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3 years ago

What is the general aerobic activity recommendation for an exercise frequency that is at least five days per week?

Physics

1 answer:

Fantom [35]3 years ago

4 0

Explanation:

Minimum of aerobic activity 150 minutes, or a mix of moderates and intensive exercise 75 minutes of vigorous aerobic activity a week. We can extend this practice throughout the week with the instructions. Mild aerobic workouts might include practices like quick strolling or swimming, while activity like running can include strong aerobic activity.

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Which of the following speeds is greatest? (2.54 cm = 1.00 in.)

Mama L [17]

Answer:

Hi, There!

O 10 m/s

O 10 yd/s

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Have a great day!

7 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10

sashaice [31]

Answer:

Time interval;Δt ≈ 37 seconds

Explanation:

We are given;

Angular deceleration;α = -1.6 rad/s²

Initial angular velocity;ω_i = 59 rad/s

Final angular velocity;ω_f = 0 rad/s

Now, the formula to calculate the acceleration would be gotten from;

α = Change in angular velocity/time interval

Thus; α = Δω/Δt = (ω_f - ω_i)/Δt

So, α = (ω_f - ω_i)/Δt

Making Δt the subject, we have;

Δt = (ω_f - ω_i)/α

Plugging in the relevant values to obtain;

Δt = (0 - 59)/(-1.6)

Δt = -59/-1.6

Δt = 36.875 seconds ≈ 37 seconds

6 0

4 years ago

A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of

PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box. There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0

3 years ago

The standard measure used to compare sound intensities is the ______ .

uranmaximum [27]

The answer is Decibels. <span />

4 0

3 years ago

Read 2 more answers