All categories

4 years ago

What is a Rift Valley ?

2 answers:

7 0

a steep-sided valley formed by the downward displacement of a block of the earth's surface between nearly parallel faults or fault systems.

Naddika [18.5K]4 years ago

6 0

a large elongated depression with steep walls formed by the downward displacement of a block of the earth's surface between nearly parallel faults or fault systems.

You might be interested in

A cylinder of mass mm is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

Zinaida [17]

Answer:

$y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}$

Explanation:

Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles, The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder

work done by the spring force = ¹/₂k(y₀² - y²)

work done by friction = - f(y - y₀)

work done by gravity = mg(y - y₀)

kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)

So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)

Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0

¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0

¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy

Let y₀ = 0, then the left hand side of the equation equals zero. So,

0 = ¹/₂ky² + fy - mgy

¹/₂ky² + fy - mgy = 0

Using the quadratic formula

$y = \frac{-f +/- \sqrt{f^{2} - 4 X\frac{k}{2} X -mg}}{2 X \frac{k}{2} }\\ y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}$

4 0

3 years ago

A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I

melamori03 [73]

(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
$v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s$

(b) The initial speed of the skater is
$v_i = 4 m/s$
while the final speed is
$v_f = 5.3 m/s$
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
$a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2$

(c) The initial speed of the skater is
$v_i = 13.0 m/s$
while the final speed is
$v_f=0$
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
$2aS=v_f^2 -v_i^2$
from which we find
$a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2$
where the negative sign means it is a deceleration.

4 0

4 years ago

A golfer imparts a speed of 29.0 m/s to a ball, and it travels the maximum possible distance before landing on the green. the te

Roman55 [17]

Explanation:

It is given that,

Initial speed of a golfer, u = 29 m/s

If it travels the maximum possible distance before landing. It means that it is projected at an angle of 45 degrees.

(a) We need to find the time spent by the ball in the air. It can be calculated by using second equation of motion.

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

Here,

a = -g

s = 0 (it is displacement and it is equal to 0 as the ball lands on the green).

So,

$0=29\sin(45)t-\dfrac{1}{2}\times 9.8t^2\ (\text{Initial vertical component of velocity is taken})\\\\-4.9t^2+29\times \dfrac{1}{\sqrt2}t=0\\\\-4.9t^2+20.5t=0\\\\t=0,4.184\ s$