If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:

An elevator of massmis initially at rest onthe first floor of a building. It moves upward,and passes the second and third floors

Varvara68 [4.7K]

To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

$W = mgh$

Where,

m = mass

h = height

g = Gravitational constant

When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

$h = 3h$

Replacing in our equation we have to

$W = mgh\\W = mg(3h)\\W = 3mgh$

The correct answer is 4.

7 0

3 years ago

Resistance of a light bulb with 0.33 A of current flowing from a 12V battery?

Sergio039 [100]

Resistance = (voltage) / (current)

Resistance = (12v) / (0.33 A)

Resistance = (12/0.33) ohms

<em>Resistance = 36.4 ohms</em>

8 0

3 years ago

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$