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3 years ago

What is the difference between catastrophism and uniformitarianism

Physics

1 answer:

kkurt [141]3 years ago

4 0

Explanation:

Catastrophism is a geological concept or ideology that was formerly in place. It suggests that the earth crust has evolved through only drastic and violent geological events.

Uniformitarianism suggests that the earth carefully evolved over a period of time and that the processes that are occurring today have occurred in times past.

Catastrophism presents a drastic and rapid evolution of the crust.
Uniformitarianism takes a gradual approach to events and suggests that we can use the present to unravel the past.

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A boat displaces a volume V of water as it floats on a fresh-water lake. When the boat moves into the ocean, what volume of salt

Mamont248 [21]

Answer:

Option (b)

Explanation:

let the weight of boat is W. In equilibrium condition, the weight of boat is equal to the buoyant force acting on the boat.

The buoyant force acting on the boat is equal to the weight of water displaced by the boat.

In case of fresh water:

Weight of the boat = weight of fresh water displaced by the boat

W = Volume of fresh water displaced x density of fresh water x g

W = V x 1 x g

W = V x g ....... (1)

In case of salt water:

Let the volume of salt water displaced is V'.

Weight of the boat = weight of salt water displaced by the boat

W = Volume of salt water displaced x density of salt water x g

W = V' x 1.02 x g ..... (2)

Equate equation (1) and equation (2), we get

V x g = V' x 1.02 x g

V' = 0.98 V

Thus, option (b) is true.

7 0

4 years ago

To 3 significant digits, what is the period of the pendulum using the equation for the period of a simple pendulum? Hint, the in

netineya [11]

Answer:

1.42 seconds

Explanation:

T=2*pi*(sqrt length over gravity)

T=2*pi*(sqrt 0.5/9.8)

T=1.4189

Round to 3 significant digits

T=1.42 seconds

3 0

3 years ago

What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.

djyliett [7]

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity $g=9.8m/sec^2$

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that $v^2=u^2+2gh$

$v^2=0^2+2\times 9.8\times 150$

$v=54.22m/sec$

So the velocity of the ball when its hit the ground will be 54.22 m/sec

7 0

3 years ago

A daring 510N swimmersdives off a cliff with a running horizontal lead.what must be her mimimum speed just as she leaves the top

goldfiish [28.3K]

Answer:

The running horizontal speed should be larger than 1.29 m/s.

Explanation:

In order for the swimmer to just miss the bone-breaking ledge, her horizontal speed must be

$v > \frac{1.75m}{t_{fall}}$

in which we need to know how long we she be diving through the air. To determine that, recall the formula for the distance made by an object with acceleration (in this case it is the gravitational acceleration) with no initial (vertical) velocity:

$s = \frac{1}{2}gt^2$

from which it follows that (for non-negative t)

$t = \sqrt{\frac{2\cdot9m}{9.8\fram{m}{s^2}}}\approx1.36s$

This result can be used in the initial inequality:

$v > \frac{1.75m}{t_{fall}}=\frac{1.75m}{1.36s}=1.29\frac{m}{s}$

The diving lady better gets a speed larger than 1.29 m/s to avoid landing on the ledge.

8 0

3 years ago

Read 2 more answers

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0

4 years ago