elastic, what is the total momentum of
the train cars after the collision?
a
-10 kg·m/s
b
10 kg·m/s
c
70 kg·m/s
d
50 kg·m/s
Answer:
A: -10 kg•m/s
Explanation:
6•5= 30
-2•20=-40
-40+30= -10
-10 kg•m/s
Two
vectors Å and °B are such that A = 2
and A-B-C find the angle between them
Answer:
hjg567 A 567
Explanation:
There is a disease known as formula fixation that is common among physics students. It particularly infects those who perceive physics as an applied math course where numbers and equations are simply combined to solve algebra problems. However, this is not the true nature of physics. Physics concerns itself with ideas and concepts that provide a reasonable explanation of the physical world. When students divorce the mathematics from the ideas, formula fixation takes root and even mathematical problem solving can become difficult. Do you have formula fixation? Test your health by trying these computational problems. 12. A force of 800 N causes an 80-kg fullback to change his velocity by 10 m/s. Determine the impulse experienced by the fullback. PSYW 13. A 0.80-kg soccer ball experiences an impulse of 25 N•s. Determine the momentum change of the soccer ball. PSYW 14. A 1200-kg car is brought from 25 m/s to 10 m/s over a time period of 5.0 seconds. Determine the force experienced by the car. PSYW 15. A 90-kg tight end moving at 9.0 m/s encounters a 400 N•s impulse. Determine the velocity change of the tight end. PSYW
Answer:
12) I = 800 kg m/ s , 13) Δv = 31.25 m / s, 14) F = -3600 N , 15) Δv = 4.44 m/s
Explanation:
In this strange exercise you are asked to answer a series of questions
12) force F = 800N applied mass m = 80 kg has a speed change of DV = 10 m / s, calculate the momentum
I = Δp
I = m Δv
I = 80 10
I = 800 kg m / s
13) a mass of m = 0.80 kg, has an impulse of 25 N s, determines the change in speed
I = m Δv
Δv = I / m
Δv = 25 / 0.8
Δv = 31.25 m / s
14) a mass m + 1200 kg is carried from 25 m / s to 10 m / s in a period of 5.0 s, determine the force
I = m ([tex]v_{f}[/tex] -v₀)
the proemd force for this impulse
I = F t
we substitute
F t = m (v_{f} -v₀)
F = m (v_{f} -v₀ / t
F = 1200 (10-25) / 5
F = -3600 N
the negative sign indicates that the force is in the opposite direction of the movement
15) a mass = 90 kg with velocity v = 9.0 m has a momentum of I = 400N s, determines the change in velocity
I = m Δv
Δv = I / m
Δv = 400/90
Δv = 4.44 m / s
if they asked for the final speed it would be.
I = m (v_{f} -v₀)
v_{f} = v₀ + I / m
v_{f} = 9 + 400/90
v_{f} = 13.4 m / s
6. How do the data support your claim above? Explain your statement above. Be sure to refer to specific pieces of data from your experiment that support your argument.
Answer:
hope this will help
Explanation:
if your title doesn't have enough information, make a list of the key words ... If you are not sure what should be included in each summary sentence, use ... Often you can present the hypothesis and the supporting reasoning in one paragraph. ... Describe specific parts of the procedure or data that contributed to your learning.
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pls help, i have no idea where to start
Answer:
Use equation 1/2 * m * v²
Explanation:
where m = mass in kg
and v = velocity in m/s
plug accordingly.
F eeeeeeeereeeeeee points
Answer:
Yayyyyyy thank you ≥﹏≤.!!!!!!!!!!!!!
A box moves 5\,\text m5m5, start text, m, end text horizontally when a force, F =10\,\text NF=10NF, equals, 10, start text, N, end text is applied at an angle, \theta =150\degreeθ=150°theta, equals, 150, degree as shown below. A dark grey box slides along a light grey horizontal surface. the box has a force vector F pulling up and to the left on it. A displacement vector d extends rightwards form the center of the box. The angle between vectors F and d is labeled theta. A dark grey box slides along a light grey horizontal surface. the box has a force vector F pulling up and to the left on it. A displacement vector d extends rightwards form the center of the box. The angle between vectors F and d is labeled theta. What is the work done on the box by FFF during the displacement?
