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olga_2 [115]
3 years ago
13

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra

ting section of the string is 1.90 m long.What tension is needed to tune this string properly?Express your answer with the appropriate units.
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

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iren2701 [21]

From the choices provided, the better answer is ' T ' .

8 0
3 years ago
Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experie
Stella [2.4K]

Answer:

Angle of incline is 20.2978°

Explanation:

Given that;

Gravitational acceleration on a planet a = 3.4 m/s²

Gravitational acceleration on Earth g = 9.8 m/s²

Angle of incline = ∅

Mass of the stone = m

Force on the stone along the incline will be;

F = mgSin∅

F = ma

The stone has the same acceleration as that of the gravitational acceleration on the planet.

so

ma = mgSin∅

a = gSin∅

Sin∅ = a / g

we substitute

Sin∅ = (3.4 m/s²) / (9.8 m/s²)

Sin∅ = 0.3469

∅ = Sin⁻¹( 0.3469 )

∅ = 20.2978°

Therefore, Angle of incline is 20.2978°

8 0
3 years ago
A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a
iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

3 0
3 years ago
A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara
otez555 [7]

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

r=2.3*sin70=2.161m (ball rotates in a circle)

If the system is in equilibrium, the tension is:

Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}

Replacing:

\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}

Replacing:

v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s

7 0
3 years ago
A 2.31 kg rope is stretched between supports 10.4 m apart. If one end of the rope is tweaked, how long will it take for the resu
zlopas [31]

Answer:

t = 0.657 s

Explanation:

First, let's use the appropiate equations to solve this:

V = √T/u

This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.

Where:

V: Speed of the disturbance

T: Tension of the rope

u: linear density of the rope.

The density of the rope can be calculated using the following expression:

u = M/L

Where:

M: mass of the rope

L: Length of the rope.

We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:

u = 2.31 / 10.4 = 0.222 kg/m

Now, replacing in the first equation:

V = √55.7/0.222 = √250.9

V = 15.84 m/s

Finally the time can be calculated with the following expression:

V = L/t ----> t = L/V

Replacing:

t = 10.4 / 15.84

t = 0.657 s

4 0
3 years ago
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