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Sladkaya [172]
4 years ago
13

Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?

Physics
1 answer:
Assoli18 [71]4 years ago
6 0

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion s=ut+\frac{1}{2}at^2

40=0\times 10+\frac{1}{2}a10^2

a=0.8944m/sec^2

Now from first equation of motion v=u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So v=u+at=0+0.8944\times 10=8.944m/sec

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A wooden block meauring 40cm x 10cm x 5cm has a mass 850gm . find the density of wood?
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Answer:

Explanation:

Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3

5 0
3 years ago
A car travels 30 km north in 25 min. and 40 km east in 35 min. What is the total time in hours? Be careful to carry over the pro
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25+35=60mins so 1.0 hours
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What distance will a truck trvael in 3 hours at an average speed of 50 per hour​
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Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​
Alinara [238K]

The volume of the room is the product of its dimensions:

10\times 12 \times 4 = 480\text{ m}^3

Now, from the equation

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So, multiply the density and the volume to get the mass of air in the room.

7 0
4 years ago
Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv
erastovalidia [21]

Answer:

V_2=159.9\ m/s

T_2=290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}

We know that for air

h= 1.010 x T  KJ/kg

1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}                               -----1

Now from mass balance

\rho_1A_1V_1=\rho_2A_2V_2

We also know that

\rho=\dfrac{P}{RT}

\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2

\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2

T_2=1.81V_2                  ----------2                                                                                                

Now from equation 1 and 2

V_2^2+3673.749V-612916.49=0

So we can say that

V_2=159.9\ m/s

This is the outlet velocity.

Now by putting the values in equation 2

T_2=1.81\times 159.9  

T_2=290.6K

This is the outlet temperature.

4 0
4 years ago
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