Answer:
??
Explanation:
not completely sure hold on
Answer:
-43J
Explanation:
What is the force of gravitational attraction between an object with a mass of 2kg and another object that has a mass of 2kg and a distance between them of 2.5m
Answer:
1.7x10^-10 N
Explanation:
F = G [(m_1)(m_2)]/(r^2)
F = force
G = Gravitational constant 6.67433x10^-11 (N*m^2)/kg^2
m_1 = mass first object
m_2 = mass second object
r = radius between the 2 objects
F = G[(2 kg*2 kg)/(1.25 m)^2]
F = 1.7x10^-10 N
Two similarities between balanced and unbalanced forces.
Explanation:
For balanced forces, the severity of the 2 factors is equal, while the magnitude of the 2 factors is unequal in the situation of unbalanced forces. The three separate forces operate in opposing ways in balanced forces. In unbalanced forces, on the other hand, independent forces either behave in the same or reverse direction.
Numerical Problems
A bus is moving with the initial velocity 10 m/s. After 2 seconds, the velocity
becomes 20 m/s. Find the acceleration and distance moved by the bus.
Answer:
a = 5 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 10 [m/s]
t = time = 2 [s]
a = acceleration [m/s²]
Now replacing:
[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]
Fill in the blanks in the following paragraph to correctly identify the properties of devices that use electromagnetism. (1 point)
A(n) ____________________ changes mechanical energy into ____________________. A(n) ____________________ changes electric energy into ____________________. A(n) ____________________ changes the voltage of an alternating current. A(n) ____________________ transformer has more loops in the primary coil than in the secondary coil. A(n) ____________________ transformer has ____________________ loops in the primary coil than in the secondary coil.
Answer:
this answer you question
Answer:
generatorelectricalmotormechanicaltransformerstep-downstep-upfewerExplanation:
A(n) generator changes mechanical energy into electrical. A(n) motor changes electric energy into mechanical. A(n) transformer changes the voltage of an alternating current. A(n) step-down transformer has more loops in the primary coil than in the secondary coil. A(n) step-up transformer has fewer loops in the primary coil than in the secondary coil.
Hope it's helpful!
how does a switch work in a circuit
An electric switch is a device that interrupts the electron flow in a circuit. Switches are primarily binary devices: either fully on or off and light switches have a simple design. When the switch is turned off, the circuit breaks and the power flow is interrupted. Circuits consist of a source of power and load.
A north magnetic pole brought near a south magnetic pole will...
A)
cancel the south pole.
B)
attract the south pole.
C)
repel the south pole.
What is the net force?
60N left
4ON, right
60N, right
0N
Answer:
60n left
Explanation:
Do
what effects do coil of wire on a compass after it has been connected to a battery
Explanation:
The compass needle moved when the wire was connected to the battery. The important point here is that the needle is affected by the wire only when both ends of the wire are connected to the battery because only at this time is current flowing through the circuit.
1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Answer:
Momentum, p = 23250 kg m/s
Explanation:
Given that
Mass of a car, m = 1550 kg
Speed pf car, v = 15 m/s
We need to find the momentum of the car. The formula for the momentum of an object is given by :
p = mv
Substituting all the values in the above formula
p = 1550 kg × 15 m/s
p = 23250 kg m/s
So, the momentum of the car is 23250 kg m/s.
Please help me!!
