Ask question

Ask question

All categories

Mandarinka [93]

3 years ago

15

A plane’s average speed between two cities is 800 km/h. If the trip takes 3 hours, how far does the plane fly? Type your answer

in the space below including units! (Hint: s=d/t)

2 answers:

jok3333 [9.3K]3 years ago

8 0

Answer:

266 1/3 km

Explanation:

Speed= distance divided by time.

800km/h = distance/ 3 hours

800/3= 266.67

266 1/3 km.

AlladinOne [14]3 years ago

3 0

Distance= Vxt=800km/h x 3hours=2,400km

You might be interested in

At what displacement of a sho is the energy half kinetic and half potential?

Illusion [34]

Answer:

Displacement = 0.707A

Explanation:

To solve for the displacement we know that

Potential energy PE = 1/2Total energy (Etotal)

Therefore 1/2kx^2 = 1/2(1/2KA^2)

Solving for x we have

x^2 = √A^2/2

x = A/√2

x= 0.707A

5 0

3 years ago

Why is it less air pressure up in the sky and why does it affect our ears

IgorLugansk [536]

Answer:

When you in a airplane you are in high altitude and thats where air density drops. Thats where ear pops come from.

Explanation:

8 0

2 years ago

A car is traveling at 13 m/s and accelerates at - 0.95

riadik2000 [5.3K]

Answer:

v = 12.3 m / s

Explanation:

This is an exercise in kinetics in one dimension

v² = v₀² + 2 a x

In this exercise they tell us that the initial velocity is (v₀ = 13 m / s), the acceleration is a = -0.95 m / s2 and the distance x = 9.2 m

we substitute

v = √ (13 2 - 2 0.95 9.2)

v = 12.3 m / s

note that as the acceleration is negative the vehicle is stopping

3 0

3 years ago

A person places his hand palm downward on a scale and pushes down on the scale until it reads 96.0 N. The triceps muscle is resp

Ivahew [28]

Answer:

the force exerted by the triceps muscle is 1190.4 N or 1.2 kN

Explanation:

given data

F = 96 N

D = 2.50 cm

L = 31 cm

solution

we will apply here equilibrium condition that is

∑r = about the elbow joint ............................1

so force exert by the muscle is

F(tri) × D = F × L .................2

put here value

F(tri) × 2.50 = 96 × 31

solve it we get

F(tri) = 1190.4 N

so the force exerted by the triceps muscle is 1190.4 N or 1.2 kN

3 0

3 years ago

low-velocity steam (with negligible kinetic energy) enters an adiabatic nozzle at 300 C and 3 MPa. the steam leaves the nozzle a

Zanzabum

Answer:

the quality of the steam = 1.05

the temperature at $P_2 = 2 \ MPa$ = $380.733^0 \ C$

the exit area of the nozzle = $146.33 \ mm^2$

Explanation:

Given that:

the inlet temperature of steam $T_1$ = $300^0 \ C$

Inlet pressure of steam $P_1= 3 \ MPa$

Initial Velocity $v_1$ at the inlet = 0 m/s

Exit pressure of the steam $P_2 = 2 \ MPa$

Exit velocity of steam $v_2 \ = 400 \ m/s$

Mass flow rate m = 0.4 kg/s

Now, from steam tables at $T_1$ = $300^0 \ C$ and $P_1= 3 \ MPa$

$h_1 = 2992. 35 \ kJ/kg\\s_1 = 6.53535 \ kJ/kg\\$

At $P_2 = 2 \ MPa$

$sf_2 = 2.4474 kJ/kgK\\s_g = 6.3409 \ kJ/kg$

To determine the condition of the steam at exit ;

$s_1 = s_2$

therefore,

$6.53535 = s_{f2} +x_2sg-sf_2$

$6.53535 = 2.4474 +x_2(6.3409-2.4474)$

$x_2 = 1.05$

Thus , the quality of the steam = 1.05

However, By using the energy balance equation to determine the temperature of the steam; we have:

$h_1 + \frac{v^2_1}{2}= h_2 + \frac{v^2_2}{2}$

$h_1-h_2 = \frac{v^2_2}{2}$

$h_2 = h_1 - \frac{v^2_2}{2}$

$h_2 = 2992.35 - \frac{250^2}{2000}$

$h_2 = 2912.35 \ kJ/kg$

From steam tables ; at enthalpy $h_2$ and $P_2 = 2 \ MPa$ ; the corresponding temperature $T_2 = 380.733^0 \ C$

Thus, the temperature at $P_2 = 2 \ MPa$ = $380.733^0 \ C$

Finally, to calculate the exit area of the nozzle in mm²

we use the mass flow rate relation:

$m = \frac{A_2v_2}{V_2}$

Making A the subject of the formula; we have:

$A_2 = \frac{mv_2}{V_2}$

From the superheated steam tables at pressure $P_2 = 2 \ MPa$ ; specific volume $v_2 = 0.14633 \ m^3/kg$

We have:

$A_2 = \frac{mv_2}{V_2}$

$A_2 = \frac{0.4*0.14633}{400}$

$A_2 = 1.4633*10^{-4} m^2$

$A_2 = 146.33 \ mm^2$

Thus, the exit area of the nozzle = $146.33 \ mm^2$

7 0

3 years ago