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ira [324]
4 years ago
9

look around you to find an object in motion .describe the objects motion by discussing its position and direction of motion in r

elation to a refrences point . explain how you could determine the objects speed.
Physics
1 answer:
VARVARA [1.3K]4 years ago
7 0
An example you could use for this would be if you threw a pencil across the room 10m away. the pencil would have travelled 10m(N). you could determine the objects speed by using the formula speed=distance/time. you would have had to time how long it took for the pencil to hit the 10m marking and divide 10 by however many seconds it took.
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An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 2
ozzi

Answer:

The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have

The temperature T = 602.64K and the final pressure P = 110.24MPa

Explanation:

Numbeer of moles of gas = 3.00 mol  

initial volume = 230.8×10−6 m3  

final volume = 133.4×10−6 m3 .

released energy =  8240 J

Temperature = Constant = T

Pressure =  p_{f} =unknown

From the relation the combined ideal gas law, PV = nRT

Where R = 8.314 4621.JK−1mol−1

we have The release energy from compression P1V1 -P2V2

-qrev = -nRTln(\frac{V_{2} }{V_{1} }) = 8240J

n = 3

Hence -nRTln(\frac{V_{2} }{V_{1} }) =  3×8.314 462×ln(\frac{133.4}{230.8}) × T=  -8240 J

or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K

The Final pressure is given by

PV = n×R×T from where we have V = final volume thus

P = (n×R×T)/V = (3×8.134×602.64)÷(133.4×10^{-6}) = 110237041.1 N/m^{2} = 110.237MPa

8 0
3 years ago
1The density of an of an object with a volume of 60.0 and mass of 400.0g is______
Kobotan [32]
D is the answer
To get the density you divide mass by volume
So the equation is 400/60=d
4 0
3 years ago
A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft>s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
      = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
3 years ago
Find the weight required to stretch a steel road by 2 mm, if the length of the road is 2 m and diameter 4 mm. Y= 200 GPa and g -
svetlana [45]

Answer:

Option D is the correct answer.

Explanation:

We equation for elongation

   \Delta L=\frac{PL}{AE}      

Here we need to find weight required,

We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m

A=\frac{\pi d^2}{4}=\frac{\pi ({4\times 10^{-3}})^2}{4}=1.26\times 10^{-5}m^2

E = 200GPa = 2 x 10¹¹ N/m²

L=2m

Substituting

0.002=\frac{P\times 2}{1.26\times 10^{-5}\times 2\times 10^{11}}\\\\P=2520N=256.4kg

Option D is the correct answer.

6 0
3 years ago
One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
Masteriza [31]

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0
4 years ago
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