1 a) Compare the speed that light waves travel in air to the speed that sound waves travel in the air. (Show results in meters/second)
1 b) How many times faster do light waves travel in the air in comparison to sound waves in air? (show working out)
2) Compare the speed of light in water to the speed of sound in water. (Show results in meters/second)
Answer:A sound wave is a pressure disturbance that travels through a medium by means of particle-to-particle interaction.As one particle becomes disturbed,it exerts a force on the next adjacent particle, disturbing that particle from rest and transporting the energy through the medium.Like any wave,the speed of a sound wave refers to how fast the disturbance is passed from particle to particle.While frequency refers to the number of vibrations that an individual particle makes per unit of time, speed refers to the distance that the disturbance travels per unit of time.
Since the speed of a wave is defined as the distance that a point on a wave (such as a compression or a rarefaction)travels per unit of time,it is often expressed in units of meters/second (abbreviated m/s). In equation form, this is speed = distance/time
The faster a sound wave travels,the more distance it will cover in the same period of time.If a sound wave were observed to travel a distance of 700 meters in 2 seconds,then the speed of the wave would be 350 m/s. A slower wave would cover less distance - perhaps 660 meters - in the same time period of 2 seconds and thus have a speed of 330 m/s.Faster waves cover more distance in the same period of time.
The speed of any wave depends upon the properties of the medium through which the wave is traveling.Typically there are two essential types of properties that affect wave speed - inertial properties and elastic properties.Elastic properties are those properties related to the tendency of a material to maintain its shape and not deform whenever a force or stress is applied to it.A material such as steel will experience a very small deformation of shape and dimension when a stress is applied to it.Steel is a rigid material with a high elasticity.On the other hand,a material such as a rubber band is highly flexible;when a force is applied to stretch the rubber band,it deforms or changes its shape readily.A small stress on the rubber band causes a large deformation.Steel is considered to be a stiff or rigid material,whereas a rubber band is considered a flexible material. At the particle level,a stiff or rigid material is characterized by atoms and/or molecules with strong attractions for each other.The phase of matter has a tremendous impact upon the elastic properties of the medium.For this reason,longitudinal sound waves travel faster in solids than they do in liquids than they do in gases.Even though the inertial factor may favor gases,the elastic factor has a greater influence on the speed (v) of a wave,thus yielding this general pattern:solids > liquids > gases The density of a medium is an example of an inertial property.The greater the inertia (i.e.mass density) of individual particles of the medium,the less responsive they will be to the interactions between neighboring particles and the slower that the wave will be. However,within a single phase of matter,the inertial property of density tends to be the property that has a greatest impact upon the speed of sound.Like any liquid,water has a tendency to evaporate.As it does,particles of gaseous water become mixed in the air.The temperature will affect the strength of the particle interactions an elastic property.At normal atmospheric pressure,the temperature dependence of the speed of a sound wave through dry air is approximated by the following equation:
v = 331 m/s + (0.6 m/s/C)•T
where T is the temperature of the air in degrees Celsius.Using this equation to determine the speed of a sound wave in air at a temperature of 20 degrees Celsius yields the following solution.
v = 343 m/s
(The equation itself does not have any theoretical basis;it is simply the result of inspecting temperature-speed data for this temperature range.Other equations do exist that are based upon theoretical reasoning and provide accurate data for all temperatures.The speed of light as it travels through air and space is much faster than that of sound;it travels at 300 million meters per second or 273,400 miles per hour.Visible light can also travel through other things besides through air and through space.The speed of light in water is approximately 2.26×108 2.26 × 10 8 meters per second.
Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acting on the flange is a minimum.
Answer:
The unknown force will be 18.116 lb.
Explanation:
Given that,
Three forces act on a flange as shown in figure.
The net force acting on the flange is a minimum.
[tex]\dfrac{dF_{net}}{df}=0[/tex]
We need to calculate the unknown force
Using formula of net force
[tex]\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}[/tex]
Put the value into the formula
[tex]\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}[/tex]
[tex]\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}[/tex]
The magnitude of net force,
[tex]F_{net}=\sqrt{F_{x}^2+F_{y}^2}[/tex]
[tex]F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}[/tex]
[tex]F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}[/tex]
[tex]F_{net}=\sqrt{F^2+4899.78+36.232F}[/tex]
On differentiating w.r.to F
[tex](\dfrac{dF_{net}}{dF})^2=2F+36.232[/tex]
[tex]0=2F+36.232[/tex]
[tex]F=-\dfrac{36.232}{2}[/tex]
[tex]F=-18.116\ lb[/tex]
Negative sign shows the direction of force which is downward.
