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3 years ago

To improve your cardiovascular fitness:

Physics

2 answers:

RoseWind [281]3 years ago

4 0

C.Perform push ups and curl-ups daily

matrenka [14]3 years ago

4 0

Answer:

A. Include a minimum of 20-30 minutes of aerobic activity 3-4 times per week

Explanation:

Your cardiovascular fitness, also called cardiovascular fitness, says a lot about your health and how much other conditions can affect it. Basically, cardiovascular fitness measures how well your body absorbs oxygen and distributes it to your muscles and organs during prolonged periods of exercise. The best way to improve your cardiovascular fitness is through exercise that forces and stimulates the heart to work harder. Aerobic exercise is the best alternative for this.

For this reason, we can say that the best answer to your question is the letter A.

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Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

3 years ago

The iron nail’s mass is 16 grams and its temperature drops 650 C when dropped into the water. How much heat energy did the iron

Mice21 [21]

The heat energy transferred by the iron nail is 4680 J

Explanation:

The thermal energy transferred by a substance to another substance is given by the equation

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is its change in temperature

For the iron nail in this problem, we have:

m = 16 g

$C=0.450 J/g^{\circ}C$

$\Delta T = -650^{\circ}C$

So, the amount of heat energy given off by the nail is

$Q=(16)(0.450)(-650)=-4680 J$

where the negative sign indicates that the heat is given off.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

5 0

3 years ago

Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi

Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

P₁ = m v

P₁ = 0.0575 x 26.7

P₁ = 1.535 kg.m/s

b) momentum of ball after collision

P₂ = m v'

P₂ = 0.0575 x (-29.5)

P₂ = -1.696 kg.m/s

c) change in momentum

Δ P = P₂ - P₁

Δ P = -1.696 -1.535

Δ P = -3.231 kg.m/s

d) using conservation of momentum

initial speed of racket = 0 m/s

M u + m v = Mu' + m v

M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

0.331 u' = 3.232

u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0

3 years ago

An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o

borishaifa [10]

Hi there!

Recall the equation for centripetal force:
$F_c = \frac{mv^2}{r}$

We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
$r * F_c = mv^2$

Divide both sides by Fc:
$\boxed{ r= \frac{mv^2}{F_c}}$

7 0

3 years ago

What is the difference between weight and mass? Give an example of how they are different.

marusya05 [52]

Mass is the amount of matter it has, while weight <span>is a measurement of the force placed on an object by gravity. An example of how they are different is that mass is always constant while weight varies depending on the location.</span>

8 0

3 years ago