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lina2011 [118]
3 years ago
5

Help please its due tomorrow :(

Physics
1 answer:
Solnce55 [7]3 years ago
8 0
Traditional is the old way of doing stuff, Command is a system ruled by one person, Market is having buyers and sellers make economic decisions

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Evaluate ( 64800 ms ) 2 to three significant figures and express the answer in Si units. Express your answer using three signifi
Sergeu [11.5K]

Answer:

42.0×10² second²

Explanation:

Here, time is given in milisecond

(64800 ms)²

= 4199040000 ms²

The SI unit is seconds

1 second = 1000 milisecond

1\ milisecond=\frac{1}{1000}\ second

\\\Rightarrow 1\ milisecond^2=\left(\frac{1}{1000}\right)^2\ second^2

4199040000\ ms^2=4199040000\times \left(\frac{1}{1000}\right)^2\ second^2=4199.04\ second^2

42.0×10² second²

6 0
4 years ago
PLEASE ANSWER NEED HELP!!!!!!!! PLEASE THE CORRECT ANSWER!!!!!!
Lesechka [4]

Answer:

C. unbalanced force

Explanation:

it is called unbalanced force because it has changed direction of the moving object

6 0
3 years ago
How much force is needed to accelerate a bicycle and rider with a total mass of 80 kg at a rate of 2.0 m/s2?
snow_tiger [21]
If I am not mistaken I think it is C. 320 N
5 0
4 years ago
Read 2 more answers
Brandy has built a computer model of a weather system. She has included information about bodies of water, the Sun, and the atmo
Mashcka [7]

Answer:

City A will have a higher temperature overall than City B (D)

Explanation:

If all factors other than elevation are the same for both cities, then City A will have a higher temperature overall than City B. This is because there are fewer air molecules at higher elevations. Fewer air molecules mean less heat is absorbed by the atmosphere, and temperatures are lower. This is why areas high in the mountains are often much cooler than areas down on the plains.

5 0
4 years ago
Read 2 more answers
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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