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Veseljchak [2.6K]
3 years ago
8

Given a volume of 1000. Cm^3 of an ideal gas at 300 k, what volume would iy occupy at a temperature of 600 k

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0

Answer:2000 cm³

Here, pressure remains constant.

So, b the gas law

V/V' = T/ T'

1000 / V' = 300 / 600

V' = 2000 cm³

Explanation:also pls mark brainliest

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Vitek1552 [10]
First we need to know the equation:
F= Mass times acceleration
F = 2.0 kg times 5.0 m/s^2 
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3 years ago
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A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial
Mashcka [7]

Answer: 0.076 m

Explanation:

Magnetic force= centripetal force;

F(B) = F(c) --------------------------(1)

Bqv= mv^2/R---------------------(2)

Where B= magnetic field,q= charge, = mass, R= radius of the circular path.

Radius of the circular path,R= mv/qB.

Mass of oxygen-16 is 2.66e−26kg, and they are singly charged = 18/16(mass ratio= 18:16)

18/16 × (2.26 × 10^ -26 kg)

= 2.99 × 10^26 kg

∆d=2r(18)-2r(16)-----------------(3)

r(18) is the radius of mass of oxygen-18,

r(16) is the radius of mass of oxygen-16.

From the question, v= 3.7×10^6 m/s, q= 1.6×10^-9 C.

From equation (2) above we have;

R= MV/qB------------------------(4)

This equation (4) is used in solving for the distance between the two.

Solving;∆d=2r(18)-2r(16)

∆d = 2[m(18) - m(16)] v/ qB

M(18)-m(16) =(2.99-2.66)× 10^-26:

qB= 1.6×10^-19×2 = 3.2×10^-19

= (2.99-2.66)× 10^-26/ 3.2× 10^-19

= 0.076 m

7 0
3 years ago
An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 k
vichka [17]

Answer:

1.) 0 kgm/s

2) 6.3 kg

3) -0.0978 m/s

4)

5)

6)

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion

Explanation:

Since explosions and collisions follow the law of conservation of Momentum.

1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system

Since the body was initially at rest,

Magnitude of the initial momentum of the system = 0 kgm/s

Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.

2) mass of the third piece

Sum of all the masses = 16.2 kg

4.7 + 5.2 + c = 16.2

c = 6.3 kg

3) doing an x-component balance on momentum

(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0

-0.6163 + 6.3v = 0

v = -0.0978 m/s

Hope this Helps!!!

4 0
3 years ago
Determine the energy required to accelerate an electron from (a) 0.500c to 0.900c and (b) 0.900c to 0.990c
Lera25 [3.4K]

Answer:

0.582 MeV

2.45 MeV

Explanation:

E_r=0.511\ MeV=Electron rest energy

(a) 0.500c to 0.900c

W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.9^2}}-\frac{1}{\sqrt{1-0.5^2}}\right)0.511\\\Rightarrow W=0.582\ MeV

Energy required is 0.582 MeV

(b) 0.900c to 0.990c

W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.99^2}}-\frac{1}{\sqrt{1-0.9^2}}\right)0.511\\\Rightarrow W=2.45\ MeV

Energy required is 2.45 MeV

6 0
3 years ago
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