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4 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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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago

A train passes a stationary observer. Which of

Lyrx [107]

Explanation:

Sorry but thee is none options for me to choose

4 0

3 years ago

What advantage do space telescopes have over telescopes used on earth?

Elina [12.6K]

Earth's atmosphere blocks many types of light including gamma, x-rays most ultraviolet and infrared. So optical telescopes that use visible light and ultraviolet telescopes that are used to study very hot stars are much less effective on Earth.

4 0

3 years ago

Read 2 more answers

In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having

Orlov [11]

Answer:

R = 5.28 103 km

Explanation:

The definition of density is

ρ = m / V

V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

V = 4/3 π r³

Let's replace

4/3 π r³ = m / ρ

R =∛ ¾ m / ρ π

The mass of the planet is

M = 5.5 Me

R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

ρ = 1.76 10³ kg / m³

Let's calculate

R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

R = ∛ 0.14723 10²¹

R = 0.528 10⁷ m

R = 0.528 104 km

R = 5.28 103 km

8 0

3 years ago

Boy or girl and age you can still comment if its answered

Ksenya-84 [330]

I am a girl and i have a virgina and im 13 tho so dont touch me lolz mark brainly

6 0

3 years ago