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3 years ago

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A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo

n in cm^3
a. 100cm^3

b. 1,000,000cm^3

c. 10cm^3

d. 1,000cm^3

1 answer:

cestrela7 [59]3 years ago

4 0

Answer:

Volume of balloon = 1000 cm^3

Explanation:

The head of a normal person can be assumed as a sphere with radius 10 cm.

Volume of sphere $=\frac{4}{3} \pi r^3$ , where r is the radius.

We have approximate radius = 10 cm.

Approximate volume of head $=\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3$

In the given options the closest value to the approximate volume is 1000 cm^3.

So, volume of head = Volume of balloon = 1000 cm^3

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Question 10 of 34

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Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

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2 years ago

Which change would cause the largest increase in a mountain climber’s gravitational potential energy?

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<span>carrying twice the weight and climbing twice as high</span>

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What can radiation do to our cells if we are exposed to it too much?

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it can kill you cell and win you kill you no cell no life

Explanation:

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Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 25.0 m from each other. Bein

Lerok [7]

Answer:

A) 199.78 J

B) 9.292x10^14 J

C) 4.2x10^7 m/s

D) 0.65 m

E) 1.13x10^-8 sec

D) 2.94x10^-9 sec

Explanation:

mass of ball = 0.0580 kg

A)

If smashed at v = 83.0 m/s, KE is

KE = 0.5mv^2

= 0.5 x 0.0580 x 83.0^2

= 199.78 J

B) if returned at v = 1.79×10^8 m/s, KE will be

KE = 0.5mv^2

= 0.5 x 0.0580 x (1.79×10^8)^2

= 9.292x10^14 J

C) during Einstein's return, velocity of rabbit relative to players is

Vr = 2.21×108 m/s

Rabbit's velocity relative to ball = 2.21×10^8 - 1.79×10^8

= 4.2x10^7 m/s

D) the rabbit's speed approaches the speed of light so we consider relativistic effect. The rabbit's measured distance is

l = l°( 1 - v^2/c^2)

= 2.5(1 - 2.21/3)

= 2.5 x 0.26

= 0.65 m

E) according to the players, the time taken by the rabbit is

t = d/v = 2.5/ 2.21×10^8

= 1.13x10^-8 sec

F) the time for rabbit as measured by rabbit is relativistic

t = t°( 1 - v^2/c^2)

= 1.13x10^-8 (1 - 2.21/3)

= 1.13x10^-8 x 0.26

= 2.94x10^-9 sec

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3 years ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

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Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

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3 years ago