Which of the following is not a major function of the roots?

What are the conditions under which electricity is conducted

dolphi86 [110]

There must be a conducting wire and electromotive force or free electrons

3 0

3 years ago

How does a gas do work?

Triss [41]

Your answer would be C, "By conserving energy"! A gas can do work through something called the First Law of Thermodynamics. Heat that energy used by the gas during this "work" cannot be created or destroyed however it can be transferred and converted. This energy can only be stored during these processes, not removed. An example of this would be gas in a cylindrical object such as a piston. When the gas is heated, it expands and moves the piston upwards. When it expands, it stores that energy it gains through the heat given.

I hope this helps! (:

8 0

3 years ago

Read 2 more answers

A 2.26 cm tall object is placed in 18.6 cm in front of a convex lens. The focal length

Gnesinka [82]

Answer:

the image is -38.7 cm from the lens

Explanation:

We use the lens equation which says

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

where $u$ is the distance to the object, $v$ is the distance to the image, and $f$ is the focal length.

Now, in our case

$u =18.6cm$ ,

$v= 35.8cm$ ,

$f = 35.8cm$ ;

therefore,

$\frac{1}{18.6}+\frac{1}{v}=\frac{1}{35.8}$

$\frac{1}{v}=\frac{1}{35.8}-\frac{1}{18.6}$

$\boxed{v=-38.7cm}$

where the negative sign indicates that the image is virtual and on the same side of the lens as the object.

7 0

2 years ago

A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He

Taya2010 [7]

a) 73.5 N

b) 551.3 J

c) -551.3 J

d) 0 J

e) 0 J

f) 0 J

g) 0 J

Explanation:

a)

There are two forces acting on the crate:

- The push of the worker, F, in the forward direction

- The frictional force, $F_f=\mu mg$ , in the backward direction, where

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:

$F-F_f=ma$

However, the crate here is moving with constant velocity, so its acceleration is zero:

$a=0$

So the previous equation becomes:

$F-F_f=0$

And we can find the magnitude of the applied force:

$F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N$

b)

The work done by the applied force on the crate is

$W_F=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here we have:

F = 73.5 N

d = 7.5 m

$\theta=0^{\circ}$ (the force is applied in the same direction as the displacement)

Therefore,

$W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J$

c)

The work done by friction on the crate is:

$W_{F_f}=F_f d cos \theta$

where in this case:

$F_f=73.5 N$ is the magnitude of the force of friction

d = 7.5 m is the displacement of the crate

$\theta=180^{\circ}$ , because the displacement is forward and the force of friction is backward, so they are in opposite direction

Therefore, the work done by the force of friction is:

$W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J$

d)

To find the normal force, we analyze the situation of the force along the vertical direction.

We have two forces on the vertical direction:

- The normal force, N, upward

- The force of gravity, $mg$ , downward, where

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:

$N-mg=0\\N=mg=(50)(9.8)=490 N$

The work done by the normal force is:

$W_N=Nd cos \theta$

In this case, $\theta=90^{\circ}$ , since the normal force is perpendicular to the displacement of the crate; therefore, the work done is

$W_N=(490)(7.5)(cos 90^{\circ})=0$

e)

The work done by the gravitational force is:

$W_g=F_g d cos \theta$

where:

$F_g=mg=(50)(9.8)=490 N$ is the gravitational force

d = 7.5 m is the displacement of the crate

$\theta=90^{\circ}$ is the angle between the direction of the gravitational force (downward) and the displacement (forward)

Therefore, the work done by gravity is

$W_g=(490)(7.5)(cos 90^{\circ})=0 J$

f)

The total work done on the crate can be calculated by adding the work done by each force:

$W=W_F+W_{F_f}+W_N+W_g$

where we have:

$W_F=+551.3 J$ is the work done by the applied force

$W_{F_f}=-551.3 J$ is the work done by the frictional force

$W_N=0$ is the work done by the normal force

$W_g=0$ is the work done by the force of gravity

Substituting,

$W=+551.3+(-551.3)+0+0=0 J$

So, the total work is 0 J.

g)

According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:

$W=\Delta E_K$

where

W is the work done on the crate

$\Delta E_K$ is the change in kinetic energy of the crate

In this problem, we have:

$W=0$ (total work done on the crate is zero)

Therefore, the change in kinetic energy of the crate is:

$\Delta E_K = W = 0$

5 0

3 years ago

How many years does one month symbolize on the cosmic calendar

Maslowich

It is about two years on the cosmiccalender

7 0

2 years ago