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Sauron [17]
3 years ago
10

Find the acceleration if a 32.5 N force is used on an object that has a mass of 128.6 kg.

Physics
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

Acceleration=3.95

Explanation:Use the formula a=m/f

a=128.6/32.5

a=3.95

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A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.
Galina-37 [17]

Answer:

Change in specific internal Energy=250\ \rm Btu/lb

Explanation:

Given:

  • Mass of the gas, m=0.4 lb
  • Initial pressure and volume are p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\
  • Final pressure and temperature are p_1=480\ \rm lbf/in^2
  • Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\

So the final volume of the gas is calculated.

Work in any isotropic is given by w

w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\

According to the first law of thermodynamics we have

Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\

So the Specific Internal Change is given by

\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb

So the specific Change in Internal energy is calculated.

6 0
3 years ago
How to create or build a homemade golf club. is this possible? If so, what suggestion did you find? Do you have an original idea
goldenfox [79]

Answer:

All you need is golfing equipment make a small hole in the ground and have like something to cover the sun off of you and boom you got a golfing spot and a golf club no skills needed. Oh and you need a small flag for your gold hole but then your good and you can have fun

Explanation:

8 0
2 years ago
Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t
inna [77]

Answer:

Explanation:

a. The equation of Lorentz transformations is given by:

x = γ(x' + ut')

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

x' = 0

t' = 5.00 s

u =0.800 c,

c is the speed of light = 3×10⁸ m/s

Then,

γ = 1 / √ (1 - (u/c)²)

γ = 1 / √ (1 - (0.8c/c)²)

γ = 1 / √ (1 - (0.8)²)

γ = 1 / √ (1 - 0.64)

γ = 1 / √0.36

γ = 1 / 0.6

γ = 1.67

Therefore, x = γ(x' + ut')

x = 1.67(0 + 0.8c×5)

x = 1.67 × (0+4c)

x = 1.67 × 4c

x = 1.67 × 4 × 3×10⁸

x = 2.004 × 10^9 m

x ≈ 2 × 10^9 m

Now, to find t we apply the same analysis:

but as x'=0 we just have:

t = γ(t' + ux'/c²)

t = γ•t'

t = 1.67 × 5

t = 8.35 seconds

b. Mavis reads 5 s on her watch which is the proper time.

Stanley measured the events at a time interval longer than ∆to by γ,

such that

∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)

c. According to Stanley,

dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m

which is the same as in part (a)

7 0
3 years ago
If a planet's orbital speed is 20 km/s when it's at its average distance from the sun which is most likely orbital speed when it
Alona [7]

Answer:25km/s

Explanation:

8 0
3 years ago
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