All categories

3 years ago

Why doesn't cloud formation take place until the dew point temperature is reached

1 answer:

6 0

<span>Clouds form when air reaches its dew point, the temperature when the air is saturated. This can happen in two ways. First, the air temperature can stay the same while the humidity increases. This is common in locations that are warm and humid.</span>

You might be interested in

What are the tree types of thermal transfer

avanturin [10]

Convection, conduction, and radiation.

4 0

3 years ago

Read 2 more answers

Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing ind

olya-2409 [2.1K]

Answer: 43 degrees

Explanation:

The force of the crutch can be broken into components. The horizontal component of F is the static friction force keeping the crutch from sliding. The vertical component is opposing the weight and is the Normal force. Using the orientation of the angle q, we have the following

fs = Fx = F sin (angle (tita))

N = Fy = F cos (angle(tita))

Maximum angle implies maximum static friction

Therefore,

fsmax = UsN = Us x F cos(angle tita)

Where U = miyu

F sin(angle tita) = Us x Fcos (angle tita)

Sin (angle tita) / cos (angle tita) = Us

Therefore, tan (angle tita) = Us

Angle tita = tan^-1(Us) = tan^-1 (0.931) = 42.95 degrees = 43 degrees

There for Angle tita Max = 43 degrees

7 0

3 years ago

Which process best describes part of a scientific investigation?

lara31 [8.8K]

the answer is c because it is the 3rd step in the scientific analysis steps

5 0

3 years ago

Read 2 more answers

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

7 0

3 years ago

What is the net force acting on the buggy. ?The net force is pointing to the ?

Papessa [141]

Answer: 390, right

explanation: The net force is just the sum of all of these forces acting on an object. ... This equation is the sum of n forces acting on an object. The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula.

6 0

3 years ago