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3 years ago

Calculating the Mass of a Granite Monument

Physics

1 answer:

Volgvan3 years ago

8 0

Answer:

m = 684,865,8 g

Step-by-step explanation

V = 25,365.4 cm^3 Is volume

r = 27g/cm^3 Is density

To calculate mass you use formula:

m= V*r

m = 25,365.4 x 27

m = 684,865,8 g

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H H H O H O N C C N C C H O H H O H H C H H H ALANINE GLYCINE

Trava [24]

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2 years ago

A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it mai

Sloan [31]

Answer: Hello the missing piece of your question is attached

question : Determine mass of steam that has entered ( in kg )

answer : 0.206 kg

Explanation:

V1 = 0.1 m^3 ,

v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg

V2 = 0.2 m^3

using the steam tables

at ; P = 1000 kPa, v' = 0.167 m^3/kg

U1 = 2321 KJ/kg

at ; P = 1000 kPa , T2 = 280°C

v'2= 0.2481 m^3kg

U2 = 2760.6

at ; P = 5MPa , T = 500°C

h1 = 3434.7 KJ/Kg

calculate final mass ( m2 )

M2 = V2 / v'2

= 0.2 / 0.2481 = 0.806 kg

therefore the mass added = m2 - m1

= 0.806 - 0.6 = 0.206 kg

3 0

3 years ago

What is the speed of an electron traveling 32 cm in 2 ns?

abruzzese [7]

Answer:

Speed, $v=1.6\times 10^8\ m/s$

Explanation:

Given that,

Distance covered by the electron, d = 32 cm = 0.32 m

Time, t = 2 ns

We need to find the speed of an electron. Speed is equal to distance covered divided by time. So,

$v=\dfrac{0.32}{2\times 10^{-9}}\\\\v=1.6\times 10^8\ m/s$

So, the speed of the electron is $1.6\times 10^8\ m/s$ .

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3 years ago

Suppose a new star cluster is born with one o star, 10 a stars, 100 g stars, and 1000 m stars. which stellar type dominates the

uranmaximum [27]

Stars are classified by their spectra (the elements that they absorb) and their temperature. There are seven main types of stars. In order of decreasing temperature, O, B, A, F, G, K, and M. Type O stars are the stars with the highest temperature and biggest luminosity. So, if there is a new born star cluster : 1. with one O star,
2. with 10 A stars,
3. with 10 G stars,
4. with 1000 M stars

the stellar type O will dominate the light output from the cluster.

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3 years ago

A wildebeest calf is cruising at its top speed of v= 10 m/s when it passes over a sleeping cheetah. By the time the cheetah stan

Gre4nikov [31]

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

The given parameters;

speed of the wildebeest calf, Vw = 10 m/s

distance traveled by the calf before the cheetah stands up = 7 m

constant acceleration of the cheetah, a = 9.5 m/s²

Let the speed of the cheetah = Vc

let the time the cheetah catches up with the wildebeest = t

$a = \frac{V_c}{t}$

$V_c = at$

Apply relative velocity formula to determine the time when the cheetah catches up with the widebeest;

Assuming the wildebeest and the cheetah are running in the same direction;

$(V_c - V_w) t = 7 \\\\(at - V_w) t = 7\\\\(9.5t - 10)t = 7\\\\9.5t^2 - 10t = 7\\\\9.5t^2 -10t-7 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 9.5 \ b= -10 \ , c =-7\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-10)\ \ +/- \ \ \sqrt{(-10)^2-4(9.5\times -7)} }{2(9.5)}\\\\t = \frac{10 \ \ +/- \ \ 19.13}{19} \\\\t = 1.53 \ s$

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero;

$(V_c - V_w) t = 0\\\\(at - V_w) t = 0\\\\(9.5t - 10)t = 0\\\\9.5t^2 - 10t = 0\\\\9.5t^2 = 10t\\\\9.5t = 10\\\\t = \frac{10}{9.5} = 1.05 \ s$

Thus, if the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

Lear more here: brainly.com/question/24430414

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3 years ago