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3 years ago

Which of the following is not an intensive physical property?

Physics

2 answers:

Nataliya [291]3 years ago

6 0

I would say freezing point im not positive about it but it sounds like it goes with the question

Kaylis [27]3 years ago

5 0

the answer is mass. hoped this helped

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Naming covalent compounds <br> P4S5

Andrew [12]

Answer:

what the heck is sakurfa

Explanation:

3 0

2 years ago

A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i

dezoksy [38]

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

$h=ut+\frac{at^2}{2}$

here Initial vertical velocity is 0

$8.3=\frac{gt^2}{2}$

$t^2=1.69$

$t=1.301 s$

Horizontal distance

$R=u\times t$

$R=30\times 1.301=39.04 m$

7 0

3 years ago

horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes

lbvjy [14]

Answer:

The tension is $T = 103.96N$

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

The distance between the two poles is $D =14 m$

The mass tied between the two cloth line is $m = 3Kg$

The distance it sags is $d_s = 1m$

The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline

Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium

And this can be mathematically represented as

$\sum F_y = 0$

To obtain $\theta$ we apply SOHCAHTOH Rule

So $Tan \theta = \frac{opp}{adj}$

$\theta = tan^{-1} [\frac{opp}{adj} ]$

$= tan^{-1} [\frac{1}{7}]$

$=8.130^o$

$=> \ \ \ \ \ \ \ \ 2T sin\theta -mg =0$

$=> \ \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}$

$=> \ \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }$

$=> \ \ \ \ \ \ \ \ T =\frac{29.4}{2sin(8.130)}$

$=> \ \ \ \ \ \ \ \ T = 103.96N$

5 0

2 years ago

Read 2 more answers

A runner achieves a velocity of 11.1 m/s, 9 sec after he begins. What is his

Elena-2011 [213]

Answer:

1.23 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 11.1 m/s

t = 9 s

Find: a

Equation:

v = at + v₀

Plug in:

11.1 m/s = a (9 s) + 0 m/s

a = 1.23 m/s²

The runner's acceleration is 1.23 m/s².

7 0

3 years ago

We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu

liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

$\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}$

$\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}$

f = 30 cm

using lens formula

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})$

$R_1 = R\ and\ R_2 = -R$

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})$

$R = (n -1)\ f$

$R = 2(1.5 -1)\ 30$

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0

3 years ago