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DENIUS [597]
3 years ago
7

You are doing an experiment to determine how many passengers would fit into a full-sized plane. If a scale model plane has space

for 87 scaled down passengers, how many would fit into the actual plane?
Physics
2 answers:
romanna [79]3 years ago
5 0
The real place should theoretically have space for 87 passengers if it is an exact model and doesn't have modifications in the seat numbers.
amm18123 years ago
5 0
If the "scales" for the plane model and the people models were the same, then the experiment indicates that the real plane would hold 87 real people.
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Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas
Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}

For 4 particles:

I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}

c)

I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

8 0
3 years ago
Write relationship between hertz and megahertz​
dusya [7]

Explanation:

1 mega Hertz = 1000000 hertz

8 0
3 years ago
3. What exerts a greater force on the table of 2 kg book lying flat or a 2 kg book on its
Darina [25.2K]

Answer:

A book on its side exerts a greater force.

Explanation:

Pressure = Force / Area

Assuming that 1kg = 10N

2kg = 20N

Area of book lying flat = 0.3m × 0.2m

                                     = 0.6m²

Pressure of book lying flat = 20N / 0.6m²

                                            = 30Pa (1 s.f.)

Area of book on its side = 0.2m × 0.05m

                                        = 0.01m²

Pressure of book on its side = 20N / 0.01m²

                                               = 2000Pa (1 s.f.)

Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.

5 0
2 years ago
A block of mass 2 kg slides down a frictionless ramp of length 1.3 m tilted at an angle 25o to the horizontal. At the bottom of
marin [14]

Answer:

Diagrams in pictures

Explanation:

Using energy I can get

m g h = 1/2 m v^2

So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.

(H= 1,3m sin25)

V=2,32m/s

V= a t

And

X= v t +1/2 a t^2

Knowing v=2,32 m/s and x= 1,3 m

I can get

a= 6,21m/s2

F= m a

I can get the force of the box when it collides with the spring

F= 12, 425 N

The force the spring makes on the box then is

F = -12,425N = -k d

Then the spring's constant is k= 51,75N/m

To make the two diagrams I need the functions of time when the box slows down

I use the same two equations

V= a t

And

X= v t + 1/2 a t^2

Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.

I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).

Then,

V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec

X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.

for time between 0 and 0,689 sec

Diagrams and equations are in the pictures

7 0
3 years ago
The formula shown below is used to calculate the energy released when a specific quantity of fuel is burned. Calculate the energ
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Answer: 8400 J

Explanation:

The formula referenced in the question is:

Q=m. c. \Delta T  

Where:

Q  is the thermal energy

m=100g \frac{1 kg}{1000 g}=0.1 kg is the mass  of the water sample

c=4200 \frac{J}{kg\°C}  is the specific heat capacity of  water

\Delta T=20\°C  is the variation in temperature

Solving:

Q=(0.1 kg)(4200 \frac{J}{kg\°C})(20\°C)  

Q=8400 J  This is the thermal energy released

8 0
3 years ago
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