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3 years ago

Which forces are capable of affecting particles or objects from large distance

Physics

1 answer:

Ganezh [65]3 years ago

8 0

Answer:

only long-range force that affects all particles is the gravitational force.

Explanation:

In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.

The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.

Consequently the only long-range force that affects all particles is the gravitational force.

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In a particular region, there is a uniform magnetic field with a magnitude of 2.00 T. You take a particle with a charge of +6.00

expeople1 [14]

Answer:

a) 1.68 N b) 0 c) 14.5º

Explanation:

The force on a charge moving in a magnetic field, is a vector perpendicular to the plane defined by the velocity of the charge and the magnetic field.
The magnitude of the force F is given by the following expression:

$F = q*v*B*sin\theta (1)$

where q= magnitude of the charge of the particle, v=velocity of the particle, B= magnitude of the magnetic field, and θ= angle between v and B.
Replacing by the known values in (1) we have:

$F = q*v*B*sin\theta (1) = 6.00e-6C*1.40e5m/s*2T*sin\theta=\\ \\ F= 1.68N*sin\theta$

The maximum possible value of F happens when sin θ =1
This means that v and B are perpendicular each other (in the same plane)
So, Fmax = 1.68 N

If the maximum possible value of F happens when sin θ = 1, the minimum possible is when sin θ = 0.
In this case, when v and B are parallel each other, the force is just 0.

If the force is 0.25 of the maximum possible, (which is when sin θ =1), this means that sin θ = 0.25, as it can be seen below:

$F_{\theta} = q*v*B*sin \theta = 0.25*Fmax\\ F_{\theta} = q*v*B*sin\theta= 0.25 (q*v*B*sin 90) \\ sin \theta = 0.25$

$\theta = sin^{-1} (0.25) =14.5 deg$

The angle between the velocity v and the magnetic field B is 14.5º.

3 0

4 years ago

What clues should you look for to determine if an authors purpose is to make you think deeply

8090 [49]

If the context is difficult to understand they want you to think hard to get what their trying to say and make you feel a certain way.

7 0

3 years ago

Which change is the best example of a physical change?

murzikaleks [220]

The answer would be a nail rusting, but really it could be any of these.

3 0

4 years ago

When current first leaves the battery, how much voltage does it have

rjkz [21]

Answer:

depends on the voltage of battery

Explanation:

for example if you connect a battery of 6V,6V will be provided

8 0

3 years ago

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$

Solving for initial velocity (when the boulder left the volcano):

$v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}$

$v_i=101.44 \frac{m}{s^{2}}$

3 0

3 years ago