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3 years ago

Xplain the difference between aerobic and anaerobic exercise and give 1 example of each?

Physics

2 answers:

gregori [183]3 years ago

7 0

Answer:

An example of an aerobic exercise is swimming. An example of an anaerobic exercise is sprinting.

Explanation:

Lilit [14]3 years ago

6 0

Answer:

The term "aerobic" refers to the body's ability to provide energy through the use of oxygen. The term anaerobic refers to the body's ability to provide energy without the use of oxygen.

An example of an aerobic exercise is swimming. An example of an anaerobic exercise is sprinting.

Explanation:

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Calcula el peso aparente de una bola de aluminio de 50 cm3, cuando se encuentra totalmente sumergida en alcohol. Datos: la densi

artcher [175]

Answer:

W_apparent = 93.1 kg

Explanation:

The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.

W_apparent = W - B

The push is given by the expression of Archimeas

B = ρ_fluide g V

ρ_al = m / V

m = ρ_al V

we substitute

W_apparent = ρ_al V g - ρ_fluide g V

W_apparent = g V (ρ_al - ρ_fluide)

we calculate

W_apparent = 980 50 (2.7 - 0.8)

W_apparent = 93100 g

W_apparent = 93.1 kg

7 0

2 years ago

The distance between two stations is 180 km. A train takes 2 hours to cover this distance. The speed of the train in m/sec is...

Mamont248 [21]

Answer:

Av = 25 [m/s]

Explanation:

To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.

$Av=\frac{distance}{time}$

where:

Av = speed [km/h] or [m/s]

distance = 180 [km]

time = 2 [hr]

Therefore the speed is equal to:

$Av = \frac{180}{2} \\Av = 90 [km/h]$

Now we must convert from kilometers per hour to meters per second

$90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]$

4 0

2 years ago

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

5 0

2 years ago

True or False? 13. All living things are made of cells 14. All cells have DNA within their nucleus 15. The cell is the basic uni

Leni [432]

Answer:

1: True

2: False

3: True

4: False

5: True

3 0

2 years ago

Choose the situation below in which the force applied is the greatest.

Gnesinka [82]

Answer:

Explanation:

We know the formula for Work to be:

W = f * d

Where W is work done

f is force

d is the distance

Work = 50

Distance = 50

So, Force is:

Force = 50/50 = 1

Work = 400

Distance = 80

Force = 400/80 = 5

Work = 365

Distance = 73

Force = 365/73 = 5

Work = 144

Distance = 16

Force = 144/16 = 9

Hence, D is the situation in which the force applied is the greatest.

6 0

3 years ago