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2 years ago

Short-range forecasts tends to ________ longer-range forecasts.

1 answer:

5 0

<span>Short-range forecasts are more accurate than longer range ones. Short term forecasts may use mathematical techniques such as moving averages and, exponential smoothing. Longer term forecasts not only use different methodologies, such as qualitative vs. quantitative, they also tend to consider different issues.</span>

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Dao makes a table to identify the variables used in the equations for centripetal acceleration. A 2 column 5 rows. The first col

Zanzabum

Answer:

Column X. Tangential Speed

Column Y. radius

Explanation:

The equation for centripetal acceleration is

$a_{c}$ = v² / r

Where v is the tangential velocity of the body and the radius of curvature.

To analyze this equation you must place the tangential velocity in one column and in the other the turning radius

Let's check the answers

Column X. Tangential Speed

Column Y. radius

This is the correct answer.

5 0

1 year ago

Read 2 more answers

13. An aircraft heads North at 320 km/h rel:

AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

$v' = v + v_a$

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

$v_a$ is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

$v_a = -80 km/h$ (since the air is moving from North)

Therefore, we find

$v'=+320 + (-80) = +240 km/h$ (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0

2 years ago

In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

1 year ago

The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp

Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0

1 year ago

A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.

Umnica [9.8K]

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

8 0

1 year ago