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Arada [10]
3 years ago
15

A 10.0 L balloon contains helium gas at a pressure of 660 mmHg . What is the final pressure, in millimeters of mercury, of the h

elium gas at each of the following volumes, if there is no change in temperature and amount of gas?
Physics
1 answer:
Ksivusya [100]3 years ago
6 0

Answer:

(a)= 264mmHg

(b)= 2000mmHg

(c)474.82mmHg

(d)= 511.63mmHg

Explanation:

the question deals with boyles law, which states that the volume of a given mass of gas at constant temperature is inversely proportional to its pressure

V ∝ 1/P

P₁V₁ = P₂V₂

making V₂ as the subject of formular

P₂ = P₁V₁/ V₂

with a volume of  25.0L

P₂ = 660×10 / 25

= 264mmHg

with a volume of 3.30 L

P₂  = 660 × 10 / 3.30

= 2000mmHg

with a volume of 13900 mL

= 13.9L

P₂  =660× 10 / 13.9

474.82mmHg

with a volume of 12900 mL

P₂ =660×10 / 12.9

= 511.63mmHg

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Your friends sit in a sled in the snow. If you apply a force pf 75 N to them, they have an acceleration of 0.9 m/s ^ 2. What is
Elza [17]

Answer:

Mass of the sled in the snow 83.33 kg.

<u>Explanation</u>:

Given that,  

Force applied to move the sled in the snow (F) = 75N

\text { Acceleration }(a)=0.9 \mathrm{m} / \mathrm{s}^{2}

We know that

Newton's second law of motion is  

\text { Force }=\text { mass } \times \text { acceleration }

F = ma (Or "force" is equal to "mass" times "acceleration".)

So if we move this around we can isolate mass and get mass

\text { Mass }=\frac{\text { force }}{\text { accelearation }}

\mathrm{M}=\frac{75}{0.9}

M = 83.33 kg

Mass of the sled in the snow <u>83.33 kg.</u>

3 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
Will give you brainliest pls help
ExtremeBDS [4]

1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].

2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].

3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].

4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].

4 0
2 years ago
Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp
In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the  final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M  = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0
3 years ago
An object is located 70 cm from a concave mirror with a focal length of 15 cm. What is the image
Stels [109]

(a) The distance of the image formed by the concave mirror is 19.1 cm.

(b) The image formed is diminished and real.

<h3>Image distance </h3>

The distance of the image formed by the concave mirror is calculated as follows;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/15 - 1/70

1/v = 0.05238

v = 1/0.05238

v = 19.1 cm

The image distance is smaller than object distance, thus the image formed is diminished and real.

Learn more about concave mirror here: brainly.com/question/13164847

#SPJ1

5 0
1 year ago
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