Which musical instrument uses resonance?
flute
drums
bongos
electric keyboard

Answers

Answer 1

Answer:

It's A: Flute!


Related Questions

A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb

Answers

Voltage=V=120VPower=P=60W

Current=I

[tex]\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A[/tex]

The resultant of two vectors is of magnitude 3 units and 4 units is 1 units, what is the value of their dot product? ​

Answers

Answer:

A dot B = C     is the vector equation for this expression

A · B = A B cos θ

3 * 4 cos θ = 1       the value 1 is their dot product

cos θ = 1 / 12 = .083       θ = 85.2 deg

help me get a answer ​

Answers

Answer:

I think D sorry if I'm wrong

If a transverse wave osculates 7 times every second and the speed of the wave is 27 m/s what is the wavelength of the wave

Answers

Explanation:

formula is ˠ=vf

f=1/T

1/7

f=0.14

wavelength=27Ⅹ0.14

=3.78m

OR

7Ⅹ27

=189m

Two identical charges are located 1 m apart and feel a 1 N repulsive electric force. What is the charge of each particle.

Answers

The charge on each particles which are 1 m apart and feeling a repulsive force of 1 N is 1.05×10¯⁵ C

Assumption

Let the charge on each particles be q

How to determine the charge Final force (F) = 1 NDistance apart (r) = 1 mElectrical constant (K) = 9×10⁹ Nm²/C²Charge on 1st particle (q₁) = q =? Charge on 2nd particle (q₂) = q =?

The charge on each particle can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = Kq² / r²

1 = (9×10⁹ × q²) / 1²

1 = 9×10⁹ × q²

Divide both side by 9×10⁹

q² = 1 / 9×10⁹

Take the square root of both side

q =  √(1 / 9×10⁹)

q = 1.05×10¯⁵ C

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What are the three symbols used in Ohm's law. Explain what each symbol represents and give the units for each of the variables.

Answers

Answer:

Step by step explanation:

An elastic string of length 20cm stretch by 7mm under a load 50N . The strain in the spring is?

Answers

Answer:

0.035

Explanation:

7÷1000

=0.007m

0.007/0.2

=0.035

Mechanical strain of an elastic substance is the ratio of its change in length to the original length. The strain of the elastic string of length 20 cm when stretched by 7 mm is 0.035.

What is strain ?

Strain is a physical quantity used to express the change in dimension of an elastic material. It is the ratio of change in length to the original length. Stress is the force that causing the deformation.

According to Hook's law of elasticity, the strain is directly proportional to the strain. They show a linear relation. Let L be the original length of a string and its change in length be ΔL.

Thus, strain = ΔL / L

Given that, the change in length = 7 mm = 0.7 cm

original length of the string = 20 cm

strain = 0.7 cm/ 20 cm

         = 0.035

Therefore, the strain of the elastic string of length 20 cm when stretched by 7 mm is 0.035.

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16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current drawn b) The quantity of electricity in Coulombs taken in 5 minutes c) The energy absorbed in 100 hours.

Answers

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t    = power X time

W / t = P   power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec  = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c)  E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

If something weighs 2891N on earth, what’s it’s mass?

Answers

Answer:

289.1kg

Explanation:

because W=M*G, and M=W/G, so 2891N/10m/s²=289.1kg

5. Apply Concepts A heat engine has an energy input of 300 J. Its output of use-
ful energy is 150 J. What is the total output of non-useful energy from the heat
engine? Explain how you know.

Answers

Given the data from the question, the total output of non-useful energy from the heat engine is 150 J

What is a heat engine?

A heat engine is a device / equipment that can convert or transfer thermal energy into useful-work. Some examples of heat engines include

refrigeratorsInternal combustion Thermal power stationfirearms  heat pumps

How to determine output non-useful energy

From the question given above, the following data were obtained:

Input energy = 300 JOutput (useful) = 150 JOutput (non-useful) =?

Output (non-useful) = Input – Output (useful)

Output (non-useful) = 300 – 150

Output (non-useful) = 150 J

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A curve is banked at an angle of 29.1 degrees above the horizontal and the road surface has a coefficient of static friction of 0.4. What must the radius of curvature be for the safe minimum speed of 27.1 m/s?

Answers

Hi there!

We can begin by summing the forces acting on the car.

Along the axis of the incline, the forces of friction and gravity are working. The force of friction points towards the top of the ramp, while the force of gravity works towards the bottom.

We can use trigonometry to determine the force due to gravity along the ramp.

