Which option is an example of a physical property?

If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long wil

AfilCa [17]

Answer:

Explanation:

Let the volume of air be V. at atmospheric pressure, that is 10⁵ Pa

At 20 m below surface pressure will be

atmospheric pressure + hdg

10⁵ + 20 x 9.8 x 1000 = 2.96 x 10⁵Pa

At this pressure volume V becomes V/ 2.96

This volume will last 1/2.96 times time that is 60/2.96 = 20.27 minutes.

8 0

2 years ago

For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro

serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf = 37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

a)

h = vot + 1/2gt²

71.3 = 0t + 1/2(9.81)t²

2(71.3) = 9,81t²

t² = 2(71.3)/9.81

t² = 14.53

t = 3.81 s

b)

vf = 0 + (9.81)(3.81)

vf = 37.4 m/s

3 0

2 years ago

Sound travels at about 360 meters/sec. What is the wavelength of sound with a

ollegr [7]

<u>Answer:</u>

0.24 m

<u>Explanation:</u>

Given:

Wave velocity ( v ) = 360 m / sec

Frequency ( f ) = 1500 Hz

We have to calculate wavelength ( λ ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > λ = v / f

Putting values here we get:

= > λ = 360 / 1500 m

= > λ = 36 / 150 m

= > λ = 0.24 m

Hence, wavelength of sound is 0.24 m.

6 0

2 years ago

Which graph accurately shows the relationship between kinetic energy and speed as speed increases?

mixer [17]

Answer:

B

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = $\frac{1}{2}$ m $v^{2}$

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE = $\frac{1}{2}$ x 2 x $3^{2}$

= 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE = $\frac{1}{2}$ x 2 x $4^{2}$

= 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 0

2 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

2 years ago