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Evgen
11 days ago
11

Which of these nebulae is the odd one out?

Physics
2 answers:
Softa [2K]11 days ago
5 0

The correct choice is D!

Clarification:

kicyunya [2.2K]11 days ago
3 0

Response:

Clarification:

A

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One day you are driving your friends around town and you drive quickly around a corner without slowing down. You are going at a
Ostrovityanka [2198]

Result:

1.60 g

Elucidation:

Based on the attached document:

we can infer that:

v = v_x =v_y = 20 \ m/s \\t = 2s

The distance covered in 2 seconds will be:

x = vt

x = 20 m/s × 2 s

x = 40 m

The segment corresponds to a quarter of a circle with radius r,

therefore, if 2 πr = 4 x

Then the radius (r) can be calculated as:

r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}

r = 25.5 m

Centripetal acceleration can be expressed as:

a = \frac{v^2}{r}

thus;

a = \frac{(20 \ m/s^2)}{25.5 \ m}

a = 15.7 m/s²

The acceleration magnitude suffered by your passengers in relation to the acceleration due to gravity can be calculated using:

a' = \frac{a}{g} g

a' = \frac{15.7 \m/s ^2}{9.81 \ m/s^2}\ g

a' = 1.60 \ g

∴ The acceleration magnitude experienced by your passengers while turning = 1.60 g

6 0
19 days ago
Read 2 more answers
Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =
Sav [2217]

Response:

Magnitude of the electrostatic force acting on the +32 µC charge, F_{net} = 12 N

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }

F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }

F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N

The resultant electrostatic force on the 32 µC charge is F_{net} = |F_{2} + F_{3}|

F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N

7 0
21 day ago
Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric fi
Ostrovityanka [2198]
Let A represent the area of each plate. According to Gauss's Law, the electric field present between the plates can be derived.
8 0
3 days ago
A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.
Keith_Richards [2248]

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

m = 100 * 11.34 = 1134 g

m = 1.134 kg

Therefore, its weight in air is noted as

W = mg = 1.134* 9.8 = 11.11 N

Next, the buoyant force acting on the lead is defined as

F_B = W - F_{net}

F_B = 11.11 - 11 = 0.11 N

We know that

F_B = \rho V g

0.11 = 1000* V * 9.8

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.


#2

The buoyant force acting on the lead block counterbalances its weight

F_B = W

\rho V g = W

13* 10^3 * V * 9.8 = 11.11

V = 87.2 cm^3

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.


#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

3 0
4 days ago
myron is almost late for class and he is running quickly to arrive before the professor begins lecturing as he listen to the pro
inna [2205]
No established theory exists here.
Myron has presented a strong hypothesis to clarify his observations.
Alternative hypotheses could be:

-- An infected mosquito might have bitten him during his sleep, causing symptoms to manifest.

-- He may have consumed something for dinner that was a bit spoiled.

-- He might have had excessive alcohol at the fraternity party last night.

-- The air in the classroom could contain elevated levels of Carbon Dioxide.

-- His body might be responding to the physical exertion of rushing to class.

Currently, Myron has merely formulated a hypothesis.
He cannot draw any "conclusion" until he tests his hypothesis and demonstrates that similar outcomes consistently result from the same conditions. Testing his hypothesis may prove challenging, but unless he does so, he lacks a comprehensive theory.

In my view, while his hypothesis may indeed be valid, the most probable explanation for his experience is the recent physical strain from running to class. It’s crucial to note that I cannot convince anyone of this conclusion; my perspective is merely another hypothesis. Its validity holds no significance unless it undergoes testing.
6 0
14 days ago
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