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bazaltina [42]
2 years ago
10

A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration

of 10 cm/s2 for another 5.0 s. The particle moves in the same direction along a straight line. The average speed over the whole time interval is?
Physics
1 answer:
wel2 years ago
8 0

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

V=X/T

First stage:

T1=5s

v_{f}  =v_{o} - at

But, v_{f}  =0   (decelerates to rest)

then: a =v_{o} /t=0.3/5=0.06m/s^{2}

on the other hand:

x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m

X1=75cm

Second stage:

T2=5s

x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

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According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
Bond [772]

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0
3 years ago
Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b
zmey [24]

Answer:

The maximum amount of work is  W = 1563.289 \ J

Explanation:

From  the question we are told that

   The temperature of the environment is  T = 280\ K

    The volume of container A is  V_A = 2 m^3

    Initially the number of moles  is  n = 1.2 \ moles

     The volume of container B is V_B = 3.5 \ m^3

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             W =  P_A V_A  ln[ \frac{V_B}{V_A} ]

Now from the Ideal gas law

          P_A V_A =  nRT

So substituting for P_A V_A in the equation above

          W =  nRT ln [\frac{V_B}{V_A} ]

Where R is the gas constant with a values of  R =  8.314 \  J/mol

Substituting values we have that

            W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]

          W = 1563.289 \ J

8 0
3 years ago
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