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LUCKY_DIMON [66]
3 years ago
9

If you added 5 x more mass into a wheelbarrow, how much greater would

Physics
1 answer:
aev [14]3 years ago
7 0

Answer:

5 x biger

Explanation:

becuze if it ia 5× more it nids 5×more force to be =

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According to Newton’s Third Law, action and reaction are
Komok [63]
<span>B. equal and in opposite directions</span>
5 0
3 years ago
Read 2 more answers
An open organ pipe 30 cm long and a closed organ pipe 23 cm long, both of same diameter , are each sounding its first overtone ,
cupoosta [38]

First overtone of open organ pipe is given as

f_{1o} = \frac{v}{L_1 + 2e}

first overtone of closed organ pipe is given as

f_{1c} = \frac{3v}{4(L_2 + e)}

now they are in unison so we will have

\frac{v}{L_1 + 2e} = \frac{3v}{4(L_2 + e)}

\frac{1}{30 + 2e} = \frac{3}{4(23 + e)}

90 + 6e = 92 + 4e

e = 1 cm

so end correction of both pipes is e = 1 cm

8 0
3 years ago
A passenger jet travels from Los angles to Bombay, India, in 22 h. The return flight takes 17 h. The difference in flight times
Hoochie [10]

Answer:

70.5 mph

Explanation:

A passenger jet travels from Los Angeles to Bombay, India, in 22h.

The return flight takes 17 h.

The difference in flight times is caused by winds over the Pacific Ocean that

blow primarily from west to east.

If the jet's average speed in still air is 550 mi/h what is the average speed

of the wind during the round trip flight? Round to the nearest mile per hour.

Is your answer reasonable?

:

Let w = speed of the wind

:

Write a distance equation (dist is the same both ways

17(550+w) = 22(550-w)

9350 + 17w = 12100 - 22w

17w + 22w = 12100 - 9350

39w = 2750

W = 2750/39

w = 70.5 mph seems very reasonable

:

Confirming if the solution by finding the distances using these value

17(550+70.5) = 10549 mi

22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph

6 0
3 years ago
During convection of air currents,
pav-90 [236]

During convection of air currents,  cool air sinks. <em>(b)</em>

4 0
3 years ago
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A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.
Keith_Richards [23]

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

= 28.4 kg

The mass finally contained in the tank is

m2 =V2/v

= 0.85 m^3 /0.0422 m^3 /kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

4 0
3 years ago
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