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just olya [345]
2 years ago
10

Calculate the speed of the car at each checkpoint by dividing the distance between each checkpoint, in meters, by the change in

time. Record your answers in Table E of your Student Guide.
The speed at the first quarter checkpoint is __ m/s.

The speed at the second quarter checkpoint is __ m/s.

The speed at the third quarter checkpoint is __ m/s.

The speed at the finish line is __ m/s.
Physics
2 answers:
Ivenika [448]2 years ago
8 0

Answer:

1.09

1.95

2.37

2.80

Explanation:

just did it

qwelly [4]2 years ago
6 0

Answer:

1.09

1.95

2.37

2.80

Explanation:

I took the test on Edginuity and got a 100% on it

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A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficie
Rudiy27

Answer:

a) xf = 5.1 m

b) u = 0.304

c) x = 10.3 m

Explanation:

we will use the following formula:

u = 0.1 + A*x

Si x = 12.5 m, u = 0.6

Clearing A:

A = 0.5/12.5 = 0.04 m^-1

a) we have to:

W = Ekf - Eki

where Ekf = final kinetic energy

Eki = initial kinetic energy

9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)

Clearing xf, we have:

xf = 5.1 m

b) Replacing values for u:

u = 0.1 + (0.04*5.1) = 0.304

c) Wf = Ekf - Eki

-u*m*x*g = 0 - (m*v^2)/2

Clearing x:

x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m

4 0
2 years ago
Read 2 more answers
The total energy of a 0.050 kg object travelling at 0.70 c is
quester [9]
Would presume the energy as kinetic energy.

E = (1/2)*mv²

But m = 0.05kg, velocity here = 0.70c, where c is the speed of light ≈ 3* 10⁸ m/s

Ke =  (1/2)*mv² = 0.5*0.05*(0.7*<span>3* 10⁸)</span>² = 1.1025 * 10¹⁵ Joules


There is no exact match from the options.
4 0
2 years ago
Steam at 700 bar and 600 oC is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is t
Setler79 [48]

Answer:

Final temperature of the steam = 304.29 K = 31.14°C

Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T = 0 J/K.

But 0.01 J/k.s was obtained though, which is approximately 0.

Explanation:

For an adiabatic system, the Pressure and temperature are related thus

P¹⁻ʸ Tʸ = constant

where γ = ratio of specific heats. For steam, γ = 1.33

P₁¹⁻ʸ T₁ʸ = P₂¹⁻ʸ T₂ʸ

P₁ = 700 bar

P₂ = 10 bar

T₁ = 600°C = 873.15 K

T₂ = ?

(700⁻⁰•³³)(873.15¹•³³) = (10⁻⁰•³³)(T₂¹•³³)

T₂ = 304.29 K = 31.14°C

b) Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T

To prove this

Entropy of the process

dQ - dW = dU

dQ = dU + dW

dU = mCv dT

dW = PdV

dQ = TdS

TdS = mCv dT + PdV

dS = (mCv dT/T) + (PdV/T) =

PV = mRT; P/T = mR/V

dS = (mCv dT/T) + (mRdV/V)

On integrating, we obtain

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

To obtain V₁ and V₂, we use PV = mRT

V/m = specific volume

Pv = RT

R for steam = 461.52 J/kg.K

For V₁

P = 700 bar = 700 × 10⁵ Pa, T = 873.15 K

v₁ = (461.52 × 873.15)/(700 × 10⁵) = 0.00570 m³/kg

For V₂

P = 10 bar = 10 × 10⁵ Pa, T = 304.29 K

v₂ = (461.52 × 304.29)/(10 × 10⁵) = 0.143 m³/kg

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

m = 2 kg/min = 0.0333 kg/s, Cv = 1410.8 J/kg.K

ΔS = [0.03333 × 1410.8 × In (304.29/873.15)] + (0.03333 × 461.52 × In (0.143/0.00570)

ΔS = - 49.56 + 49.57 = 0.01 J/K.s ≈ 0 J/K.s

3 0
2 years ago
Which of the following substances will show the smallest change in temperature when equal amounts of energy are absorbed?
SSSSS [86.1K]
It would be water because if you freeze it than you will still be able to see it and if you boil it than you will be able to see it disappear.
3 0
2 years ago
Read 2 more answers
Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^
ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

T \approx e^{-2CL}

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)

Next, you divide the equation (3) into (4), and finally, you solve for L':

\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0
2 years ago
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