A track star runs 100 meters in 10 seconds. What is the star's average speed?

Please help im stuck and can't find the answer sheet ANYWHERE :( PLEASEEEEEEEEEEE

Vladimir [108]

Answer:

All steps are 20 * 100 (break the rest into appropriate pieces)

You can multiply as follows

(2000) * ((3 * 60) + (2 * 60) + 60)

V = 2000 * 6 * 60) = 720,000 cm^3 = .72 m^3

.72 m^3 * 2400 kg / m^3 = 1728 kg

6 0

2 years ago

The pressure of air pushing down on a house roof is extremely large. Why is the roof not crushed?

Degger [83]

There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.

7 0

3 years ago

One gallon of paint (volume = 3.79 X 10-???? m3) covers an area of 25.0 m2. What i the thickne s of the fresh paint on the wall?

Drupady [299]

Answer:

Explanation:

Given

Volume of paint is $V=3.79\times 10^{-3}\ m^3$

Area of cover $A=25\ m^2$

Suppose paint to be a rectangular box with thickness t and volume V

therefore we can write as

$V=A\times t$

$t=\frac{V}{A}$

$t=\frac{3.79\times 10^{-3}}{25}$

$t=1.516\times 10^{-3}\ m$

$t=1.516\ mm$

6 0

3 years ago

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

a train car of mass 444 kg moving at 5 m/s bounces into another car on the same tracks of mass 344 king. of the second car was m

Kamila [148]

Answer:

3.7 m/s

Explanation:

M = 444 kg

U = 5 m/s

m = 344 kg

u = - 5 m/s

Let the velocity of train is V and the car s v after the collision.

As the collision is elastic

By use of conservation of momentum

MU + mu = MV + mv

444 x 5 - 344 x 5 = 444 V + 344 v

500 = 444 V + 344 v

125 = 111 V + 86 v .... (1)

By using the formula of coefficient of restitution ( e = 1 for elastic collision)

$e = \frac{V-v}{u-U}$

-5 - 5 = V - v

V - v = - 10

v = V + 10

Substitute the value of v in equation (1)

125 = 111 V + 86 (V + 10)

125 = 197 V + 860

197 V = - 735

V = - 3.7 m/s

Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.

4 0

3 years ago