Hence, The unknown force will be 18.116 lb.
Following are the solution to the given question:
Calculating the Net force on flange:
[tex]\overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 70 \cos 30^{\circ} -40 ) \hat{i}+(70 \sin 30^{\circ} - F \sin 45^{\circ} )\hat{y} \\\\ \overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 20.62 ) \hat{i}+(35 -F\sin 45^{\circ})\hat{y} \\\\ [/tex]
Calculating the magnitude:
[tex]= \sqrt{f_{X}^2 +f_{y}^{2}}\\\\ = \sqt{( F \cos 45^{\circ} + 20.62 )^2+(35 -F\sin 45^{\circ})^2} \\\\ = \sqt{( F^2+20.62^2+ 35^2+ 41.24 F \cos 45^{\circ} -70 F\sin 45^{\circ})}\\\\ = \sqt{( F^2+20.33F +1650.1844)} \\\\[/tex]
Differentiate the value
[tex]\to \frac{F_{net}}{df}=0 [/tex]
[tex]\to 2F-20.33=0\\\\ \to 2F=20.33\\\\ \to F=\frac{20.33}{2}\\\\ \to F= 10.165\ lb [/tex]
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What does the Big Bang theory explain?
A. How time began
B. How the universe began
C. Both A and B
D. Neither A nor B
Answer:
how the universe began
Explanation:
time never stops or starts, we only think it starts when we start recording it
1. A sailor pulls a boat along a dock using a rope at an angle of 60.0º with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3.00 m?
W = FdcosO =255 x 3 x cos 60 =
2. An elephant pushes with 2000 N at an angle of 33o above the horizontal on a load of trees. It then pushes these trees for 150 m. How much work did the elephant do?
3. Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35o above the horizontal to drag his backpack a horizontal distance of 129 m. Determine the work done upon the backpack.
4. If 100 N force has 30o angle pulling on a 15 kg block for 5 m. What’s the work?
Answer:
(1)Work done by snail is 382.5 N
(2)Work done by ELEPHANT is 25160 N
(3)Work done by HANS is 2420 N
(4)Work done on block is 433 N
Explanation:
The work done due to force F applied at an angle θ from the horizontal to the body is give by
Work done = Fscosθ where, s is distance traveled by body
Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]
Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N
thus work done by snail is 382.5 N
Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]
Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N
thus work done by elephant is 25160 N
Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]
Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N
thus work done by Hans is 2420 N
Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]
Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N
thus work done on block is 433 N
(1) The work done by the sailor is 382.5 J.
(2) The work done by elephant is 25160 J.
(3) The work done by Hans upon the backpack is 2420 J.
(4) The required work done on block is 433 J.
Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,
[tex]W =F \times s \times cos \theta[/tex]
Here, s is the distance covered by the body.
(1)
Given data:
F= 255N , s= 3.00m and θ = 60.
The work done is calculated as,
W = Fscosθ
W= 255 x 3.00 x cos60 = 382.5 J.
Thus, the work done by the sailor is 382.5 J.
(2)
Given data:
F= 200N , s= 150m and θ = 33.
The work done by the elephant is calculated as,
W = Fscosθ
W= 200 x 150 x cos33. = 25160 J
Thus, the work done by elephant is 25160 J.
(3)
Given data:
F= 22.9N , s= 129m and θ = 35.
Then the work done upon the backpack is calculated as,
W= Fscosθ
W= 22.9 x 129 x cos35. = 2420 J
Thus, the work done by Hans upon the backpack is 2420 J.
(4)
Given data:
F= 100N , s= 5m and θ = 30.
The work done is calculated as,
W = Fscosθ
W= 100 x 5.00 x cos30. = 433 J
Thus, the required work done on block is 433 J.
