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4 years ago

What information can be learned from bloodstains at a crime scene?

Chemistry

1 answer:

tangare [24]4 years ago

3 0

you can learn who's blood it is, you can also see the blood patterns and how the crime maybe happened.

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La leche se considera un alimento basico por su elevado contenido de nutrientes su densidad es de 1,4g/ml. Si se hace una compra

noname [10]

Answer:

8.82 kg es la masa total de leche

Explanation:

Consideraremos que el volumen de cada bolsa de leche es de 900 mL.

Sabiendo que tenemos en total 7 bolsas, el volumen final en litros que tenemos disponible es de:

900 mL . 7 = 6300 mL

Utilizando la densidad podemos calcular la masa de leche que tenemos en el total de las 7 bolsas

Densidad = masa / volumen

Densidad . volumen = masa → 1.4 g/mL . 6300 mL = 8820 g

Convertimos la masa de gramos a kg

8820 g . 1 kg /1000g = 8.82 kg

5 0

3 years ago

What is it called when water drops form something

GarryVolchara [31]

It is the gravitational force pulling on something

4 0

4 years ago

What is your "Resting Heart Rate" for 1 minute? (it will be a number)

taurus [48]

A resting heart rate is usually around 40 beats per minute (Thats the normal amount)

8 0

3 years ago

Read 2 more answers

A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a l

elixir [45]

Answer:

$T_{C}$ = -4.2°C

$T_{H}$ = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

| $\frac{Q_{C} }{Q_{H} }$ | = $\frac{T_{C} }{T_{H} }$

Similarly,

| $Q_{H}$ | = | $Q_{C}$ | + | $W_{S}$ |

During winter, the value of | $T_{H}$ | = 20°C = 273.15 + 20 = 293.15 K and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{H}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{H} }$ | = 1 - $\frac{T_{C} }{T_{H} }$

1.5/0.75*(293.15- $T_{C}$ ) = 1 - ( $T_{C}$ /293.15

Further simplification,

$T_{C}$ = -4.2°C

During summer, $T_{C}$ = 25°C = 273.15+25 = 298.15 K, and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{C}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{C} }$ | = $\frac{T_{H} }{T_{C} }$ - 1

1.5/0.75*( $T_{H}$ - 298.15) = ( $T_{H}$ /298.15

Further simplification,

$T_{H}$ = 49.4°C

4 0

4 years ago

Please help ASAP!<br> Question and answers below in the picture.

Ber [7]

Acceleration i am pretty sure

4 0

3 years ago

Read 2 more answers