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3 years ago

What is the name of the meteorologist who proposed the concept of continental drift?

Physics

2 answers:

viva [34]3 years ago

7 0

The correct answer is Alfred Wegener. I am absolutely positive.

yaroslaw [1]3 years ago

5 0

Alfred wegener Im pretty sure

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3. Qué tipo de alcohol en bebidas

Keith_Richards [23]

El etanol (o alcohol etílico) es el tipo de alcohol que se usa en las bebidas.

4 0

3 years ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

Lisa [10]

Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒ 0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0 and x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If v₀x = 15 m/s = vx and ymax = 24 m

y = ? when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒ y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒ y = 18.4 m

7 0

3 years ago

Read 2 more answers

The time constant of an RC circuit is 2.7 s. How much time t is required for the capacitor (uncharged initially) to gain 0.63 of

gayaneshka [121]

Answer:

2.7s

Explanation:

The solution of time required is shown below:-

In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time

Now, the

63% that is equal to 0.63 which is full equilibrium charge

Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s

So, the correct answer is 2.7s

8 0

3 years ago

A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr

irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

$a = \dfrac{T}{m_2} - g$ .......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

$a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$ .........(2)

from equations 1 and 2

$\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$

$\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

$T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

$T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}$

$T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}$

$T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}$

T = 90.61 N

from equation (1)

$a = \dfrac{90.61}{8} - 9.8$ .......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

$t = \sqrt{2.63}$

t = 1.62 s

5 0

3 years ago

At which position is the roller coaster car's amount of potential energy about equal to the amount of kinetic energy?

meriva

It will be potential because the roller coaster is going down so it’s not giving off energy to the roller coaster.

7 0

3 years ago

Read 2 more answers