A block of cheese is pulled on by a string and slides to the right along a rough surface.

When the net force of opposite forces is zero , the forces are

lorasvet [3.4K]

The answer is balanced

4 0

3 years ago

Which of the following is constant during uniform circular motion?

tatuchka [14]

An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction. So I'm thinking velocity

8 0

3 years ago

Man made objects that spread an impulse over a large amount of time

lyudmila [28]

Answer:

Examples of man-made objects that spread an impulse over a large amount of time are "airbags" in vehicles and "arrestor beds" (for trucks).

Explanation:

The question above is highly related to the topic about "Impulse" in Physics.

"Impulse" refers to an object's change in momentum (the amount of motion in an object) when a force acts upon it for an interval time. When it comes to providing safety to people when it comes to vehicular crashes, impulse plays a vital role.

Let's take the example of airbags in vehicles. Once a vehicle collides with another object, the driver is carried by a forward motion. Without airbags, the time is normally shorter for the driver to be stopped by the windshield. This results to a greater force. However, with the presence of air-bags, the driver will hit the airbag, instead of the windshield. This will lengthen the time of the impact, thus reducing the force.

Another example are the arrestor beds for trucks. Arrestor beds have been designed in order for trucks to stop, since it's hard to maneuver them. With the help of arrestor beds, trucks are able to come to a stop with a longer time interval, but decreased force.

5 0

3 years ago

At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen

Misha Larkins [42]

Answer:

The specific mechanical energy of the air in the specific location is 40.5 J/kg.
The power generation potential of the wind turbine at such place is of 2290 kW
The actual electric power generation is 687 kW

Explanation:

The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: $\frac{1}{2} V^2$ where V is the velocity of the air.
The specific kinetic energy would be: $\frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}$ .
The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: $\frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}$
Then the mass flow is obtain from the volumetric flow times the density of the air: $m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}$
Then, the Power generation potential is: $40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW$
The actual electric power generation is calculated using the definition of efficiency: $\eta=\frac{E_P}{E_I}}$ , where η is the efficiency, $E_P$ is the energy actually produced and, $E_I$ is the energy input. Then solving for the energy produced: $E_P=\eta*E_I=0.30*2290kW=687kW$

6 0

3 years ago

The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o

Anuta_ua [19.1K]

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

$E_n=\frac{-13.6eV}{n^2}$

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

$E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV$

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

$E_{k}=11.1eV-10.2eV=0.9eV$

6 0

3 years ago