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MariettaO [177]

2 years ago

9

An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and

the second is 155 km 60.0° east of north. What is the total displacement of the airplane?

1 answer:

torisob [31]2 years ago

5 0

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

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Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions about a fixed axis. Compare the linear

dezoksy [38]

Answer:

Explanation:

Wheel completes four revolution.

The linear displacement is zero.

The angular displacement is 4 x 2π = 8π radian.

So, option (c) is correct.

4 0

2 years ago

A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic

andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0

2 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

2 years ago

The net force on a boat causes it to accelerate at 1.55 m/s2. The mass of the boat is 215 kg. The same net force causes another

jekas [21]

Answer:

2666 kg

0.11567 m/s²

Explanation:

m = Mass of boat

a = Acceleration of boat

From Newton's second law

Force

$F=ma\\\Rightarrow F=215\times 1.55\\\Rightarrow F=333.25\ N$

Force on the first boat is 333.25 N

$F=ma\\\Rightarrow m=\frac{F}{a}\\\Rightarrow m=\frac{333.25}{0.125}\\\Rightarrow m=2666\ kg$

Hence, mass of the second boat is 2666 kg

Combined mass = 2666+215 = 2881 kg

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{333.25}{2881}\\\Rightarrow a=0.11567\ m/s^2$

The acceleration on the combined mass is 0.11567 m/s²

6 0

2 years ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = $\frac{4}{3} \pi (\frac{D}{2})^3$

Therefore $\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3$

r of the disc = $1.15(\frac{ D}{2} )$

Using conservation of angular momentum;

The $M_i$ of the sphere = $\frac{2}{5} m \frac{D}{2}^2$

$M_i$ of the disc = $m*\frac{ \frac{1.15*D}{2}^2 }{2}$

$\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }$

= 0.6

5 0

2 years ago