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2 years ago

Figure 13 shows a child’s toy . The toy hangs from a hook in the ceiling.

Physics

1 answer:

Grace [21]2 years ago

6 0

The number of oscillations of the toy in a second is 1.25.

<h3>What is frequency?</h3>

This is the number of complete oscillation of an object is a given period.

The given parameter:

Frequency of the toy, F = 1.25 Hz

The frequency of an object is calculated as follows;

$f = \frac{n}{t} \\\\$

where;

n is the number of oscillations
t is the time of motion

The number of oscillations of the toy in a second is calculated as follows;

$1.25 = \frac{n}{1} \\\\n = 1.25$

Learn more about frequency here: brainly.com/question/10728818

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A bag of sugar weighs 1.50 lb on earth. what would it weigh in newtons on the moon, where the free-fall acceleration is one-sixt

Fittoniya [83]

0.25 is the answer :)

8 0

4 years ago

When falling through the air, which of the following objects will hit the ground first?

ipn [44]

I think the answer is a penny

5 0

3 years ago

Read 2 more answers

Can someone help me convert these?

fomenos

Assuming you are supposed to write each conversion in scientific notation:

(2) 1 m = 100 cm, so

(67 cm) × (1/100 m/cm) = 67/100 m = 0.67 m = 6.7 × 10 ⁻¹ m

(3) 1 km = 1,000 m, so

(1.2 km) × (1000 m/km) = 1200 m = 1.2 × 10³ m

(4) 1 m = 1,000 mm = 10³ mm, so

(6.2 × 10 ⁻³ m) × (10³ mm/m) = 6.2 mm

(5) 1 m = 1,000,000,000 nm = 10⁹ nm, so

(4.05 × 10³ nm) × (1/10⁹ m/nm) = 4.05 × 10 ⁻⁶ m

(6) 1 g = 1,000,000 µg = 10⁶ µg, so

(3200 µg) × (1/10⁶ g/µg) = 3200 × 10 ⁻⁶ g = 3.2 × 10 ⁻³ g

4 0

3 years ago

An 18 gauge copper wire (the size usually used for lamp cords), with a diameter of 1.02 mm,1.02 mm, carries a constant current o

larisa86 [58]

Answer:

J = 2.044x10⁶ A/m²

v = 1.50x10⁻⁴ m/s

Explanation:

The current density (J) of the copper wire is giving by:

$J = \frac {I}{A}$

<em>where I: electric current and A: cross-sectional area of the copper wire</em>

<u>The cross-sectional area of the copper wire can be calculated by:</u>

$A = \frac {\pi d^{2}}{4} = \frac {\pi (1.02 \cdot 10^{-3} m)^{2}}{4} = 8.17 \cdot 10^{-07} m^{2}$

<u>Substituting the calculated area in the equation (1) we have:</u>

$J = \frac {1.67 A}{8.17 \cdot 10^{-7} m^{2}} = 2.044 \cdot 10^{6} \frac {A}{m^{2}}$

Hence, the current density is 2.044x10⁶ A/m².

To find the drift speed (v), we need to use the next equation:

$v = \frac {J}{n q}$

<em>where n: the free-electron density, q: module of the charge of the electron </em>

$v = \frac {2.044 \cdot 10^{6} \frac {A}{m^{2}}}{(8.5 \cdot 10^{28} {m^{-3}}) (1.6 \cdot 10^{-19} C)}$

$v = 1.50 \cdot 10^{-04} \frac {m}{s}$

So, the drift speed is 1.50x10⁻⁴ m/s.

Have a nice day!

4 0

4 years ago

Read 2 more answers

1. A note has a wavelength of 0.77955 m. If the speed of sound is 343.00 m/s, what pitch is this note? 2. A note has a wavelengt

AVprozaik [17]

Answer:

1.) 440 Hz

2.) 659.3 Hz

Explanation:

1.) Given parameters are:

wavelength = 0.77955 m.

speed of sound = 343.00 m/s

Frequency = speed/ wavelength

Substitute speed and wavelength into the formula

Frequency = 343/ 0.77955

Frequency = 439.99

Frequency = 440 Hz approximately

2.) The parameters given are:

wavelength = 0.52028 m.

speed of sound = 343.00 m/s

Using the same formula

Frequency = speed/wavelength

Substitute all the parameters into the formula

Frequency = 343 / 0.52028

Frequency = 659.3 Hz approximately

The pitch of a note depends on the frequency of the sound waves.

The pitch of a sound increases as the frequency of the sound waves increases.

5 0

3 years ago

Read 2 more answers