[tex]F_g = Mg sin\theta[/tex]

The force due to friction is equal to:
[tex]F_f = \mu N[/tex]

The normal force is the vertical component of the force due to gravity, so:
[tex]F_f = \mu Mgcos\theta[/tex]

Now, the combination of these two forces produces a component of the centripetal force. Drawing a diagram, we see that the true centripetal force is the HYPOTENUSE, while these forces sum up to its horizontal component along the ramp.

Therefore:
[tex]F_c sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]

The centripetal force is equivalent to:
[tex]F_c = \frac{Mv^2}{r}[/tex]

m = mass (kg)

v = velocity (m/s)
r = radius (m)

Rewrite:
[tex]\frac{Mv^2}{r}sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]

Cancel out 'M'.

[tex]\frac{v^2}{r}sin\theta = gsin\theta - \mu gcos\theta[/tex]

Rearrange to solve for 'r'.

[tex]r = \frac{v^2sin\theta}{gsin\theta - \mu gcos\theta}[/tex]

Plug in values and solve.
[tex]r = \frac{(27.1^2)sin(29.1)}{(9.8)sin(29.1) - 0.4(9.8)cos(29.1)} = \boxed{266.365 m}[/tex]

Please answer and show formula

Answers

Voltage = 5 volts

[tex]\sf \dfrac{number \ of \ turns \ in \ primary \ coil}{voltage\ in \ primary \ coil} = \dfrac{number \ of \ turns \ in \ secondary\ coil}{voltage\ in \ secondary\ coil}[/tex]

[tex]\hookrightarrow \sf \dfrac{400}{100} = \dfrac{20}{V_2}[/tex]

[tex]\hookrightarrow \sf 400V_2}{} = 20*100[/tex]

[tex]\hookrightarrow \sf V_2 = 5 V[/tex]

define key terms related to the muscular, skeletal, and nervous systems

Answers

Answer:

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static frictional Force​

Answers

Answer:

The static frictional force is the force between the magnetic forces statically and contains only two objects during this physical reaction. It results in electricity and a slight shock.

What is the solution of this? ​

Answers

Answer:

1.2 is the correct answer due to the graph

Explanation:

math teacher

The answer is the third one 0.8s

Alex (31kg) and Cassie (19Kg) sit on a 10kg metre-long see-saw at the local park. The pivot of the see-saw is in the middle of its length. If Cassie sits at one end of the see-saw, where relative to the other end must Alex sit so the net torque is balanced? (unit:metres)

Answers

Answer:

M1 g L1 = 19 kg * 9.8 m/s^2 * 5 m = counter clockwise torque - Cassie at left end

M1 g L1 = M2 g L2        for torques to balance

L2 = M1 L1 / M2 = 19 * 5 / 31 = 3.06 M

Alex should sit at 3.1 m from the fulcrum (at 5 m from each end)

An object of mass m is oscillating with a period T. The position x of the object as a function of time is given by the equation x(t)=Acosωt . The maximum net force exerted on the object while it is oscillating has a magnitude F. Which of the following expressions is correct for the maximum speed of the object during its motion?

Answers

What the equation given?

x(t)=Acos[tex]\omega[/tex]t

Maximum velocity occurs at the equilibrium position

So

x=0

Now

x(0)=Acos0[/tex]x(0)=A

Now

As we know the formula

[tex]\\ \rm\rightarrowtail V_max=A\omega[/tex]

These expressions can be used

The maximum speed of the oscillating object will be given by [tex]V_{max}=Aw[/tex]

What is oscillation?

An oscillation is defined as the repitative periodic motion of any object about its mean or equilibrium position.

The given equation is as follows:

x(t)=Acost

Maximum velocity occurs at the equilibrium position x=0

x(0)=Acos0

x(0)=A

Hence the maximum velocity will be  [tex]V_{max}=Aw[/tex]

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Which number represents runoff on the hydrologic cycle diagram?


1


2


3


4

Answers

Answer:

number 3

Explanation:

3 represents runoff


Runoff is the water collected after rainfall that drains from the surface area of land. It will be tricky to choose 4 which is likely a smaller waterbody draining to a bigger waterbody


hope it helped

Who is the father of the humans

Answers

Answer:

Adam Was the first person

Explanation:

Answer:

God is

Explanation:

Adam was the first human, Yes but God is the one who man Adam in his image and likeness, that is God looks just like the human

because we are made in God's image

state the formula to compute average speed acquired by a vehicle to cover a distance between two points.

ty! :)​

Answers

The general speed average formula for an object is given as [Average Speed = Total Distance Traveled ÷ Total Time Taken]. SI unit of average speed is m/s.

Who came up with the 3 Laws of Motion?

Answers

Answer:

Isaac Newton came up with the 3 Laws of Motion

Answer:

Isaac Newton came up with three laws of motion

How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 20 N/m?

