Explain why is the temperature of a hot tea higher than the temperature of iced tea?

How old is a bone if it still has 50% of its carbon-14 content?

Degger [83]

Answer:

5730 years

Explanation:

The half life of carbon-14 is 5730 years. If 50% of the carbon-14 remains, then exactly 1 half life has passed.

The half-life equation is:

A = A₀ (½)^(t / T)

where A is the remaining amount,

A₀ is the initial amount,

t is time,

and T is the half life.

In this case, A = ½ A₀ and T = 5730.

½ A₀ = A₀ (½)^(t / 5730)

½ = (½)^(t / 5730)

1 = t / 5730

t = 5730

6 0

2 years ago

Which color in the rainbow has the shortest wavelength?

klio [65]

Answer:

violet

Explanation:

violet has shortest wavelength

8 0

2 years ago

What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long water tank that can hold 50.0 kg of water by the weight

svetoff [14.1K]

Answer:

1088.9N/m2

Explanation:

Calculation for What pressure is exerted

First step is to find the area of bottom of the tank using formula

Area=Width*breadth

Let plug in the formula

Area=0.5*0.9

Area=0.45m2

Now let calculate what pressure is exerted using this formula

Pressure=Force/Area

Where,

Force=Mass *Gasoline

Area=Width of the tank* Length of the tank

Let plug in the formula

Pressure=50*9.8/0.5*0.9

Pressure=490/0.45

Pressure=1088.9N/m2

Therefore What pressure is exerted is 1088.9N/m2

3 0

2 years ago

A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does an equally thick wire 6 m long, made

shusha [124]

Answer:

2.4 mm

Explanation:

Given that:

Initial Original length of the wire L = 3 mm

The stretch of the first wire ΔL= 1. 2 mm

The length of the second wire L'' = 6 mm

The stretch of the second wire ΔL'' = ???

Considering the Tension of the system; the Young modulus and the cross sectional remains constant ; as such:

$\frac{Y}{Y''} = \frac{FL}{A \Delta L} *\frac{A \Delta L''}{FL''}$

$1= \frac{L \Delta L''}{L'' \Delta L}$

$\Delta L''= \frac{L'' \Delta L }{L}$

$\Delta L''= \frac{6 \ m * 1.2 \ mm }{3 \ m}$

$\Delta L''=2.4 \ mm$

Thus, the same material under the same tension stretches 2.4 mm

8 0

3 years ago

Read 2 more answers

Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard

podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0

3 years ago