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andre [41]
3 years ago
12

A guitarist finds that the pitch of one of her strings is slightly flat—the frequency is a bit too low. Should she increase or d

ecrease the tension of the string? Explain.
Physics
1 answer:
Yuri [45]3 years ago
6 0

Answer:

The guitarist should increase the tension of the string.

Explanation:

The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.

Therefore it is advisable that she increase the tension of the string.

I hope it helps, please give brainliest if it does

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Al considerar 51 m la distancia entre dos puntos situados realmente a 52.06 m cual es el error absoluto y relativo.
Tamiku [17]

Answer:

Absolute error = 1.06 m

Relative error = 0.02

Explanation:

When considering 51 m the distance between two points actually located at 52.06 m which is the absolute and relative error.

Given that the actual distance between the two points is 52.06 m

Considered distance = 51 m

Error = Actual value - Measured value

=52.06m - 51m

=1.06m

The value is already positive, so the absolute error is also 1.06m.

Relative error =( error)/(actual value)

=1.06/52.06=0.02

Hence the relative error is 0.02.

7 0
3 years ago
Please help it’s easy will give brainlist !! Thank youu worth a lot be serious pls
Genrish500 [490]

Answer:

Part 1

Stationary

Part 2

20 Newtons

Part 3

Force

Part 4

4.0 m/s²

Part 5

Normal

Part 6

Cart m

Part 7

The gravitational force is less than magnetic force

Explanation:

Part 1

The position time graph of the object is an horizontal straight line passing across the top of the position 3 boxes vertically up from the origin

As the time increases by the units of number boxes to the left, the position of the object does not change and remains at the 3 boxes up above the origin, therefore, the object is stationary

Part 2

By Newton's third law of motion, the action action obtained from a force is equal to the reaction given to the force, therefore, we have;

The force exerted by the student on the scale = The force exerted by the scale on the student = 20 N

Part 3

A force is a the directional push on an object or pull from the object as a form of interaction with another object which tends to alter or maintain the motion of the object

Part 4

The given parameters are'

The mass of block A = 1.0 kg

The mass of block B = 2.0 kg

Both blocks, "A" and "B" are initially at rest

The applied horizontal force, F = 12-N

The nature of the surface over which the blocks move = Smooth surface

Force, F = Mass, m × Acceleration, a

F = m × a

The blocks two blocks experience a common acceleration, a

The combined mass of the two blocks, m = 1.0 kg + 2.0 kg = 3.0 kg

m = 3.0 kg

Therefore, a = F/m = 12-N/(3.0 kg) = 4 m/s²

Part 5

A normal force is a force acting perpendicularly to a surface that supports the weight of an object

Part 6

The given parameters are;

M = 2 kg

The mass of the left cart attached to the spring = M = 2 kg

The mass of the right cart attached to the spring = 2·M = 2 × 2 kg = 4 kg

Therefore, given that the force exerted by one cart on the other after the spring is removed, we have;

Force, F = M × a₁ = 2·M × a₂

Where;

a₁ = The average acceleration of the cart with mass, M

a₂ = The average acceleration of the cart with mass, 2·M

M × a₁ = 2·M × a₂

∴ a₁ = 2·a₂

The acceleration, a₁, of car M = 2 × The acceleration, a₂, of car 2·M

The acceleration of cart M is two times the acceleration of cart 2·M

Cart M will experience a greater average acceleration

Part 7

For the top magnet of two magnets placed two magnets arranged so that one is on top the other, to be held in the air with a space between the two magnets, the gravitational force pulling the top magnet down, is less than the repelling magnetic force between the two magnets.

7 0
3 years ago
If i click on a mouse does that show energy transfer?
Radda [10]

Answer:

Yes

Explanation:

You are using energy to click the mouse, and the energy moves from your fingers to the mouse clicker.

8 0
2 years ago
Read 2 more answers
A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball
RideAnS [48]

Answer:

C=2.32\times 10^{-4}\ Ns^2/m^2

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, F_r=Cv^2

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

Cv^2=mg

C=\dfrac{mg}{v^2}

C=\dfrac{0.046\times 9.8}{(44)^2}

C=2.32\times 10^{-4}\ Ns^2/m^2

Hence, this is the required solution.

4 0
3 years ago
What is an electronic signal
rosijanka [135]

where are the answer choises

6 0
3 years ago
Read 2 more answers
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