The periodic table charge between night and day caused by

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

mel-nik [20]

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

4 0

3 years ago

Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i

Tanzania [10]

The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
$v= \frac{d}{t}$ ⇒ $d=v*t$

It is given:
$v = \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s} = 3.2 m/s$
t = 10 s
d = ?

So:
$d= v*t=3.2m/s*10s = 32m$

3 0

3 years ago

How do climate differences affect the movement at the Mariana Trench

vovangra [49]

It pushes the currents to opposite sides

8 0

3 years ago

At Hoover Dam, the distance the water effectively falls before encountering the electric generators depends on the water levels

pashok25 [27]

Answer:

Explanation:

recall that power is energy carried out or work done per time

P=W/t

P=2*10^6*35

t=6*60=420S

W=Energy

E=2*10^6*35*360S

E=25200000000

Energy stored by water from rest is called potential energy. Since the water is falling from a height , we calculate potential energy as thus

E=M*g*h

Assume that the water intakes are effectively 175 m above the electric generators. How much water must pass through the generators to power 2 million 35-W Las Vegas light bulbs for 6.0 minutes?

M=mass of water

g=acceleration due to gravity 9.81m/s^2

h=height ,175m

25200000000=M*9.81*175

M= $\frac{25200000000}{175*9.81}$

M=1716.75kg

3 0

3 years ago