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2 years ago

The specific heat of water is 4.18 J/gºC. How many Joules of

Physics

1 answer:

densk [106]2 years ago

8 0

Answer:

Q = 59565 [J]

Explanation:

In order to calculate the amount of thermal energy needed we must use the following equation.

$Q=m*C_{p}*(T_{i}-T_{f})$

where:

m = mass = 150 [g]

Cp = 4.18 [J/g*°C]

Tfinal = 5 [°C]

Tinitial = 100 [°C]

Now replacing:

$Q=150*4.18*(100-5)\\Q=59565[J]$

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A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w

Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

4 0

3 years ago

Which microbes can be Eukaryotic? A. bacteria B. virus C. protists D. fungi

MissTica

Hey there,

Question : Which microbes can be Eukaryotic?

Answer : A, Bacteria

Hope this helps :D

<em>~Top♥</em>

3 0

3 years ago

Read 2 more answers

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

6 0

3 years ago

Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from a

riadik2000 [5.3K]

Answer:

Check the explanation

Explanation:

The beat frequency is

df = f2 - f1

the wavelength is

lamda1 = (v/f1)

and lamda2 = (v/f2)

where v = 340 m/s,f1 = 25.0 kHz and f2 = 20.0 kHz

8 0

3 years ago

Molly does not have many activities after school, so she has plenty of time to exercise. She has never been very athletic and pl

elixir [45]

The answer is D No, smaller goals would be better

4 0

3 years ago