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4 years ago

How does heat play a role in both the formation of folded mountains and the formation of volcanic mountains?

1 answer:

5 0

Among the most striking of geologic features are mountains, created by several types of tectonic forces, including collisions between continental masses. Mountains have long had an impact on the human psyche, for instance by virtue of their association with the divine in the Greek myths, the Bible<span>, and other religious or cultural traditions. One does not need to be a geologist to know what a mountain is; indeed there is no precise definition of mountain, though in most cases the distinction between a mountain and a hill is fairly obvious. On the other hand, the defining characteristics of a volcano are more apparent. Created by violent tectonic forces, a volcano usually is considered a mountain, and almost certainly is one after it erupts, pouring out molten rock and other substances from deep in the </span>earth<span>.
does it help?</span>

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

The weather map shows some conditions in the atmosphere at noon on a

earnstyle [38]

The geographical region where I would expect the warmest weather is: C. in the southeast.

<h3>What is a weather map?</h3>

A weather map can be defined as a type of chart that is typically used to provide information about the average atmospheric condition of a particular geographical region over a specific period of time.

Based on the weather map shown in the image attached below, we can infer and logically deduce that the geographical region where the warmest weather is expected is in the southeast due to its very high atmospheric pressure.

Read more on weather here: brainly.com/question/24730207

#SPJ1

5 0

2 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

3 years ago

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0

Tema [17]

Answer:

841.5 Hz

Explanation:

Given

y = 50 cm = 0.5 m

d = 5.00 m

L = 12.0 m away from the wall

v = speed of sound = 343 m/s

The image of the scenario is presented in the attached image.

When destructive interference is being experienced from 50 cm (0.5 m) parallel to the wall, the path difference between the distance of the two speakers from the observer is equal to half of the wavelength of the wave.

Let the distance from speaker one to the observer's new position be d₁

And the distance from the speaker two to the observer's new position be d₂

(λ/2) = |d₁ - d₂|

d₁ = √(12² + 3²) = 12.3693 m

d₂ = √(12² + 2²) = 12.1655 m

|d₁ - d₂| = 0.2038 m

(λ/2) = |d₁ - d₂| = 0.2038

λ = 0.4076 m

For waves, the velocity (v), frequency (f) and wavelength (λ) are related thus

v = fλ

f = (v/λ) = (343/0.4076) = 841.5 Hz

Hope this Helps!!!

7 0

4 years ago

Read 2 more answers

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago