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Kisachek [45]
3 years ago
15

Two charged particles are projected into a magnetic field that is perpendicular to their initial velocities. If the charges are

deflected in opposite directions, what does this tell you about them? (Ignore the interaction between these two particles.)
Physics
1 answer:
Svetllana [295]3 years ago
7 0

Answer:

The two charged particles are opposite charge (one positive, the other negative)

Explanation:

This can be explained through Fleming's left hand rule. As we know in Fleming's LH rule

Thumb: represent motion of the charge

Fore finger: direction of magnetic field

Centre finger: direction of current

In this case, the direction of magnetic field is in one way (perpendicular to direction of the particles motion)

If the charge is positive, equivalently it's direction is the same as the direction of current. If the charge is negative, it's direction is opposite the direction of current.

Using the fleming left hand rule, we'll see that as we change the direction of current without changing the direction of field, the direction of motion will change towards the opposite way.

For example, let's say the particles is moving away from you and the magnetic field is from left to right:

If the charge is positive (hence direction of current is away from you), the particles will move downward.

If the charge is negative (hence direction of current is toward you), the particles will move upward.

With these evidence, no doubt that the two particles have different charges.

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A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
3 years ago
Refer the attached photo
Fofino [41]

Answer:

A

Explanation:

since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet

8 0
3 years ago
The temperature on the moon is
Nuetrik [128]
B.

It can go from very hot to very cold, it depends on the area of the moon and where the sunlight hits.
4 0
3 years ago
Read 2 more answers
While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the
BartSMP [9]

In the experiment of free fall bob released a bag of mass 1 lb

so here we can say that initial speed of the bag is Zero

time taken by the bag to free fall is given as

t = 1.5 s

also the acceleration of free fall is given as

a = 9.8 m/s^2

now we will use kinematics equation here for finding the distance of free fall

d = v_i * t + \frac{1}{2} at^2

d = 0 + \frac{1}{2}*9.8* 1.5^2

d = 4.9 * 2.25

d = 11.025 m

so the bag will fall down by total distance of 11.025 m from its initial released position.


3 0
3 years ago
You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on
just olya [345]

Answer:

a)   v_{f} = 0.25 m / s  b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

     m₁ = 0.40 kg

     v₁₀ = 9.0 m / s

     m₂ = 14 kg

     v₂₀ = 0

Initial

     po = m₁ v₁₀

Final

     p_{f} = (m₁ + m₂) vf

     po = pf

     m₁ v₁₀ = (m₁ + m₂) v_{f}

      v_{f} = v₁₀ m₁ / (m₁ + m₂)

      v_{f} = 9.0 (0.40 / (0.40 +14)

      v_{f} = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

      u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

     K₀ = ½ m₁ v₁₀² +0

     K₀ = ½ 0.40 9 2

     K₀ = 16.2 J

Final

     K_{f}= ½ (m₁ + m₂) v_{f}2

      K_{f} = ½ (0.4 +14) 0.25 2

    K_{f} = 0.45 J

   

    ΔK = K₀ -  K_{f}

    ΔK = 16.2-0.445

    ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

      v₁₀’= v₁₀ -u

      v₁₀’= 9 -.025

      v₁₀‘= 8.75 m / s

      v₂₀ ‘= v₂₀ -u

      v₂₀‘= - 0.25 m / s

      v_{f} ‘=   v_{f} - u

      v_{f} = 0

Initial

    K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

    Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

    Ko = 15.31 + 0.4375

    K o = 15.75 J

Final

   k_{f} = ½ (m₁ + m₂) vf’²

  k_{f} = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0
3 years ago
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