Learn more about the work done here:
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A 20N cart is being pulled to the right with an applied force of 20N at an angle of 40°. The normal force on the cart is 7N and the friction is 3N. Draw the free body diagram. What is the net force in the x and y direction of the cart?
Answer:
F_{net-x} = 12.32 N , F_{net-x} = -0.14 N
Explanation:
For this case we apply Newton's second law on each axis, see attachment
X axis
F cos 40 - fr = m a
Y Axis
F sin 40 +N - W = 0
the net force is
X axis
[tex]F_{net-x}[/tex] = F - fr
F_{net-x} = 20 cos 40 - 3
F_{net-x} = 12.32 N
Y Axis
F_{net-y} = F sin 40 + N - W
F_{net-x} = 20 sin 40 + 7 - 20
F_{net-x} = -0.14 N
we see that the force in this axis is approximately zero
At the surface of a certain planet, the gravitational acceleration g has a magnitude of 20.0 m/s^2. A 22.0-kg brass ball is transported to this planet.
What is (a) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?
Answer:
a, 22 kg and 22 kg
b, 215.8 N and 440 N
Explanation:
a
The mass of the ball remains constant and unchanged irrespective of where it has been to, need to go or is going. So, basically the mass of the ball on earth is as the same mass of the ball on the said planet, 22 kg
b
The weight of any object factors in the acceleration due to gravity of the said area(or planet).
W = mg, with m being the mass and g being the acceleration due to gravity.
On earth
W = 22 * 9.81 = 215.8 N
On the said planet,
W = 22 * 20 = 440 N
Do therefore, the weight is 215.8 N on earth and 440 N on the planet
What are the characteristics of supernovae?
release of matter and energy
provide new material for future solar systems
move on to form a black hole or neutron star
might be seen from earth
Answer:
c
Explanation:
I NEED HELP WITH 8 I WILL GIVE U BRAINLIEST
Answer:
u will need 1/2 cells
Explanation:
Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B moves forward at 10 m/s and car A is at rest. What fraction of the initial kinetic energy is lost in the collision?
Given :
Initial speed of car A is 15 m/s and initial speed of car B is zero.
Final speed of car A is zero and final speed of car B is 10 m/s.
To Find :
What fraction of the initial kinetic energy is lost in the collision.
Solution :
Initial kinetic energy is :
[tex]K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}[/tex]
Final kinetic energy is :
[tex]K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}[/tex]
Now, fraction of initial kinetic energy loss is :
[tex]Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25[/tex]
Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .
what is the force of gravity attraction between an object with a mass of 0.5 kg and another object has a mass of 0.33 kg and a distance between Dam of 0.002 M
Answer:
[tex]from \: newton \: law \: of \: gravitation \\ F = \frac{GMm}{ {r}^{2} } \\ G = 6.67 \times {10}^{ - 11} \\ M = 0.5 \: kg \\ m = 0.33 \: kg \\ r = 0.002 \: m \\ substitute \\ F = \frac{(6.67 \times {10}^{ - 11}) \times (0.5) \times (0.33)) }{ {(0.002)}^{2} } \\ = 2.75 \times {10}^{ - 6} N[/tex]
The ____ involves relations between microsystems or connections between contexts.
Answer:
mesosystem
Explanation:
The mesosystem involves relations between microsystems or connections between contexts.
A circus performer walked up and to the right for a total displacement of 10\,\text m10m10, start text, m, end text along a diagonal tightrope angled 30 \degree30°30, degree above the ground.
What was the vertical displacement of the circus performer in \text mmstart text, m, end text?
Answer:
5m
Explanation:
khan academy
Vertical displacement of the circus performer is 5 m.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
A circus performer walked up and to the right for a total displacement of 10 m, start to end along a diagonal tightrope angled 30° degree above the ground.
Vertical displacement = 10 sin 30° = 5 m
Vertical displacement of the circus performer is 5 m.
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