7. What does the changing colour perceived by the person as the filter changes indicate to you

What is the mass of a truck in grams of it produces a force of 1500N while accelerating at a rate of 6 m/s²?

aleksley [76]

Answer:

250,000

Explanation:

<h2>formula = ( F=ma </h2>

F=1500N
a=6m/s^2
F= ma
m=?
1500/6 = m
m=250 kg
1kg =1000gm so 250kg =250,000gm
m =250×10^3 gm

5 0

3 years ago

An iron nail goes through a chemical change when iron in the nail combines with oxygen in the air to form iron oxide. This compo

Mazyrski [523]

It is like that, except most nails are steel or stainless steel, slowing to rusting process to about 5 years.

6 0

2 years ago

What is the sequence of energy transformations associated with a hydroelectric dam?

Hatshy [7]

Gravitational potential energy -> Kinetic energy -> Mechanical energy -> Electrical energy.

The water starts up (potential) and flows down (kinetic), the flowing water turns a big wheel (mechanical) which creates electricity (electrical).

3 0

2 years ago

Read 2 more answers

If the speed of sound in air is 343 m/s and the wavelength of this note is 2.62 m, what is the frequency of this C3 note? (Round

snow_lady [41]

Answer:

The frequency of of the note = 131 Hz.

Explanation:

Frequency: Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)

v = λf ............................ Equation 1

Making f the subject of the equation,

f = v/λ .......................... Equation 2

Where v = Speed, λ = wavelength, f = frequency

Given: v = 343 m/s, λ = 2.62 m.

Substituting these values into equation 2

f = 343/2.62

f = 131 Hz

Thus the frequency of of the note = 131 Hz.

8 0

2 years ago

You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g

Vlada [557]

Answer:

(a) L = 128.75 m

(b) L = 182.56 m

(c) L = 114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

L = 128.75 m

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

L = 182.56 m

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

L = 114.28 m

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

Mass of Gold = 7.68 kg = 7680 gram

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

Cost of Gold Wire = $ 307040

7 0

3 years ago