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2 years ago

Which of the following determines the natural frequency of an object?

Physics

1 answer:

Vinvika [58]2 years ago

3 0

Answer:

Explanation:

The longer an object, the slower it vibrates and since frequency is the number of oscillations an object makes per second, then you can say the longer the object, the lower the frequency, and the inverse is true.

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2. Express the following in SI units a) 0.009 cm? b) 25 km

pentagon [3]

Answer:

a).

S.I units of length is meters. So we've to change cm → m

${ \rm{1 \: cm \: \dashrightarrow \: 0.01 \: m}} \\ { \rm{0.009 \: cm \: \dashrightarrow \: (0.009 \times 0.01)}} \\ \\ = { \underline{ \rm{ \: \: 0.00009 \: m \: \: }}}$

b).

km → m

${ \rm{1 \: km \: \dashrightarrow \: 1000 \: m}} \\ { \rm{25 \: km \: \dashrightarrow \: (25 \times 1000)}} \\ \\ = { \underline{ \rm{ \: \: 25000 \: m \: \: }}}$

4 0

2 years ago

Need help on this thanks<br>

lisov135 [29]

Answer:

increase

Explanation:

5 0

2 years ago

Engineers design a system for heating the air in a house. They want the

tino4ka555 [31]

The answer would be c

4 0

2 years ago

Which one of the following scenarios accurately describes a condition in which resonance can occur? A. A column of air has a hei

ollegr [7]

The correct option is D.
Resonance is a phenomenon in which a vibrating system or an external force induces another system to vibrate at greater amplitude at particular frequencies.
When a tuning fork is struck with another object, the tuning fork will start to vibrate as a result of the force that is exerted by the object used to struck it. This type of vibration, without direct excitation by a man is an example of resonance.

8 0

3 years ago

A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m

konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, $\gamma _0$ = 0.875

mass of the object in oil, $M_o$ = 0.013 kg

mass of the object in water, $M_w$ = 0.012 kg

let the mass of the object in air = $M_a$

weight of the oil, $W_0 = M_a - 0.013$

weight of the water, $W_w = M_a - 0.012$

The relative density of the oil is given as;

$\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg$

Therefore, the mass of the body is 0.02 kg.

6 0

2 years ago