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2 years ago

How must a force be applied to cause resonance?

Physics

1 answer:

Aleonysh [2.5K]2 years ago

8 0

Answer:

At its most natural frequency. ... A forceful voice, exquisite control of frequency, and oscillating

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Rewire each of the following using the correct prefix using 2 decimal places where applicable.

Ede4ka [16]

Answer:

a. 1.2×10^-6

b. 0.42×10^9

c. 246.8×10^3

d. 88

3 0

2 years ago

Gold, which has a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin

Dennis_Churaev [7]

Answer:

area = 5733.33 cm²

length = 5.47 × $10^{7}$ cm

Explanation:

Given data

density = 19.32 g/cm³

mass = 33.16 g

thickness = 3.000 µm = 3 × $10^{-4}$ cm

radius r = 1.000 µm = 1 × $10^{-4}$ cm

to find out

area of the leaf and length of the fiber

solution

we know volume formula that is

volume = mass / density

volume = 33.16 / 19.32

volume = 1.72 cm³

we know that volume = thickness × area

so area

area = volume / thickness

area = 1.72 / 3 × $10^{-4}$

area = 5733.33 cm²

and

we know volume = πr²L

so L = volume / πr²

length = 1.72 / π(1× $10^{-4}$ )²

length = 5.47 × $10^{7}$ cm

3 0

3 years ago

What causes a wave to begin

kakasveta [241]

Wind or gravitational pull. A wave is made by friction between wind and the water’s surface. Waves can also be made by the sun and moon’s gravitational pull on the earth.

5 0

3 years ago

A 65-kg swimmer pushes on the pool wall and accelerates at 6 m/s^2. The friction experienced by the swimmer is 100 N. How many N

Helga [31]

Answer:

490N

Explanation:

According Newton's second law!

\sum Force = mass × acceleration

Fm - Ff = ma

Fm is the moving force

Ff s the frictional force = 100N

mass = 65kg

acceleration = 6m/s²

Required

Moving force Fm

Substitute the given force into thr expression and get Fm

Fm -100 = 65(6)

Fm -100 = 390

Fm = 390+100

Fm = 490N

Hence the force that will cause two cart to move is 490N

5 0

2 years ago

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

Musya8 [376]

Answer:

$n = 1.266\times 10^{12}$

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is $\theta = 13 degree$

thickness of string 300 mm = 0.3 m

$sin\theta =\frac{2}{0.3 m}$

r =0.3 sin 13 = 0.067 m

$Fe = \frac{ kq_1 q-2}{d^2}$

$Fe = \frac{kq^2}{(2r)^2} = mg tan\theta$

$q^2 = mg tan\theta \frac{(2r)^2}{k}$

$= 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}$

$q^2 = 4.10\times 10^{-14}$

$q = 2.026 \times 10^{-7} C$

q = ne

$n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}$

$n = 1.266\times 10^{12}$

3 0

2 years ago