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3 years ago

Why does a typhoon hit a location twice?

1 answer:

7 0

The reason a location usually doesn't get a storm within a few weeks of another one is more because of the general odds. They're low. And they're lower in Texas than in Florida. ... There are many locations a storm can develop and track

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A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele

artcher [175]

Answer:

the net work done after starting from rest is = 2.4 × 10⁵ J

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J), The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)

Substituting these values into equation 2

v² = 0² +2×1×30

v² = 60

v = √60

v = 7.75 m/s

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

Substituting these values into equation 1,

W = 1/2(8.0×10³)(7.75)²

W = (4.0×10³)(60)

W = 240 × 10³ J

W = 2.4 × 10⁵ J

Therefore the net work done after starting from rest is = 2.4 × 10⁵ J

4 0

3 years ago

Is frequent anger a problem for you? If so, what strategies could you try to overcome it?

erastova [34]

Answer:

I tend to slap thicc trees a#ss

Explanation:

3 0

3 years ago

A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin

Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0

3 years ago

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

4 years ago

A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?

Anastaziya [24]

Answer: $1.089\times 10^5\ A$