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3 years ago

Which accounts for an increase in the temperature of a gas that is kept a constant volume

Physics

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer:

An increase in pressure

Explanation:

The ideal gas law states that:

$pV=nRT$

where

p is the gas pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature of the gas

in the equation, n and R are constant. For a gas kept at constant volume, V is constant as well. Therefore, from the formula we see that if the temperature (T) is increase, the pressure (p) must increase as well.

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A bat with a mass of 0.35 kg travels at a rate of 15 m/s. What is the animal's momentum?

Tanya [424]

Answer:

momentum= mass x velocity = 0.35x15= 5.25kgm/s

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3 years ago

The charge density of a uniformly charged disk 0.420 m in diameter is 2.92 ✕ 10−2 C/m2. What is the magnitude of the electric fi

iragen [17]

Answer:

E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}

R is radius = Diameter/2 = 0.210m.

At z = 0.2m,

Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,

E can be calculated at distance 0.2m away from the centre of the disk.

Put z = 0.3m and all other values in the equation,

E can be calculated at distance 0.3m away from the centre of the disk

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3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

John, paul, and george are standing in a strawberry field. paul is 14.0 m due west of john. george is 36.0 m from paul, in a dir

sdas [7]

Position of paul with respect to john is given as

14 m due west of john

$r_{pj} = r_p - r_j = -14\hat i$

position of George with respect to Paul is given as 36 m in direction 37 degree south of east

$r_{GP} = r_G - r_p = 36cos37\hat i - 36 sin37\hat j$

now we need to find the position of George with respect to John

$r_{GJ} = r_G - r_j[\tex]now for the above equation we can add the two equations[tex]r_{Gj} = -14\hat i + 36 cos37\hat i - 36sin37\hat j$

$r_{Gj} = 14.75\hat i - 21.67\hat j$

so the magnitude is given as

$r = \sqrt{14.75^2 + 21.67^2} = 26.2 m$

and direction is given as

$\theta = tan^{-1}\frac{21.67}{14.75}= 55.75 degree$

<em>so it is 26.2 m at an angle 55.75 degree South of east</em>

7 0

3 years ago

Read 2 more answers

A uniform beam of arbitrary, unsymmetrical cross-section and length 2???? is built-in at one end and simply supported in the ver

leonid [27]

Answer:

the bending moment will be W from either sides

Explanation:

bending moment= force (load) * perpendicular distance, if I understand the question the distance will be 1/2 of the length

=> f x 1/2(l) =W*1/2(2) =W

5 0

3 years ago