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cluponka
20 hours ago
10

In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? Include a + or - sign to indic

ate the direction.

Physics
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Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights
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275 kPa Explanation: Here the mass of the gas equals m=1.5 kg with an initial volume of V₁=0.04 m³ and an initial pressure P₁=550 kPa. As provided, the final volume is double the original volume, making V₂ equal to 2 V₁. Since the temperature remains constant, T₁=T₂=T. By substituting the values into the equation... results in final pressure being P₂=275 kPa.
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1) Compute the travel time ratio of the other objects to the hollow cylinder. 2) Discuss how the travel time ratio of solid cyli
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Answer:

(1) Utilize the information provided in Table R2 and the error propagation principle to calculate the travel time ratio (with errors) of the other objects compared to the hollow cylinder? ℎ?. Complete Table R5 below. [6] Table R5 Solid cylinder Billiard ball Racquetball?? ℎ? ± ± ± (2) Examine how the solid cylinder's ratio to the hollow cylinder supports or contradicts the theoretical ratio in Eq. (8) stated in the manual. Compute the percentage error and discuss. [4] Answer: (3) Based on the travel time ratio, determine (i) if the billiard ball is solid or hollow, and (ii) if the racquetball is solid or hollow. Provide your reasoning. (Answers may vary if your measurements lack sufficient clarity.) [4]

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PHYS2125 Physics Laboratory I ©2018 Kuei Sun The University of Texas at Dallas 5 Answer: (4) Identify the object in Table R2 with the highest SEOM. Provide reasoning for the relatively high SEOM and suggest improvements. [3] (5) Discuss TWO potential systematic errors in measurement. [3] Answer: **Please attach your calculation details. Use as many pages as needed; calculations that reflect your understanding may earn partial credit. **Ensure your workspace and equipment are identical to how you left them.

Explanation:

6 0
2 months ago
A 230mg piece of gold is hammered into a sheet measuring 23cm x 17cm. What is the thickness of the sheet in meters? Density of g
kicyunya [3294]
The sheet's thickness is calculated to be t= 0.0003 mm. The mass provided is m = 230 mg, equivalent to 0.23 g. The area of the sheet is A= 23 x 17 cm², totaling A= 391 cm². Given the density of gold is ρ = 19.32 g/cm³, we assume the sheet's thickness is t cm. From the equation Mass = Density x Volume, we know m = ρ A t. Substituting the values results in: 0.23 = 19.32 × 391 × t, leading to t = 0.000030 cm, or equivalently, 0.0003 mm as the final thickness.
7 0
1 month ago
In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at
Ostrovityanka [3204]
The force can be determined using the equation F (force) = mass * acceleration. The unit of measurement, N (Newton), is equivalent to kilogram-meter/seconds2.
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7 0
1 month ago
Read 2 more answers
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
kicyunya [3294]

Answer:

tan \theta = \mu_s

Explanation:

Aby obiekt był w spoczynku na nachyleniu, wynikowa siła działająca na niego musi wynosić zero. Równanie sił działających w kierunku równoległym do nachylenia jest następujące:

mg sin \theta - \mu_s R =0 (1)

gdzie

mg sin \theta to składowa ciężaru równoległa do nachylenia, przy czym m oznacza masę obiektu, g oznacza przyspieszenie grawitacyjne, a \theta to kąt nachylenia

\mu_s R to siła tarcia, z \mu_s jako współczynnikiem tarcia oraz R jako reakcją normalną nachylenia

Równanie sił w kierunku prostopadłym do nachylenia to

R-mg cos \theta = 0

gdzie

R to reakcja normalna

mg cos \theta to składowa ciężaru prostopadła do nachylenia

Obliczając R,

R=mg cos \theta

I podstawiając do (1)

mg sin \theta - \mu_s mg cos \theta = 0

Rearanżując równanie,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

To jest warunek, przy którym równowaga jest zachowana: gdy tangens kąta staje się większy niż wartość \mu_s, siła tarcia nie jest w stanie zrównoważyć składowej ciężaru równoległej do nachylenia, dlatego obiekt zaczyna zsuwać się w dół.

4 0
2 months ago
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