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3 years ago

5

Airplane tickets are extremely cheap right now. You decide to take a trip to Paris for a quick weekend vacation. You hop on the

25,000 kg
airplane and your journey begins. The airplane is cruising at a velocity of 155 m/s at a height of 6,000 m. The pilot decides to glide up to a
height of 7,000 m. How fast is the airplane now traveling?
Useg - 9.8 m/s^2

1 answer:

andreyandreev [35.5K]3 years ago

6 0

Answer:

i don't exactly know but im srry for not giving the right answer

Explanation:

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A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the appara

saw5 [17]

Answer:

(a) $\alpha = 53.73 m/s^2$

(b) I =428 $kgm^2$

(c) $\tau = 428 \times 53.73 = 22996 .44Nm$

Explanation:

GIVEN

mass = 18.2 kg

radial arm length = 3.81 m

velocity = 49.8 m/s

mass of arm = 22.6 kg

we know using relation between linear velocity and angular velocity

$\omega = \frac{v}{l}$

$\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s$

for angular acceleration, use the following equation.

$\omega _{f}^2 = \omega_{i}^2+2\alpha\theta$

since $\omega _{i} = 0$

here for one circle is 2 π radians. therefore for one quarter of a circle is π/2 radians

so for one quarter $\theta = \frac{\pi }{2}$

$(12.99)^2 = 2\alpha(\frac{\pi }{2})$

on solving

$\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2$

(b)

For the catapult,

moment of inertia

$I = \frac{1}{2}MR^2$

$I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2$

For the ball,

$I = MR^2$

$I = 18.2 \times 14.51$

$I = 264 kgm^2$

so total moment of inertia = 428 $kgm^2$

(c)

$\tau = I\alpha$

$\tau = 428 \times 53.73 = 22996 .44Nm$

3 0

4 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

4 years ago

Is this a scientific model? Use complete sentences to explain why or why not.

grin007 [14]

Answer:

A scientific model is an activity that is aimed at making a specific part or feature easier to understand and define.

Modeling is an essential scientific discipline and is used in the fields of education. Modeling seems to represent empirical objects, phenomena in a logical and objective way.

The oval-shaped world map showing vegetation in green color is a 2D model of the entire earth and demarcates us the world vegetation distribution and growth.

Hence this is a scientific model as its helps in visualizing the land surface of the earth on paper.

Explanation:

Yes because a scientific models doesn't have to be complex, just like this one

8 0

3 years ago

Which statement is true about metals?

Nat2105 [25]

Metals are good conductors of electricity

7 0

3 years ago

Find the height or length of these natural wonders in km, m, and cm.A. A cave system with a mapped length of 354 miles.B. A wate

cricket20 [7]

Explanation:

Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,

1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.

Now, use these values to convert the given lengths.

A. length of cave = 5 miles (given)

From standard value 1 mile = 1.609344 km

$\Rightarrow$ 5 miles = $\times$ 1.609344 km= 8.04672 km,

$\Rightarrow$ 5 miles =8.04672 $\times$ 1000 m= 8046.72 m [ as 1 km = 1000 m]

$\Rightarrow$ 5 miles = 8046.72 $\times$ 100 cm=804672 cm

B. Height of the waterfall = 1235 ft. (given)

1 ft.= 30.48 cm [ from standard value]

$\Rightarrow$ 1235.2 ft.= 1235.2 $\times$ 30.48 cm=37648.896 cm,

$\Rightarrow$ 1235.2 ft.=37648.896 $\times$ 0.01 m =376.48896 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 1235.2 ft.=376.48896 $\times$ 0.001 km =0.37648896 km [ as 1 m = 0.001 km]

C. Height of the mountain= 21320 ft. (given)

From standard value: 1 ft.= 30.48 cm

$\Rightarrow$ 21320 ft.= 21320 $\times$ 30.48 cm=649833.6 cm,

$\Rightarrow$ 21320 ft.=649833.6 $\times$ 0.01 m =6498.336 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 21320 ft.=6498.336 $\times$ 0.001 km =6.498336 km [ as 1 m = 0.001 km].

D. Depth of canyon =6630 ft.

From standard value: 1 ft.= 30.48 cm

$\Rightarrow$ 6630 ft.= 6630 $\times$ 30.48 cm=202082.4 cm,

$\Rightarrow$ 6630 ft.=202082.4 $\times$ 0.01 m =2020.824 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 6630 ft.=2020.824 $\times$ 0.001 km =2.020824 km [ as 1 m = 0.001 km].

4 0

4 years ago