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3 years ago

What is the climate of northern Alaska?

Physics

2 answers:

sasho [114]3 years ago

5 0

Answer:

It's short, cool summers (where snow could be year round) and long cold winters.

ss7ja [257]3 years ago

5 0

Answer:

It's short, cool summers (where snow could be year round) and long cold winters.

Explanation:

You might be interested in

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t²- 2 t to + to²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to. The rock y has not left and the distance increases

For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂

3 0

3 years ago

The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from

Allushta [10]

Answer:

y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

let's substitute

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

0 = v₀ sin θ - 9.8 t

Let's write our system of equations

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

v₀ sin θ = 9.8 t

we substitute

10 = (9.8 t) t - ½ 9.8 t2

10 = ½ 9.8 t2

10 = 4.9 t2

t = √ (10 / 4.9)

t = 1,429 s

Now let's use the first equation and the last one

45 = v₀ cos θ t

0 = v₀ sin θ - 9.8 t

9.8 t = v₀ sin θ

45 / t = v₀ cos θ

we divide

9.8t / (45 / t) = tan θ

tan θ = 9.8 t² / 45

θ = tan⁻¹ ( 9.8 t² / 45 )

θ = tan⁻¹ (0.4447)

θ = 24º

Now we can calculate the maximum height

v_y² = $v_{oy}^2$ - 2 g y

vy = 0

y = v_{oy}^2 / 2g

y = (20 sin 24)²/2 9.8

y = 3,376 m

the other angle that gives the same result is

θ‘= 90 - θ

θ' = 90 -24

θ'= 66'

for this angle the maximum height is

y = v_{oy}^2 / 2g

y = (20 sin 66)²/2 9.8

y = 17 m

thisis the correct

6 0

2 years ago

Does the boat hold more weight in the regular water or the salt mixture

zmey [24]

It holds more weight in the regular water.

6 0

2 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

7 0

2 years ago

Can a metallic sphere be charged without rubbing

morpeh [17]

We need to charge a metal sphere positively without touching it. This can be achieved using electrostatic induction.

6 0

2 years ago