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larisa [96]
3 years ago
15

A scientist tested Charles’s law, which states that volume is directly related to temperature. Assume constant pressure. What wi

ll the scientist observe?
A) as volume decreases, temperature increases
B) as volume decreases, temperature decreases
C) as volume increases, temperature stays constant
D) as volume decreases, temperature can increase or decrease
Physics
2 answers:
liubo4ka [24]3 years ago
7 0

Answer:

B) as volume decreases, temperature decreases

Explanation:

As per ideal gas equation we know that

PV = nRT

now we know that

V = (\frac{nR}{P}) T

so here we know that since the pressure is constant so volume is directly proportional to the temperature

So here if we change the volume then the same change will occur to the temperature

so correct answer will be

B) as volume decreases, temperature decreases

sergeinik [125]3 years ago
5 0

Based on the options given, the most likely answer to this query is 

B) as volume decreases, temperature decreases.
Charles' law states that when volume increase or decrease, accordingly, temperature becomes similar.

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere
Tanzania [10]

Answer:

The total charge Q of the sphere is 2.094\times10^{-10}\ C.

Explanation:

Given that,

Radius = 5 cm

Charge density J= 400\ nC/m^3

We need to calculate the total charge Q of the sphere

Using formula of charge

q=\rho V

Where, \rho = charge density

V = volume

Put the value into the formula

q=\rho\times(\dfrac{4}{3}\pi r^3)

Put the value into the formula

q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3

q=2.094\times10^{-10}\ C

Hence, The total charge Q of the sphere is 2.094\times10^{-10}\ C.

6 0
3 years ago
Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
The water is reflecting light, Is this specular or diffuse reflection? explain your answer​
FinnZ [79.3K]
Yes the answer is true
8 0
3 years ago
A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

3 0
3 years ago
Against the wind a commercial airline in south america flew 784 miles in 4 hours. with a tailwind the return trip took 3.53 hour
ira [324]

First let us assign variables,

d = distance travelled

t = time it took

v = velocity of the commercial airline

In linear physics, the equation for velocity is given as:

v = d / t

Rewriting for d:

d = v t

We know that the distance to and from south America are equal therefore:

d1 (going) = d2 (return)

Let us say that velocity of air is v3. Since going to South America, the wind is against the direction of the plane and the return trip is the opposite, therefore:

(v1 - v3) t1 = (v1 + v3) t2

(v1 – v3) 4 = (v1 + v3) 3.53

4 v1 – 4 v3 = 3.53 v1 + 3.53 v3

0.47 v1 = 7.53 v3

v1 = 16.02 v3

Since we also know that:

(v1 - v3) t1 = 784

(16.02 v3 – v3) * 4 = 784

60.085 v3 = 784

v3 = 13.05 mph

Therefore the speed of the plane in still air, v1 is:

v1 = 16.02 * 13.05

<span>v1 = 209.03 mph           (ANSWER)</span>

<span> </span>

4 0
3 years ago
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