Answers

Distance=3cm=0.03mSpring constant=k=20N/mForce=F

[tex]\\ \rm\rightarrowtail F=-kx[/tex]

[tex]\\ \rm\rightarrowtail F=-20(0.03)[/tex]

[tex]\\ \rm\rightarrowtail F=-0.6N[/tex]

A vector is 0.888 m long and
points
in a
205 degree
direction.
Find the y-component of the
vector.

Answers

Answer:

Since the second quadrant ends at 180 deg,

205 will be 25 deg into the third quadrant (-x and -y)

y = -.888 sin 25 = -.375

One can also write y = .888 sin 205 = -.375

An unbalanced force acting on an object results in ______.
A. inertia
B. friction
C. acceleration
D. faster
E. slower

Answers

Answer:

friction

Explanation:

In question 8 from “Problems 2”, you calculated that an archery arrow shot with an initial velocity of 45 m/s at an angle of 10 degrees would travel 70.67 m (2 d.p.) before hitting a target at the same height from which it was released. If the archery target was moved further away from the archer, discuss some kinematic-based strategies that the archer could adopt to ensure that the arrow reaches the target. Ensure that the variables discussed are relevant to your answer. Assume air resistance is negligible.

Answers

The distance traveled by the arrow and horizontal velocity are directly proportional, thus when the distance increases, an increase in initial velocity will allow the arrow to hit the target.

Time of motion of the projectile

If the horizontal distance the projectile would travel before hitting the target is 70.67 m, the time of motion of the projectile is calculated as;

X = Vx(t)

t = X/Vx

t = X/Vcosθ

t = (70.67) / (45 x cos10)

t = 1.6 s

When the archery target is moved further away from the archer, the archer needs increase the initial velocity of the arrow assuming the angle of projection is constant.

Since the distance and horizontal velocity are directly proportional to each other, thus when the distance increases, an increase in initial velocity will allow the arrow to hit the target.

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The amount of energy released when 45 g of -175° C steam is cooled to 90° C is


A. 101,700 J


B. 317,781 J


C. 419,481 J


D. 417,600 J

Answers

Answer:

The answer should be choice B.

You are standing on roller skates if you try to throw a 20-pound bowling ball while standing on skates,
what happens. How does this show Newton 3rd law

Answers

Answer:

I'd probably drop it on my toes. That would happen.

Explanation:

You want to improve your dominant trait score, and you decide to take daily walks by yourself. What trait will you strengthen with this activity?

Answers

Muscles such as quadriceps, hamstrings, the calf muscles and the hip adductors get stronger by walking.

Which muscles get stronger due to walking?

The quadriceps, hamstrings, the calf muscles and the hip adductors are the muscles which get stronger by doing walking because impact is produced on these muscles while walking. Walking is a good exercise which have many health benefits.

So we can conclude that muscles such as quadriceps, hamstrings, the calf muscles and the hip adductors get stronger by walking.

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Answer:

introversion is the correct answer

Explanation:

God bless! Have a great day!

The electric potential, which is a scalar field, may be represented graphically by equipotential curves. The lines of electric field may be obtained from those equipotentials.

Four rectangles have been placed on the graph below. Assess the region of the graph where the magnitude of the electric field is greatest. Drag the label Emax to the rectangle in that region. Assess the region of the graph where the magnitude of the electric field is smallest. Drag the label Emin to the rectangle in that region.

Answers

The region with high electric potential  (80 V and 70 V) will have high electric field while the region with low electric potential (20 V) will have low electric field.

What is Electric field?

Electric field is the field that surrounds electrically charged particles and exerts force on all other charged particles in the field.

Relationship between electric potential and electric field

E = V/d

where;

V is electric potential (V)E is electric field (V/m)d is the distance (m)

Since electric field is directly proportional to electric potential, the region with high electric potential  (80 V and 70 V) will have high electric field while the region with low electric potential (20 V) will have low electric field.

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Need help with this question. Thirty points.

Answers

[tex]1)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2 \\\\\implies m = \dfrac{2E_k}{v^2} = \dfrac{2\times 4500}{30^2}=10~ kg\\ \\2)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12mv^2 \\\\\implies v^2 = \dfrac{2E_k}m\\\\\implies v = \sqrt{\dfrac{2E_k}{m}}=\sqrt{\dfrac{2 \times 320}{\dfrac{20}{9.8}}}= 17.709 ~ ms^{-1}\\\\3)\\\\\text{Kinetic energy,}\\\\E_k=\dfrac 12 mv^2=\dfrac 12 \times 50 \times 10^2=2500 ~J\\\\\\[/tex]

[tex]4)\\\\\text{Potential energy,}\\\\E_p =mgh = 5\times 9.8\times 1.5 = 73.5 ~J[/tex]

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