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Igoryamba
3 years ago
15

You're conducting a physics experiment on another planet. You drop a rock from a height of 2.1 m and it hits the ground 0.6 seco

nds later. What is the acceleration due to gravity on this planet?
A. 7.0 m/s2
B. 9.8 m/s2
C. 11.7 m/s2
D. 5.2 m/s2

(What formula should I use to solve this on my own?)
Physics
2 answers:
WINSTONCH [101]3 years ago
8 0

Answer:

11.7 m/s2

Explanation:

Apex ; good luck everyone! :)

umka21 [38]3 years ago
3 0

The formula to use is the one that connects the acceleration,
the distance fallen, and the time spent falling:

         Distance = 1/2 a T² .

You said  2.1 meters in 0.6 second .

                                                         2.1 m              = 1/2 a  (0.6 sec)²

Multiply each side by  2 :                  4.2 m              =       a (0.6 sec)²

Divide each side by (0.6 sec)² =     (4.2/0.36) m/s² =        a

                                                         a = (11 and 2/3) m/s²

                                        (about 19% more than Earth's gravity)

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Consider a sealed 20 cm high electronic box whose base dimensions are 40cm x 40cm placed in a vacuum chamber. The emissivity of
Genrish500 [490]

Answer:

T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

Explanation:

The formula we are going to use is:

\dot Q_{rad}=\epsilon\sigma A_s(T_s^4-T_{surr}^4)

Where:

ε is the emissivity

σ is the Stefan constant

T_s is the final temperature of surrounding surfaces

T_{surr} is the required temperature

A_s is the are of surrounding surface

Calculating The area:

A_s=(0.4)(0.4)+4(0.4)(0.2)\\A_s=0.48\ m^2

σ= 5.67*10^{-8}\ W/m^2.K^4

ε =0.95

T_s=55+273

T_s=328 K

\dot Q_{rad=100 W

100=0.95(5.67*10^{-8})(0.48)(328^4-T_{surr}^4)\\3867693926=(328^4-T_{surr}^4)\\T_{surr}^4=7706623130\\T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

8 0
3 years ago
In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo
Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

4 0
3 years ago
Characteristics, such as field name and data type, are_____ that define a field.
Aleonysh [2.5K]

Characteristics, such as field name and data type are properties that define a field.

<h3>What are the properties of a given discipline or research field?</h3>

The properties of a given discipline or research field are those that characterize the use of a particular type of information within the range of the attained field.

For example, genomics is a field characterized to make use of sequencing data, which refers to the linear order of nucleotides in the DNA sequence to provide useful information about the gene content and the presence of regulatory non-coding sequences.

Therefore, with this data, we can see that the properties of a given discipline or research field are mainly associated with the type of information that such discipline uses to make predictions in a given research and they are also related to the outcomes we can expect from such data.

Learn more about the properties of a scientific research field here:

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zmey [24]

Answer:

632.5 MPa

Explanation:

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\sigma_f = Fiber fracture strength = 2.5 GPa

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\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa

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