Mass wasting _____. creates deltas and moraines changes the land by plucking is the movement of sediment by gravity transports rock up an incline by sliding or slumping

Answers

Answer 1

Answer:

Answer: Is the movement of sediment by gravity.

Mass wasting is the movement of sediments generate due to the mass movement of rock, soil or other debris from a hilly or slopy region downwards due to the effect of gravity. This effect can be highly be observed during landslides. Agents like wind and water also promote this effect.

Answer 2

Answer:

Mass wasting is the movement of sediment by gravity transports. Meaning the earth’s outer crust is being ‘wasted’ away on a ‘massive’ scale and falling to lower elevations. It is also a type of erosion and can make changes to the side of the mountain.

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A force of 34.5 newtons is applied to an object at an angle of 45 º with the horizontal. What is the force acting on the object in the horizontal direction?

Answers

The horizontal component of the force is 34.5 cos(45) = 24.4 newtons

The vertical component of the force is 34.5 sin(45) = 24.4 newtons

Answer:

24.4

Explanation:

When you hit a nail into a board using a hammer the head of the nail gets warm. In terms of kinetic and thermal energy describe why you think this happens?

Answers

The kinetic energy is transferred to thermal energy through friction

When we hit the nail, the particulate objects are set into kinetic energy and they begin to move. However, due to the rigid structure, the particulate objects cannot move from their equilibrium position, and therefore, it is dissipated through heat.

Heat of Fusion How much thermal energy is needed change 50.0 g of ice at -20.0°C to water at 10.0°C?

Answers

We should divide the problem into 3 separate processes.

1) Bring the temperature of the ice from $-20.0^(\circ)C$ to its melting point ( $0^(\circ)C$ ): the amount of heat needed in this process is
$Q_1=mC_s \Delta T$
where
$m=50.0 g$ is the mass of the ice
$C_s = 2.108 J/g^(\circ)C$ is the specific heat capacity of ice
$\Delta T=0^(\circ)C-(-20^(\circ)C)=20^(\circ)C$ is the increase of temperature

Plugging numbers into the equation, we find
$Q_1 = (50.0 g)(2.108 J/g^(\circ)C)(20^(\circ)C)=2108 J$

2) Fusion of ice
When the ice is at melting point, we need to add a certain amount of heat in order to melt it, and this amount of it is given by:
$Q_2 = mL_f$
where
$m=50.0 g$ is the mass of ice
$L_f = 334 J/g$ is the latent heat of fusion of ice

Plugging numbers into the equation, we find
$Q_2 = mL_f = (50.0g)(334 J/g)=16700 J$
During this phase transition, the temperature of the ice/water does not change.

3) Bring the temperature of the water from $0^(\circ)C$ to $10^(\circ)C$

The amount of heat needed for this process is
$Q_3 = mC_s \Delta T$
where
$m=50.0 g$ is the mass of water
$C_s = 4.187 J/g^(\circ)C$ is the specific heat capacity of water
$\Delta T=10^(\circ)C-0^(\circ)C=10^(\circ)C$ is the increase of temperature

Plugging numbers into the equation, we find
$Q_3 = (50.0 g)(4.187 J/g^(\circ)C)(10.0^(\circ)C)=2094 J$

--> therefore, the total energy needed for the whole process is:
$Q=Q_1+Q_2+Q_3=2108 J+16700 J+2094 J=20902 J=20.9 kJ$

Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight down at a mass rate of m/?t = 0.030 kg/s and an initial velocity of v0 = -15 m/s and strikes the roof perpendicularly. Suppose that the hail bounces off the roof of the car with a velocity of +15 m/s. Ignoring the weight of the hailstones, calculated the force exerted by the hail on the roof. __________N

Answers

Answer:

0.9 N

Explanation:

The force exerted on an object is related to its change in momentum by:

$F=(\Delta p)/(\Delta t)$

where

F is the force exerted

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

The change in momentum can be rewritten as

$\Delta p = m(v-u)$

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

$F=(m(v-u))/(\Delta t)$

In this problem we have:

$(m)/(\Delta t)=0.030 kg/s$ is the mass rate

$u=-15 m/s$ is the initial velocity

$v=+15 m/s$ is the final velocity

Therefore, the force exerted by the hail on the roof is:

$F=(0.030)(+15-(-15))=0.9 N$

If a big person and a small person run up the stairs at the same time who has more force

Answers

Answer:

a big person

Explanation:

because the big person exerts more force over the same distance

Answer:

the big person exerts the largest force on the stairs because he weighs more

Explanation:

this question answer is ☝️☝️

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at rest, how fast is it moving after it travels 3 meters?

Answers

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µn

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

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