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3 years ago

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A flask that can withstand an internal pressure of 2500 torr, but no more, is filled with a gas at 21.0C and 758 torr and heated

. At what temperature will it burst?

1 answer:

marishachu [46]3 years ago

7 0

Answer : The temperature will be, 969.66 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 758 torr

$P_2$ = final pressure of gas = 2500 torr

$T_1$ = initial temperature of gas = $21.0^oC=273+21.0=294K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{758torr}{294K}=\frac{2500torr}{T_2}$

$T_2=969.66K$

Thus, the temperature will be 969.66 K

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Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3 ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$ ----- ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 = 2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0

3 years ago

An electric heater is rated at 1400 W, a toaster is rated at 1150 W, and an electric grill is rated at 1560 W. The three applian

gayaneshka [121]

Answer:

The current in the heater is 12.5 A

Explanation:

It is given that,

Power of electric heater, P₁ = 1400 W

Power of toaster, P₂ = 1150 W

Power of electric grill, P₃ = 1560 W

All three appliances are connected in parallel across a 112 V emf source. We need to find the current in the heater. We know that in parallel combination of resistors the current flowing in every branch of resistor divides while the voltage is same.

Electric power, $P_1=V\times I_1$

$I_1=\dfrac{P_1}{V}$

$I_1=\dfrac{1400\ W}{112\ V}$

$I_1=12.5\ A$

So, the current in the heater is 12.5 A. Hence, this is the required solution.

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4 years ago

Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.

elena-14-01-66 [18.8K]

Answer:

Explanation:

It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.

Let in equilibrium , tension in rope be T

For balancing Joe

T = M g

For balancing Simon

friction + T = mgsinθ

μmgcosθ+T = mgsinθ

μmgcosθ+Mg = mgsinθ

M = (msinθ - μmcosθ)

M = m(sinθ - μcosθ)

5 0

3 years ago

Two asteroids in outer space collide and stick together. The mass of each asteroid, and the velocity of each asteroid before the

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Answer:

(5) It depends on whether the collision was elastic or inelastic.

Explanation:

Due to the absence of external forces and the fact that masses and initial velocities are known, it is needed to determine if collision is elastic or inelastic. If collision is inelastic, the Momentum Principle is sufficient to calculate final momentum, but if it is elastic, the Momentum Principle should be complemented with the application of the Energy Principle.

Hence, the correct answer is (5).

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4 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

3 years ago