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3 years ago

Compare and contrast aerobic physical activity with anaerobic physical activity.

2 answers:

tatiyna3 years ago

5 0

Answer: ompare and contrast aerobic physical activity with anaerobic physical activity. ... Aerobic physical activity is better for the cardiovascular system than it is for strength endurance, while anaerobic physical activity is better at building strength endurance than it is at increasing cardiorespiratory fitness.

Explanation: hopefully this helped ;)

Natali [406]3 years ago

5 0

Answer:

Aerobic physical activity is actvity with motion. Anaerobic activity is the opposite. This means that it is an activity without motion.

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A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released

anastassius [24]

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

substituting the values of the variables into the equation, we have

k = mg/x

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

5 0

3 years ago

Math Practice: Cart A loaded with blocks (600g) moving left at 0.7m/s hits stationary cart B (200g).

Vika [28.1K]

Answer:

The velocity of cart B after the collision is 1.29 m/s.

Explanation:

We can find the velocity of cart B by conservation of linear momentum:

$p_{i} = p_{f}$

$m_{A}v_{A_{i}} + m_{B}v_{B_{i}} = m_{A}v_{A_{f}} + m_{B}v_{B_{f}}$

Where:

$m_{A}$ is the mass of cart A = 600 g = 0.6 kg

$m_{B}$ is the mass of cart B = 200 g = 0.2 kg

$v_{A_{i}}$ is the inital velocity of cart A = 0.7 m/s

$v_{A_{f}}$ is the final velocity of cart A = 0.27 m/s

$v_{B_{i}}$ is the initial velocity of cart B = 0

$v_{B_{f}}$ is the final velocity of cart B =?

Taking the left direction as the positive horizontal direction:

$0.6 kg*0.7 m/s + 0 = 0.6 kg*0.27 m/s + 0.2 kg*v_{B_{f}}$

$v_{B_{f}} = 1.29 m/s$

Therefore, the velocity of cart B after the collision is 1.29 m/s.

I hope it helps you!

5 0

3 years ago

Can uh help in in this question step by step

Luda [366]

Initial velocity=u=72km/h

Convert to m/s

$\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s$

Final velocity=v=0m/s
Time=2s=t

$\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}$

$\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}$

$\\ \sf \longmapsto Acceleration=a=-10m/s^2$

Distance be s

Using second equation of kinematics

$\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2$

$\\ \sf \longmapsto s=40+(-20)$

$\\ \sf \longmapsto s=40-20$

$\\ \sf \longmapsto s=20m$

Now

Mass=m=5000kg

Using newtons second law

$\\ \sf \longmapsto Force=ma$

$\\ \sf \longmapsto Force=5000(-10)$

$\\ \sf \longmapsto Force=-50000N$

Force is in opposite direction so its negative

$\\ \sf \longmapsto Force=50kN$

7 0

3 years ago

Read 2 more answers

A baseball is thrown directly upward at time t= 0 and is caught again at time t=5s. Assume that air resistance is so small that

babymother [125]

Answer:

Explanation:

since the ball was thrown at 0s and caought again at 5 s. applying eqaution of motion.

$s = ut-\frac{1}{2}\times g \times t^{2}$

0 = u×5 - $\frac{1}{2}$ ×10×5×5

solving the eqaution we gaet initial velocity u = 25 m/s.

there fore total energy E = $\frac{1}{2}$ ×m×25×25 J

where m is the mass of the ball according to conservation of energy E remains constant

conservation of energy:

kinetic + potential energy of the ball = E

kinetic energy = E - mgh $\rightarrow$ 1

h = $ut - \frac{1}{2}\times g\times t^{2}$

applying ot in the eqaytion 1

kinetic energy = e - $mg(25t - \frac{1}{2}\times g\times t^{2})$ 2

Therefore kinetic energy vs height will be a straight line with negative slope and kinetic energy vs time will be parabola that is open upward.

6 0

3 years ago

Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024

myrzilka [38]

Given: Mass of earth Me = 5.98 x 10²⁴ Kg

Radius of earth r = 6.37 x 10⁶ m

G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r

but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²

GMe = 4π²r³/T²

T² = 4π²r³/GMe

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds or around 1.4 Hour

3 0

3 years ago