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forsale [732]
3 years ago
9

A cylinder of gas is compressed by a piston from an initial volume of 125 liters to a final volume of 90 liters. The compression

occurs at constant pressure, and the work done on the gas by the piston is 104 joules. What is the gas pressure during the compression? (1 liter = 10-3 meters3)
Physics
2 answers:
ololo11 [35]3 years ago
7 0
Since the process occurs at constant pressure, the work done by the gas on the piston is given by
W=p \Delta V = p (V_f - V_i)
where
W is the work
p is the pressure
V_f is the final volume
V_i is the initial volume

In our problem, we have
V_i =125 L = 125 \cdot 10^{-3}m^3
V_f = 90 L = 90 \cdot 10^{-3} m^3
W=-104 J (where the negative sign means the work is done by the piston on the gas)

We can re-arrange the previous equation and use these values to find the pressure:
p= \frac{W}{V_f - V_i}= \frac{-104 J}{90 \cdot 10^{-3} m^3 - 125 \cdot 10^{-3} m^3}=2971 Pa
mrs_skeptik [129]3 years ago
4 0

Answer: 3 x 10^5 pascals

Explanation: I think you meant to put 10^4 joules instead of 104, which is why nobody likes the previous answer, using 10^4 joules gets you. the answer with the following:

Using the same equation the person above me posted....

 -10^4/(90E-3-125E-3) And then only one significant figure from the question and we round to 3 x 10^5 pascals

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Which statement best reflects the approach of Gestalt psychology?
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a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration​
Margarita [4]

Answer:

9.67\cdot 10^{-3}m/s^2

Explanation:

We can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem:

u = 0 (it starts from rest)

v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s is the final velocity

s = 10 km = 10 000 m is the displacement

Solving for a, we find:

a=\frac{v^2-u^2}{2s}=\frac{13.9^2}{2(10000)}=9.67\cdot 10^{-3}m/s^2

7 0
3 years ago
According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the
slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

A=\dfrac{\pi}{4}\times d^2

Put the value into the formula

A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2

A=1.96\times10^{-9}\ m^2

We need to calculate the magnitude of the force

Using formula of force

F=\sigma A

Put the value into the formula

F=196\times10^{6}\times1.96\times10^{-9}

F=0.38416\ N

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

strain=\dfrac{\Delta l}{l_{0}}

\Delta l=strain\times l_{0}

Put the value into the formula

\Delta l=0.380\times l_{0}

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

l=l_{0}+\Delta l

Put the value into the formula

I=l_{0}+0.380\times l_{0}

l=1.38l_{0}

l_{0}=\dfrac{l}{1.38}

l_{0}=\dfrac{12\times10^{-2}}{1.38}

l_{0}=0.0869\ m

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0
3 years ago
After robbing a bank, a criminal tries to escape from the police by driving at a constant speed of 55 m/s (about 125 mph). A pol
vfiekz [6]

Answer:

18.03 s

Explanation:

We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.

x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2

The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:

x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2

Equalizing these two equations and solving for t:

(55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s

6 0
3 years ago
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