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3 years ago

An oxygen deficit occurs

2 answers:

6 0

There is an oxygen deficit when an oxygen intake exceeds its intake in the body. During strenuous exercise, oxygen deficits occur naturally. The body works to fill up oxygen levels during the so - called regeneration period, when the drill is responsible for oxygen deficits.

Stolb23 [73]3 years ago

4 0

A second definition comes from Work Place Testing: “Oxygen deficit exists when a body's consumption of oxygen exceeds its intake.” Oxygen deficit occurs naturally during strenuous exercise. When exercise triggers an oxygen deficit, the body will work to replenish oxygen levels during what is known as a recovery period.

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A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring

topjm [15]

Here we can use the work energy theorem

$W_f + W_s = K_f - K_i$

here we know that

$K_f = 0$

as it come to rest finally

$K_i = \frac{1}{2}mv_i^2$

$K_i = \frac{1}{2}\times 1\times 2^2$

$K_i = 2 J$

now work done by friction force will be given as

$W_f = - F_f \times d = -\mu mg d$

$W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu$

Work done by spring force is given as

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$

$W_s = \frac{1}{2}(10)( 0 - 0.10^2)$

$W_s = -0.05 J$

so now plug in all data above

$- 0.05 - \mu(0.98) = 0 - 2$

$\mu = 1.99$

so above is the friction coefficient

4 0

3 years ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago

A tightrope walker more easily balances on a tightwire if his pole

cestrela7 [59]

B) droops.

Why?
To maintain balance, you do not need something short so you're balanced well... You need something long and droopy to maintain balance. The pole should be held by your waist and it should be light.

Hope this helps!~

4 0

2 years ago

HELP!!!!

kykrilka [37]

Answer:

Clouds form when below the dew point

4 0

3 years ago

30 POINTS!!! CAN U AWNSER IT?? :)

solniwko [45]

Answer:

5235.84 kg

Explanation:

There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.

$F\Delta t = m \Delta v$ As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, $\Delta v = 69.4 m/s$ ;

$45 450 \cdot 8 = 69.4 m \rightarrow m = \frac{45450\cdot 8}{69.4} = 5235.84 kg$

7 0

2 years ago