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2 years ago

7

20 points?!!!!!!please help I need this asap

1 answer:

IgorLugansk [536]2 years ago

6 0

$F - F_f = ma$

We are trying to find F.

F - 15 = 30(1.5)

F - 15 = 45

F = 60 N

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A piano tuner is tuning a piano when he discovers that the G above middle C is vibrating with a higher frequency than his G tuni

kotegsom [21]

Answer:

Tuning a piano is the process of adjusting the tension of its strings, thereby altering their pitch, or frequency of vibration, by slightly turning the tuning pins to which they're attached, so that each string sounds pleasingly in harmony with every other string.

<h2>Hopefully u will satisfy with my answer of ur question..!!</h2>

<h2>Have a nice day ahead dear..!!</h2>

6 0

2 years ago

Total mechanical energy (The sum of kinetic and potential energy

victus00 [196]

Answer:

Emechanical=mgh+ $\frac{1}{2}$ mν²

Explanation:

The equation for the total mechanical energy is:

Emechanical=Epotential+Ekinetic

In which,

Epotential=mgh; m: mass of the body, g: gravity; h: height

Ekinetic= $\frac{1}{2}$ mν²; m: mass of the body, ν: velocity of the body

So,

Emechanical=mgh+ $\frac{1}{2}$ mν²

5 0

2 years ago

A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as

Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

$T= 3600\ sec$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega=\dfrac{2\pi}{3600}$

$\omega=0.001745\ rad/s$

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

$a=r\omega^2$

Put the value into the formula

$a=0.55\times(0.001745)^2$

$a=1.675\times10^{-6}\ m/s^2$

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

3 0

2 years ago

The data table depicts an object moving with changing speed.<br><br> True<br> False

Nimfa-mama [501]

The answer is true. The table does show an object moving with changing speed.

6 0

2 years ago

Read 2 more answers

A 0.450 kg soccer ball has a kinetic energy of 119 J.

Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2 = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

8 0

2 years ago