How do you think the switch controls the flow of current to the light?

WINSTONCH [101]

Humans need a digestive system to do cellular respiration, which produces energy used for growth and movement.

Plants do not need one as they make their own energy through photosynthesis. The process produces glucose, which is a form of energy for the plants.

4 0

3 years ago

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use the instantaneous center of zero velocity to determine the angular velocity !ab of link ab and the velocity vb of collar b f

allsm [11]

Instantaneous center:

It is the center about a body moves in planer motion. The velocity of Instantaneous center is zero and Instantaneous center can be lie out side or inside the body. About this center every particle of a body rotates.

From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

Therefore, when the mass is at its equilibrium position (which corresponds to x=0), the velocity of the mass will be maximum.

To know more about velocity, refer: brainly.com/question/12413963

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7 0

1 year ago

An engine pulls a train of 20 freight cars, each having a mass of 4.900 × 104 kg, with a constant force. The cars move from rest

bixtya [17]

Answer:

The force with which the tenth car pulls the eleventh one is called tension and is equal to:

T=119715.91 N

Explanation:

The force (F) with which the tenth car pulls the eleventh one is called tension and its direction is the X-direction or horizontal. According to Newton's Second Law of motion:

$\sum F=ma$

That is, the force of the car is equal to the acceleration (a) times its mass (m). The acceleration is the change in the velocity divided by the time (i is for initial and f is for final).

$a=\frac{v_f-v_i}{t_f-t_f}$

Using Newton's second law:

To find the forces, you have to solve the equilibrium in X-direction:

$\sum F_x=T=ma_x$

Now you can substitute the accelertion in terms of velocity and time:

$\sumF_x=T=ma_x=m\frac{v_f-v_i}{t_f-t_i}$

Solve the equation using the data from the problem, remember that the mass of the object is 10 times the mass of one car because the 10th car has to pull all the other cars:

$T=m\frac{v_f-v_i}{t_f-t_i}=(10)*(4.900 \times 10^4 kg)(\frac{4.3 m/s}{17.60s})\\T=119715.91 N$

4 0

3 years ago

You are at the controls of a particle accelerator, sending a beam of 3.60 x10^7 m/s protons (mass m) at a gas target of an unkno

matrenka [14]

Answer:

a) mass of unknown nucleus = 0.04245 mp, where mp is the proton mass

b) Speed of the unknown nucleus = (7.067 x 10^7) m/s

Explanation:

Considering the initial conditions, the observed collisions are ellastic, i.e, the total kinetic energy are conserved. The proton's mass will refer as $m_{p}$ .

(a)

Total kinetic energy conservation

$\frac{1}{2}m_{p}v_{p_0}^{2}+\frac{1}{2}m_{u}v_{u_o}^{2}=\frac{1}{2}m_{p}v_{p_f}^{2}+\frac{1}{2}m_{u}v_{u_f}^{2}$

where $v_{u_o}$ represents the initial velocity of the unknown element, $m_{u}$ the mass of the unknown element, and $v_{u_f}$ the final velocity of the unknown element

Linear momentum conservation

$m_{p}v_{p_0}+m_{u}v_{u_o}=m_{p}v_{p_f}+m_{u}v_{u_f}$

Using the initial speed of the target nucleus (unknown) is negligible, i.e, its speed is zero. Thereby, using the relation of linear momentum conservation given above, it is possible to find an expression of the final speed of the unknown nucleus in terms of its mass, which can be inserted in the relation of the kinetic energy conservation to obtain the value of the mass of the unknown elements, as follows;

$m_{u}v_{u_f}=m_{p}v_{p_0}-m_{p}v_{p_f}\\\\v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}$

Substituting this expression in the relation of total kinetic energy conservation,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}v^{2}_{u}_{f}$

Then,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}\frac{m^{2}_{p}(v_{p_0}-v_{p_f})^{2}}{m^{2}_{u}}\\\\m_{u}= \frac{m_{p}(v_{p_0}-v_{p_f})^{2}}{(v^{2}_{p_0}-v^{2}_{p_f})}$

Replacing the given data

$m_{u}= \frac{m_{p}(3.6x10^{7}-3.3x10^{7})^{2}}{((3.6x10^{7})^{2}-(3.3x10^{7})^{2})}$

Then,

$m_{u}=0.04245m_{p}$

(b) Using the relation of the final speed from linear momentum conservation and the above result, the speed of the unknown nucleus is calculated

$v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}\\\\v_{u_f}=\frac{m_{p}(3.6x10^{7}-3.3x10^{7})}{0.04245m_{p}}\\\\v_{u_f}=7.067x10^{7} m/s$

5 0

3 years ago

A tennis ball is hit into the air with a racket. When is the ball’s kinetic energy the greatest? Ignore air resistance.

Lady bird [3.3K]

Answer B is the correct answer

We know that kinetic energy $E = \frac{1}{2} mv^2$ , where m is the mass of object and v is the velocity of object.

In this case only velocity is the variable, mass remains constant.

So point having higher velocity has higher kinetic energy.

When it leaves the racket, the ball will be having a certain height, but just before it reaches the ground it will not having any height. So maximum velocity of ball is at that time when it reaches just above the ground.

So option B is the correct answer.

4 0

3 years ago