Ask question

Ask question

All categories

2 years ago

7

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

opter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?

1 answer:

andreev551 [17]2 years ago

8 0

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

You might be interested in

A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the

kodGreya [7K]

Answer:

3.73994

No,

Explanation:

$P_1$ = Pressure at the bottom of the lake = 3.5 atm

$P_2$ = Pressure at the top of the lake = 1 atm

$V_b$ = Volume at the bottom of the lake

$V_s$ = Volume at the top of the lake

$T_1$ = Temperature at the bottom of the lake = 4 °C

$T_2$ = Temperature at the top of the lake = 23 °C

From ideal gas law we have the relation

$\dfrac{P_1V_b}{T_1}=\frac{P_2V_s}{T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{P_2T_1}{P_1T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{1\times 277.15}{3.5\times 296.15}\\\Rightarrow \dfrac{V_s}{V_b}=0.26738^{-1}\\\Rightarrow \dfrac{V_s}{V_b}=3.73994$

The ratio is 3.73994

As Jacques is ascending if he holds his breath his lungs acting like a bubble would expand. Hence, it is not safe to hold his breath while ascending,

8 0

2 years ago

Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

sammy [17]

Answer:

Net force acting on them is 16 N and it is acting to the right side.

Explanation:

It is given that,

Force acting by the dog, $F_1 = 32\ N$ (right side)

Force acting by Simone , $F_2 = -16\ N$ (backward)

Let backward direction is taken to be negative while right side is taken to be positive.

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

So, the net force is 16 N and it is acting to the right side.

6 0

2 years ago

Read 2 more answers

Karyotypes are done by matching up _____________________________ so that they are paired up. Question 11 options:

Rufina [12.5K]

Homologous Chromosomes

6 0

2 years ago

Read 2 more answers

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its

Katyanochek1 [597]

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

d is the length of the minute hand (d=D/2)
D is the diameter of the clock
t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

I hope it helps you!

6 0

2 years ago

A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Volgvan

Answer:

A=0.199

Explanation:

We are given that

Mass of spring=m=450 g= $=\frac{450}{1000}=0.45 kg$

Where 1 kg=1000 g

Frequency of oscillation= $\nu=1.2Hz$

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator= $E=\frac{1}{2}m\omega^2A^2$

Where $\omega=2\pi\nu$ =Angular frequency

A=Amplitude

$\pi=\frac{22}{7}$

Using the formula

$0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2$

$A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398$

$A=\sqrt{0.0398}=0.199$

Hence, the amplitude of oscillation=A=0.199

4 0

2 years ago