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4 years ago

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The volume of a gas is directly related to the temperature of the gas. at 30 degrees celsius a certain gas has a volume of 125 c

ubic centimeters. what will the new volume be at 50 degrees celsius?

2 answers:

pentagon [3]4 years ago

8 0

<u>Answer:</u> The new volume is $133.25cm^3$

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=125cm^3\\T_1=30^oC=(30+273)K=303K\\V_2=?cm^3\\T_2=50^oC=(50+273)K=323K$

Putting values in above equation, we get:

$\frac{125cm^3}{303K}=\frac{V_2}{323K}\\\\V_2=\frac{125\times 323}{303}=133.25cm^3$

Hence, the new volume is $133.25cm^3$

tigry1 [53]4 years ago

5 0

To solve this we use <span>Charles' law whitch states that V1/T1=V2/T2.
So if T1=30 C, V1=123 cm^3 and T2=50 C -> V2=(V1*T2)/T1
T2=208.33 cm^3

</span>

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If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

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If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

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1 year ago

A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside

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Answer:

The value is $B = 0.0452 \ T$

Explanation:

From the question we are told that

The number of turns is N = 1000

The length is L = 50 cm = 0.50 m

The radius is r = 2.0 cm = 0.02 m

The current is I = 18.0 A

Generally the magnetic field is mathematically represented as

$B = \mu_o * \frac{N }{L} * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

So

$B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0$

=> $B = 0.0452 \ T$

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3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

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Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

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3 years ago

A bag of marbles is hanging motionless on a spring scale. What prevents the bag from falling?

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Answer:

The vector sum of all forces acting on it is zero, its at equilibrium.

Explanation:

The bag of marbles hanging on a spring scale applies its weight downwards, which was counterbalanced by the reaction from the spring scale (obeying the Newton's third law of motion). And since no external forces are applied to the system, thus the equilibrium of the system.

If the weight of the bag is greater than the reaction from the spring scale, the scale breaks and the system would not be balanced.

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3 years ago

A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the p

maks197457 [2]

Answer:

Explanation:

Given that,

Charge q=-5.90nC

Magnetic field B= -1.28T k

And the magnetic force

F =−( 3.70×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

The force in a magnetic field is given as

F= q(v×B)

−( 3.70×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.28k

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28x i×k - 1.28y j×k - 1.28z k×k)

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( 1.28xj - 1.28y i )

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28y i + 1.28x j)

So comparing comparing coefficients

let compare x axis component

-( 3.70×10−7N )i=-1.28qy i

−3.70×10−7N = -1.28qy

y= -3.7×10^-7/-1.28q

y= -3.7×10^-7/-1.28×-5.90×10^-9)

y=-48.99m/s

y≈-49m/s

Let compare y-axisaxis

7.6×10−7N j = 1.28qx j

7.6×10−7N = 1.28qx

x= 7.6×10^-7/1.28q

x= 7.6×10^-7/1.28×-5.90×10^-9

x=-100.64m/s

a. Then, the velocity of the x component is x= -100.64m/s

b. Also, the velocity component of the y axis is y=-49m/s

c. We will compute

V•F

V=-100.64i -49j

F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =

3.724×10^-5 - 3.724×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cosx

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

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3 years ago