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2 years ago

A unicorn is running 3 meters/second.how long will it take to get at the end of a 120 meter rainbow.

Physics

1 answer:

Aneli [31]2 years ago

3 0

Answer:

40 seconds

Explanation:

A unicorn is running at rate of 3 meters/second. To find how long it'll take to get at the end of a 120 meter rainbow, we'll have to find the "time".

The distance is given to be 120 meters and speed is given as 3 meters/second.

Our formula for speed is:

$\displaystyle{v = \dfrac{s}{t}}$

where v is speed, s is distance and t is time:

$\displaystyle{3 = \dfrac{120}{t}}$

Solve for t, multiply both sides by t:

$\displaystyle{3 \cdot t= \dfrac{120}{t}\cdot t}\\\\\displaystyle{3t= 120}$

Divide both sides by 3:

$\displaystyle{\dfrac{3t}{3} = \dfrac{120}{3}}\\\\\displaystyle{t= 40 \ \, \sf{seconds}}$

Therefore, it will take 40 seconds for a unicorn to reach at the end of rainbow.

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Allisa [31]

1). I started up my car. Gasoline was spritzed into the cylinders, mixed with air, and then exploded with an electrical spark. As the gasoline vapor instantly burned in the air, several new things were formed that weren't there before, like carbon dioxide, carbon monoxide, water, and oxides of nitrogen.

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3 years ago

A skull believed to belong to an ancient human being has a carbon-14 decay rate of 5.4 disintegrations per minute per gram of ca

larisa86 [58]

Answer:

9.43*10^3 year

Explanation:

For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation

To start with, we use the formula

t(half) = In 2/k,

if we make k the subject of formula, we have

k = in 2/t(half), now we substitute for the values

k = in 2 / 5730

k = 1.21*10^-4 yr^-1

In(A/A•) = -kt, on rearranging, we find out that

t = -1/k * In(A/A•)

The next step is to substitite the values for each into the equation, giving us

t = -1/1.21*10^-4 * In(5.4/15.3)

t = -1/1.21*10^-4 * -1.1041

t = 0.943*10^4 year

8 0

4 years ago

A 2.8-carat diamond is grown under a high pressure of 58 × 10 9 N / m 2 .

Rzqust [24]

Answer:

a) ΔV = 2,118 10⁻⁸ m³ b) ΔR= 0.0143 cm

Explanation:

a) For this part we use the concept of density

ρ = m / V

As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats

m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg

V = m / ρ

V = 0.56 / 3.52

V = 0.159 cm3

We use the relation of the bulk module

B = P / (Δv/V)

ΔV = V P / B

ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹

ΔV = 2,118 10⁻⁸ m³

b) indicates that we approximate the diamond to a sphere

V = 4/3 π R³

For this part let's look for the initial radius

R₀ = ∛ ¾ V /π

R₀ = ∛ (¾ 0.159 /π)

R₀ = 0.3361 cm

Now we look for the final volume and with this the final radius

$V_{f}$ = V + ΔV

$V_{f}$ = 0.159 + 2.118 10⁻²

$V_{f}$ = 0.18018 cm3

$R_{f}$ = ∛ (¾ 0.18018 /π)

$R_{f}$ = 0.3504 cm

The radius increment is

ΔR = $R_{f}$ - R₀

ΔR = 0.3504 - 0.3361

ΔR= 0.0143 cm

4 0

3 years ago

Radar uses radio waves of a wavelength of 2.4 ${\rm m}$ . The time interval for one radiation pulse is 100 times larger than t

blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x $10^{8}$ m/s)

f= 3 x $10^{8}$ / 2.4

f=1.25 x $10^{8}$ hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x $10^{8}$

T= 8 x $10^{-9}$ s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x $10^{-9}$ => 800 x $10^{-9}$ s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

$x_{min}$ = τc/2 => (800 x $10^{-9}$ x 3 x $10^{8}$ )/2

$x_{min}$ =120m

8 0

4 years ago

What mass of a material with density Ï is required to make a hollow spherical shell having inner radius r1 and outer radius r2?

inysia [295]

<span>Yes, there are! r1 and r2 are numbers. The volume of the hollow shell is 4 π 3 ( r 3 1 − r 3 2 ) 4π3(r13−r23). Now multiply by ρ to get the mass.</span>

3 0

3 years ago