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2 years ago

3. CrossFit and cross-training exercise routines can only be performed in a class setting in a gym.

Physics

2 answers:

Marrrta [24]2 years ago

8 0

Person above is correct it is false

denis23 [38]2 years ago

5 0

Answer:

False

Explanation:

I would greatly appreciate if you can give the brainliest answer crown!

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Currently, Earth's tilt is closest to _____.

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Earth's tilt is closest to 23.5<span>°. (C)</span>

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4 years ago

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What is the net force of a falling bowling ball that weighs 55 N and has 15 N of air resistance?

serg [7]

its 55 :) have a great day queen/king/whatever gender you are

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3 years ago

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A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on the inside of a circular track. During one of t

kirill115 [55]

Answer:

w = -0.475N

Explanation:

$K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N\\$

To get Va and Vb

$F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R = N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}$

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

$v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s$

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

$F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R = N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}$

$N_{B} = 0.665N$

$v_{B}= \sqrt{0.525(0.665 /0.0350 - 9.8)} \\v_{B} = 2.198 m/s$

$K.E_{B} = 0.5* 0.035 * 2.198^{2} \\K.E_{B} = 0.085 J$

$P.E_{B} = mgh_{B} \\h_{B} = A diameter = 2R = 2 * 0.525\\h_{B} = 1.05 m\\P.E_{B} = 0.035 * 9.8 * 1.05\\P.E_{B} =0.36 J$

$from K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}$

$0.92 + w_{fr} + 0 = 0.085 + 0.36\\ w_{fr} = -0.475J$

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4 years ago

A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circu

Lyrx [107]

Answer:

The magnitude of the lift force L = 92.12 kN

The required angle is ≅ 16.35°

Explanation:

From the given information:

mass of the airplane = 9010 kg

radius of the airplane R = 9.77 mi

period T = 0.129 hours = (0.129 × 3600) secs

= 464.4 secs

The angular speed can be determined by using the expression:

ω = 2π / T

ω = 2 π/ 464.4

ω = 0.01353 rad/sec

The direction $\theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})$

$\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})$

θ = 16.35°

The magnitude of the lift force L = mg ÷ Cos(θ)

L = (9010 × 9.81) ÷ Cos(16.35)

L = 88388.1 ÷ 0.9596

L = 92109.32 N

L = 92.12 kN

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3 years ago

What does stressing an Equilibrium system mean? How is stress Applied?

horsena [70]

Stressing an equilibrium simply means that the physical properties in which already exists are balanced. Stress can be applied by either changing the pressure or the volume or temperature.

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4 years ago

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