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3 years ago

State Newtons first law of motion.

Physics

1 answer:

Fofino [41]3 years ago

8 0

Answer:

An object at rest will stay at rest and an object in motion will stay in motion unless it is acted upon by an unbalanced/external force.

You might be interested in

⦁ Match the following terms:

ki77a [65]

Answer:

Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.

Isotopes - ⦁ Atoms with the same number of protons, but different number of neutrons.

Nitrogen - ⦁ The name of the element with atomic number 7.

Atomic number - ⦁ The number of protons in the nucleus of an atom.

4 0

2 years ago

A violin string vibrates at 260 Hz when unfingered. At what frequency will it vibrate if it is fingered one fourth of the way do

Advocard [28]

Answer:

346.66 Hz

Explanation:

$l_1$ = Length of string which is unfingered = l

$l_2$ = Length of string which is vibrate when fingered = $l-\dfrac{1}{4}l=\dfrac{3}{4}l$

$f_1$ = Unfingered frequency = 260 Hz

$f_2$ = Fingered frequency

Frequency is inversely proportional to length

$f=\dfrac{1}{l}$

So,

$\dfrac{f_1}{f_2}=\dfrac{l_2}{l_1}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{\dfrac{3}{4}l}{l}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{3}{4}\\\Rightarrow f_2=\dfrac{4}{3}f_1\\\Rightarrow f_2=\dfrac{4}{3}260\\\Rightarrow f_2=346.66\ Hz$

The frequency of the fingered string is 346.66 Hz

3 0

2 years ago

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

2 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stoc

vivado [14]

Answer:

Δ $R_{e}$ = 84 Ω, $R_{e}$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_{e}$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_{e}$ = 1 / R₁ + 1 / R₂

1 / $R_{e}$ = 1/500 + 1/2000 = 0.0025

$R_{e}$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_{e}$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_{e}$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_{e}$ / $R_{e}$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_{e}$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_{e}$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_{e}$ = 0.21 400

Δ $R_{e}$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_{e}$ = (40 ± 8) 10¹ Ω

6 0

2 years ago