What is watt equivalent to in terms of kg,m and s

Determine the magnitude and direction of the force on an electron traveling 3.58 E 6 m/s horizontally to the west in a verticall

olya-2409 [2.1K]

Answer:banana

Explanation:

Because the amount of potassium is that

4 0

3 years ago

Which image represents the force on a positively charged particle caused by an approaching magnet?

Amanda [17]

Answer:

Image B represents the force on a positively charged particle caused by an approaching magnet.

Explanation:

The most fundamental law of magnetism is that like shafts repulse each other and dissimilar to posts pull in one another; this can without much of a stretch be seen by endeavoring to put like posts of two magnets together. Further attractive impacts additionally exist. On the off chance that a bar magnet is cut into two pieces, the pieces become singular magnets with inverse shafts. Also, pounding, warming or winding of the magnets can demagnetize them, on the grounds that such dealing with separates the direct game plan of the particles. A last law of magnetism alludes to maintenance; a long bar magnet will hold its magnetism longer than a short bar magnet. The domain theory of magnetism expresses that every single enormous magnet involve littler attractive districts, or domains. The attractive character of domains originates from the nearness of significantly littler units, called dipoles. Iotas are masterminded in such a manner in many materials that the attractive direction of one electron counteracts the direction of another; in any case, ferromagnetic substances, for example, iron are unique. The nuclear cosmetics of these substances is with the end goal that littler gatherings of particles unite as one into zones called domains; in these, all the electrons have the equivalent attractive direction.

4 0

2 years ago

Which of the following is not a way that astronomers can find how much dark matter there is in cluster of galaxies?

Tpy6a [65]

Answer:

b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.

Explanation:

The universe is thought to be made up of 85% dark matters. Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect. This means that option b is wrong since radio wave is an electromagnetic wave. Dark matter is a form of matter that makes up about a quarter of the total mass–energy density of the universe. Dark matter was theorized due a variety of astrophysical observations and gravitational effects that cannot be explained by accepted theories of gravity unless there were more matter in the universe than can be seen.

6 0

3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W= ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7 N

8 0

3 years ago