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Fed [463]
2 years ago
15

Please help me answer this question

Physics
1 answer:
Pavel [41]2 years ago
4 0
It can either be all of them or just 1 and 3
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Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene
sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0
3 years ago
What are some possible examples of genetic characteristics that can be passed down to child??????
elena-s [515]
They can have a close similar appearance to the parents, have close relation of child reactions.


for example, everyone born in my father's side of the family had the tendency to bump their head on something as they fall asleep up to the point when you are a toddler.
8 0
3 years ago
Read 2 more answers
Night vision glasses detect energy given off by objects as (2 points)
Allushta [10]
The answer here would be infrared waves. Hot objects and humans give off heat in the form of infrared light, thermal imaging technology in the goggles enable them to catch this light emitted by these objects 
6 0
2 years ago
Which is an example of gaining a static charge by conduction?
lianna [129]

Answer:

shuffling shoes

Explanation:

7 0
3 years ago
Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge b
Elena L [17]

Answer:

a) 0.167 μC/m^2

b) 1.887 * 10^4 V/m

Explanation:

Hello!

First let's find the surface charge density:

a)

Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:

σ = Q/A

Where A is the area of the satellite, which is:

A=4πr^2 = πd^2 = π(1.9m)^2

Therefore:

σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2

Now let's calculate the electric field

b)

Just outside the surface of the satellite the elctric field will be:

E = σ/ε0

Where      ε0=8.85×10^−12 C/Vm

Therefore:

E = (0.167*10^-6 C/m^2) / (8.85*10^-12  C/Vm) = 0.01887 *10^6 V/m

E = 1.887 * 10^4 V/m

5 0
2 years ago
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