All categories

3 years ago

Which skill involves creating representations of complex objects or processes?

1 answer:

7 0

Making models is a skill that involves creating a representation of complex objects or processes. Through the years, people have been creating models in order to make a clearer visualization and give reality to a certain object. Some example of model making are toys like plastic cars as a model representation of a real car, a plastic plane, model representation of a real plane and a doll, a model representation of us, human. One of the most famous model representation as of today is the film or movie making. This represents the lives of the people and show stories about how people face every challenge in life.
<span>
</span>

You might be interested in

How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^

arlik [135]

Answer:

How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^2)?

Explanation:

<em> Mass multiplied by acceleration gives you the amount of force needed for it.</em>

3 0

2 years ago

Read 2 more answers

g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k

Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

$(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s$

Now, we will apply the law of conservation of momentum:

$m_1v_1 = m_2v_2$

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

$(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s$

Now, we again use the third equation of motion for the upward motion of the ball:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

$(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\$

6 0

2 years ago

During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m

Nezavi [6.7K]

Answer:

<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

Explanation:

Given :

Current $I = 2.5 \times 10^{4}$ A

Time $t = 40 \times 10^{-6}$ sec

We know that the current is the rate of flow of charge.

From the formula of current,

Where $Q =$ charge transfer between cloud and earth.

$Q =I t$

$Q = 2.5 \times 10^{4} \times 40 \times 10^{-6}$

$Q = 1$ C

Hence, the charge transferred from the cloud to earth is 1 Coulomb.

3 0

3 years ago

Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

jekas [21]

Answer:

$a=\frac{mBg-mAgSin\theta}{mA+mB}$

Explanation:

Given two mass on an incline code $mA$ and $mB$ and an angle of inclination $\theta$ . $g$ . Assume that $mA$ is the weight being pulled up and $mB$ the hanging weight.