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siniylev [52]
3 years ago
13

Please help homework due tomorrow,,,

Physics
1 answer:
Ad libitum [116K]3 years ago
8 0
Cody ...

Everything on this page is solved with the SAME formula !

             Distance = (speed) x (time) .


Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.

'Velocity' is speed AND THE DIRECTION OF THE  MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.

Your teacher may not be happy if you talk about this on your homework,
but that's too bad.  Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.
 

1).  Speed = (distance covered) / (time to cover the distance)

2).  Speed = (distance covered) / (time to cover the distance)

3).  Distance  =  (average speed of travel) x (time traveling at that speed)

4).  Time to cover the distance = (distance) / (speed)

5).  Car's     speed = (distance the car covered)        / (time the car took)
      Sprinter speed = (distance the sprinter covered) / (time the sprinter took)

      Calculate the car's speed.
      Calculate the sprinter's speed.
     
      ... Look at the two speeds.
          Decide which one is faster.
     
      ... Subtract the slower one from the faster one. 
          The difference is the answer to "by how much?" .

6).  Distance  =  (speed) x (time spent moving at that speed)

7).  Average speed  =  (TOTAL distance covered)
                                      divided by
                                    (time to cover the TOTAL distance).
   

You might be interested in
At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

6 0
3 years ago
Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca
Brut [27]

Let say the two train cars are of masses m_1 and m_2

now if the speed of two cars are v_1 and v_2

then we can say that the momentum of two cars before they collide is given by

P = m_1v_1 - m_2v_2

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

P = m_1v_1 - m_2v_2

5 0
3 years ago
Read 2 more answers
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
Help on this pls. !!!!!
Luba_88 [7]

Answer:

C

Explanation:

4 0
3 years ago
The answer can not be C or D its between A and B
Murrr4er [49]
If the answer is not C D then is A bc of the way it crosses over
8 0
3 years ago
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