Which of these statements correctly describes a key difference between aerobic activity and anaerobic activity?

A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?

Anettt [7]

Answer:

The acceleration of the bag of sand is $1\ m/s^2$ .

Explanation:

We have,

Mass of a bag of sand is 100 kg

Weight of the bag of sand is 100 N

It is required to find the acceleration of the bag when it is dropped. The weight of an object is given by :

$F =ma$

When it is dropped, a = g

$g=\dfrac{F}{m}\\\\g=\dfrac{100}{100}\\\\g=1\ m/s^2$

So, the acceleration of the bag of sand is $1\ m/s^2$ .

4 0

3 years ago

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 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

Which choice has more thermal energy?

prohojiy [21]

Answer:

If thermal energy is the motion energy of the particles of a substance, which has more thermal energy—the cup of hot tea or a spoonful of hot tea? It makes sense that the more particles of a substance you have, then the more thermal energy the substance has. The cup of hot tea would have more thermal energy, even if the temperature of the tea is the same in the cup and in the spoon. But which cools down the quickest (has the highest rate of thermal energy transfer)—the tea in the cup or the tea in the spoon? If I have fewer particles of the same substance, then the rate of thermal energy transfer is faster. The tea in the spoon would lose thermal energy more rapidly. So the amount of a substance you have is one factor that affects the rate of thermal energy transfer.

Explanation:

7 0

3 years ago

A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degr

7nadin3 [17]

Answer: 1.88

Explanation

Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.

1) sin35/sin24 = n(2)/1.33
2) 1.41 = n(2)/1.33
3) n(2) = 1.41 x 1.33
4) n(2) = 1.88

Hope this helps :)

7 0

3 years ago