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3 years ago

Why are the eggs and confectioner’s sugar amounts easy to work with to make 8 brownies?

Physics

1 answer:

Orlov [11]3 years ago

6 0

Answer:

Because they are both easy to measure (?)

Explanation:

(I'm not really sure, there are no choices. If there were different options I might be able to better answer this)

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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

A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t

Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

$d = d_1 + d_2$

given that

$d_1 = 200 cos20 i + 200 sin20 j$

$d_1 = 187.94 i + 68.4 j$

$d_2 = -230 sin30 i + 230 cos30 j$

$d_2 = -115 i + 199.2 j$

now the total distance is

$d = (187.94 - 115)i + (199.2 + 68.4)j$

$d = 72.94 i + 267.6 j$

now the magnitude of the distance is given as

$d = \sqrt{72.94^2 + 267.6^2}$

$d = 277.4$

also the direction is given as

$\theta = tan^{-1}\frac{267.6}{72.94}$

$\theta = 74.7 degree$

<em>so it is 277.4 km at 74.7 degree North of East</em>

8 0

3 years ago

An RL series circuit is connected to an ac generator with a maximum emf of 20 V. If the maximum potential difference across the

Rama09 [41]

If the maximum emf of the ac generator is 20 V and the maximum potential difference across the resistor is 16 V Then the maximum potential difference across the inductor is 4 V.

Calculation:

Step-1:

It is given that the RL circuit is connected to a 20 V ac generator. The maximum potential difference across the resistor is 16 V. It is required to find the maximum potential drop across the inductor.

Step-2:

The maximum emf of the generator is equal to the sum of the maximum potential difference across the resistor and the maximum potential difference across the inductor.

Therefore,

The maximum potential difference across the inductor + Maximum maximum potential difference across the resistor = Maximum emf of the generator

Thus,

Maximum maximum potential difference across the inductor + 16 V = 20 V

Therefore,

Maximum maximum potential difference across the inductor = 20 V - 16 V = 4 V

Learn more about potential differences across resistor and inductor here,

brainly.com/question/15715072

#SPJ4

5 0

2 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$