PHYSICS COLLEGE

What is an inexpensive, portable, and common way to assess body fat in the fitness industry?DEXA
Bioelectrical impedance
Skinfold testing
Hydrostatic weighing

Answers

Answer 1
Answer:

Answer: Skinfold testing

Explanation:

Skinfold testing, is also referred to as calliper testing and it's used to know the body fat percentage. Skinfold testing is an inexpensive, portable, and common way to assess body fat in the fitness industry.

It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

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The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel wire with this length and a cross-sectional area of 3.10 ✕ 10^−6 m^2 is laid on the bridge deck with its ends attached to the towers of the bridge, on a summer day when the temperature of the wire is 43.0°C. When winter arrives, the towers stay the same distance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to −10.0°C, what is the tension in the wire? Take Young's modulus for steel to be 20.0 ✕ 10^10 N/m^2. (Assume the coefficient of thermal expansion of steel is 11 ✕ 10−6 (°C)−1.)
Define reflection of sound?​
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Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black puck is 1.5 m/s. What is the speed of the white puck immediately after the collision?

Answers

Answer:

The speed of the white puck immediately after the collision is 2.6 m/s.

Explanation:

Given that,

Two pucks are equal masses.

Speed of black puck = 1.5 m/s

According to given figure,

We need to calculate the speed of the white puck immediately after the collision

Using law of conservation of momentum

mv=m_(1)v_(1)\cos\theta+m_(2)v_(2)\cos\theta

Put the value into the formula according to figure

m*3=m* v_(1)*\cos30+m*1.5*\cos60

3m=0.866m v_(1)+0.75m

v_(1)=(3-0.75)/(0.866)

v_(1)=2.6\ m/s

Hence, The speed of the white puck immediately after the collision is 2.6 m/s.

Coach ulcer paces the sidelines. Sarting at the 30 yd. line (A), he moves to the 10 yd. line (B), back to the 50 yd. line (C) and finally to the 20yd. Line (D) in 200 seconds. Determine his average speed and velocity.

Answers

Answer:

See the answer below

Explanation:

Average speed = total distance traveled/total time taken

In order to determine the total distance traveled by the coach, consider the attached image.

Distance covered:

30 yd. line to 10 yd. line (A to B)= 20 yds

10 yd. line to 50 yd. line (B to C) = 40 yds

50 yd. line to 20 yd. line (C to D) = 30 yds

Total distance covered = 20 + 40 + 30 = 90 yds

Time taken = 200 seconds

Average Speed = 90/200 = 0.45 yd/s

Velocity = speed with direction

Hence,

His Velocity = 0.45 yd/s to the left of his starting point.

Someone claps his or her hands is an example of… motion energy to sound energy sound energy to potential energy motion energy to radiant energy

Answers

motion energy to sound energy i think

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = (4.53\ cm)/(10) = 0.00453 m

Therefore,

\lambda = ((0.00453\ m)(0.00109\ m))/(8.61\ m)

λ = 5.734 x 10⁻⁷ m = 573.4 nm

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?

Answers

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\n\Rightarrow -u^2=2as-v^2\n\Rightarrow u=√(v^2-2as)\n\Rightarrow u=√(0^2-2* -9.81* 1)\n\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\n\Rightarrow t=(v-u)/(a)\n\Rightarrow t=(0-4.45)/(-9.81)\n\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+(1)/(2)at^2\n\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\n\Rightarrow t=\sqrt{(1* 2)/(9.81)}\n\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

1. Explain the change of state from solid dry ice to carbon dioxide gas.2. The motion of the particles in dry ice and carbon dioxide gas.

3. Explain how the original mass of dry ice compares with the mass of carbon dioxide gas.

Answers

1. Since the solid carbon dioxide never become liquid on melting under normal pressure. Thus through the process of sublimation, the solid carbon dioxide changes to gas

2. The molecules in dry ice, are in caged like structure just as the normal water ice but as it melts, the CO2 molecules having high affinity for gaseous state converts into a gas.

3. Dry ice is heavier than its gaseous form. Density of dry ice = 97.6 lb/cu.ft.

Density of carbon dioxide gas = 0.1144 lb/cu.ft.

With higher density and a fixed volume, mass of dry ice is higher than the CO2 gas
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