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3 years ago

What happens to beaches over time?

Physics

2 answers:

Rama09 [41]3 years ago

7 0

the answer is d i took the test be cause i go to a k12 onlie school

vitfil [10]3 years ago

3 0

Answer:

Option (D)

Explanation:

The beaches are defined as the landform that is oriented along the side of the large water bodies such as seas and oceans. They are dominantly comprised of sand and other rock particles such as pebbles and gravels.

These beaches are often affected by factors such as, the change in the sea level, that are caused due to the tidal effect, and the occurrence of storms and tsunamis.

The shoreline of the beach shifts towards the inland areas when the sea level rises and the opposite happens when the sea level drops. They are suddenly affected when a storm occurs, eroding the sand particles towards or away from the water body.

Thus, the beaches or shorelines are continuously affected due to the above factors.

Hence, the correct answer is option (D).

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Chuyển động thẳng đều là gì

suter [353]

Question:

What is uniform rectilinear motion?

Answer:

Uniform rectilinear motion is when an object travels at a constant speed with zero acceleration.

8 0

3 years ago

A 1,450 kg car drives toward a 60 kg shopping cart that has a velocity of -1.2 m/s toward the car. The two objects collide, givi

Y_Kistochka [10]

Answer:

A) v₁ = 5.66 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum conservation, which tells us that momentum is conservation before and after a collision.

The linear momentum is equal to the mass by the product of the Velocity.

P = m*v

where:

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Now, to the right side of the equal sign will take the linear momentum before the collision and to the left side of the equal sign as after the collision.

Pbefore = Pafter

(m₁*v₁) - (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the car = 1450 [kg]

v₁ = velocity of the car before the collision [m/s]

m₂ = mass of the shopping cart = 60 [kg]

v₂ = velocity of the shopping cart before the collision = -1.2 [m/s]

v₃ = velocity of the car after the collision = 5.13 [m/s]

v₄ = velocity of the shopping cart after the collision = 11.75 [m/s]

Now replacing:

(1450*v₁) - (60*1.2) = (1450*5.13) + (60*11.75)

1450*v₁ - 72 = 7438.5 + 705

1450*v₁ = 7438.5 + 705 + 72

1450*v₁ = 8215.5

v₁ = 5.66 [m/s]

4 0

3 years ago

Read 2 more answers

Cylindrical beaker of height 0.100 mm and negligible weight is filled to the brim with a fluid of density rhorhorho = 890 kg/m3k

Nesterboy [21]

Incomplete part of the question

A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.

What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale. Calculate your answer from the quantities given in the problem and express it numerically in newtons.

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Express your answer numerically in newtons.

The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.

What weight W3 does the scale now show?

Answer:

(a) 2.94 N

(b) 1 N

(d) 3.42 N

Explanation:

(a)

From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity

The weight of the ball is $W=\rho g V$