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lutik1710 [3]
3 years ago
8

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

tor. Ignoring the particle's weight, what type of path will the particle follow
Physics
1 answer:
Ugo [173]3 years ago
3 0

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

F = V\times B

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

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A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
3 years ago
What weighs more a mole of feathers or bricks?
harkovskaia [24]
Avogadro's mole is a number, like 6.023 *10^23

your question is the equivalent of asking if 100 feathers or 100 bricks weigh more

the mole of bricks would weigh more
 
6 0
3 years ago
A child dangles a 1.50-kilogram stuffed toy 0.500 meters from the ground. What is the potential energy of the toy?
inn [45]
Since we know that
Gravitational potential energy = mass × height ×gravity

then

GPE = 1.5 kg x 0.500 m x 9.8m/s^2

therefore

GPE = 7.35 J
6 0
3 years ago
Read 2 more answers
PHYSICS, HELP PLZ??!! 100PTS
romanna [79]

Hi there!

We can begin by calculating the time the ball takes to reach the highest point of its trajectory, which can be found using the following:

t_{max} = \frac{vsin\theta}{g}

Where:

tmax = (? sec)

vsinθ = vertical comp. of velocity = 10sin(48) = 7.43 m/s)

g = acceleration due to gravity (9.8 m/s²)

We can solve for this time:

t_{max} = \frac{7.43}{9.8} = 0.758 s

When the ball is at the TOP of its trajectory, its VERTICAL velocity is equivalent to 0 m/s. Thus, we can consider this a free-fall situation.

We must begin by solving for the maximum height reached by the ball using the equation:

d = y_0 + v_{0y}t + \frac{1}{2}at^2

d = displacement (m)

vi = initial velocity (7.43 m/s)

a = acceleration due to gravity

d = displacement (m)

y0 = initial VERTICAL displacement (28m)

Plug in the values:

d = 28 + 7.43(0.758) + \frac{1}{2}(-9.8)(0.758^2) = 30.817 m

Now, we can use the rearranged kinematic equation:

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(30.817)}{9.8}} = 2.51 s

Add the two times together:

0.758 + 2.51 = \boxed{3.266 s}

7 0
2 years ago
___________________ is the change over time in populations of related organisms.
Levart [38]

YOU ARE BOBO KA TANGA KA TANGINAMO

8 0
2 years ago
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