Which property describes the temperature at which a liquid changes to a gas?

An electron (mass 9 × 10-31 kg) is traveling at a speed of 0.92 in an electron accelerator. An electric force of 1.4 × 10-13 N i

pogonyaev

Given Information:

Mass of electron = m = 9x10⁻³¹ kg

initial speed of electron = v₁ = 0.92c

Force = F = 1.4x10⁻¹³ J

Distance = d = 3 m

Required Information:

Final speed of electron = v₂ = ?

Answer:

Final speed of electron = v₂ = 2.974x10⁸ m/s

Explanation:

As we know from the conservation of energy,

E₂ - E₁ = W

E₂ = E₁ + W

Where E₂ is the final energy of electron and E₁ is the initial energy of electron

The above equation can be written in the form of particle energy

γ₂mc² = γ₁mc² + W

where γ₁ and γ₂ are given by

γ₁ = 1/√1 - (v₁/c)²

γ₂ = 1/√1 - (v₂/c)²

First calculate γ₁

γ₁ = 1/√1 - (0.92c/c)²

γ₁ = 2.55 m

Now calculate γ₂

γ₂ = (γ₁mc² + W)/mc²

First we need to find the work done

W = F*d

W = 1.4x10⁻¹³*3

W = 4.2x10⁻¹³ J

so γ₂ is

γ₂ = (2.55*9x10⁻³¹*(3x10⁸)² + 4.2x10⁻¹³)/9x10⁻³¹*(3x10⁸)²

γ₂ = 7.73

Now we can find the new speed of the electron

γ₂ = 1/√1 - (v₂/c)²

Re-arranging the above equation results in

v₂ = c*√(1 - 1/γ₂²)

v₂ = 3x10⁸*√(1 - 1/7.73²)

v₂ = 2.974x10⁸ m/s

4 0

3 years ago

A race car's velocity increases from +28 m/s to +36 m/s over a 2.0-s time interval. What is the car's

BigorU [14]

Answer:

$4 m/s^{2}$

Explanation:

Acceleration, $a=\frac {v-u}{t}$

Where v and u are the final and initial velocities of the race car respectively, t is the time taken for the race car to attain velocity of 36 m/s.

Substituting 36 m/s for v, 28 m/s for u and 2 s for t then

$a=\frac {36 m/s-28 m/s}{2}=4 m/s^{2}$

3 0

3 years ago

gas is compressed by a piston in a cylinder from an initial pressure of 1.0 bar and initial volume of 0.4m3 , to a final pressur

Ket [755]

The work was done on the gas. and internal energy of the gas will be -13459 J and -21649 J.

When energy is moved from one store to another, work is completed. Work done on the gas is taken as -ve.

Given data;

Initial pressure(P₁)=1.0 bar

The initial volume, V₁=0.4m³

Final pressure(P₂) =1.4 bar.

work is done on the gas., W=?(-ve)

Heat release= - 8.1 kJ

Change in the internal energy of the gas., ΔE

During the compression process, PV=C

P₁V₁=P₂V₂

1.0 bar × 0.4m³= 1.4 bar ×V₂

V₂=0.285 m³

$\rm W=P_1V_1ln\frac{V_2}{V_1} \\\\\ \rm W=1.0 \times 0.4 ln\frac{0.285}{0.4} \\\\\ W=-13549 \ J$

The work done on the gas will be -13459 J.

The internal energy is found as;

ΔE=q+w

ΔE= -8.1 ×10³ -13549

ΔE=-21649 J

Hence, work was done on the gas. and internal energy of the gas will be -13459 J and -21649 J.

To learn more about the work done by gas refer to;

brainly.com/question/12539457

#SPJ1

4 0

2 years ago

A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.

strojnjashka [21]

Answer:

$D=99.4665307m \approx 99.5m$

Explanation:

From the question we are told that

Mass $m=6kg$

Velocity of mass $V_m=16$

Force of Tunnel $F_t=8N$

Length of Tunnel $L_t=1.6$

Height of frictional incline $H_i=2.9$

Angle of inclination $\angle =16 \textdegree$

Acceleration due to gravity $g=9.8m/s^2$

First Frictional surface has a coeﬃcient $\alpha_1 =0.21\ for\ d_c=1$

Second Frictional surface has a coeﬃcient $\alpha _2=0.1$

Generally the initial Kinetic energy is mathematically given by

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

$K.E=\frac{1}{2}(6)(16)^2$

$K.E=768$

Generally the work done by the Tunnel is mathematically given as

$w_t=F_t*d_t$

$w_t=8*1.6$

$w_t=12.8J$

Therefore

$Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t$

$E_t=K.E+E_t\\E_t=768J+12.8J$

$E_t=780.8J$

Generally the energy lost while climbing is mathematically given as

$E_c=mgh$

$E_c=(6)(9.8)(2.9)$

$E_c=170.52J$

Generally the energy lost to friction is mathematically given as

$E_f=\alpha *m*g*cos\textdegree*d_c$

$E_f=0.21*6*9.8*cos16*1$

$E_f=11.86965942 \approx 12J$

Generally the energy left in the form of mass $Em$ is mathematically given as

$E_m=E_t+E_c+E_f$

$E_m=(768J)-(170.52)-(12)$

$E_m=585.48J$

Since

$E_m=\alpha_2*g*m*d$

Therefore

It slide along the second frictional region

$D=\frac{585.46}{0.1*9.81*6}$

$D=99.4665307m \approx 99.5m$

6 0

3 years ago

Which of the following amounts to a scientific research process that can be replicated?

STALIN [3.7K]

Answer:

D.all of the above

Explanation:

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3 0

2 years ago