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Airida [17]
3 years ago
7

What is the heat extracted from the cold reservoir for the refrigerator?

Physics
1 answer:
zaharov [31]3 years ago
7 0
What is the heat extracted from the cold reservoir for the refrigerator shown in(Figure 1) ? Assume that W1 = -123J and W2 = 88J . 

<span>Qc= _________ </span>

<span>Part B 
</span>
K=105J
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Of the following, which generally causes the greatest damage?
maksim [4K]
I guess it is a: seismic waves
3 0
3 years ago
at a certain moment, an object has an amount of 200 of motion energy and 400 of gravitational potential energyThe object is also
Radda [10]

Answer:

twice as much energy

Explanation:

8 0
1 year ago
Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?
IRISSAK [1]

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

It takes 22,800 years.

4 0
3 years ago
A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m . What is the effective spring
ANEK [815]

Answer:

29856.521 N/m

Explanation:

m=Mass\ of\ diver=70\ kg\\x=Length\ compressed\ by\ spring=2.3\times 10^{-2}\ m\\a=Acceleration\ due\ to\ gravity=9.81\ m/s^2\\F=Force\ exerted\ by\ diver=m\times a\\\Rightarrow F=70\times 9.81\\\Rightarrow F=686.7\ N\\k=spring\ constant\\F=k\times x\\\Rightarrow k=\frac {F}{x}\\\Rightarrow k=\frac {686.7}{2.3\times 10^{-2}}\\\therefore k=29856.521\ N/m

5 0
3 years ago
Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0
3 years ago
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