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3 years ago

This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.

Physics

2 answers:

ss7ja [257]3 years ago

8 0

Answer:

C-Changing the S to N

Explanation:

As we know that magnetic field lines originated from North pole of magnet and then terminate into the south pole of magnet.

Now if we put similar poles of magnet in front of each other then we can say that the magnetic field lines of two poles will repel each other and due to this two similar poles have repulsion force.

Similarly if two different poles of magnet will place in front of each other then the magnetic field lines will originate from north pole and terminate at south pole.

This will show attraction force between the poles of magnet.

Now we can see the figure that magnetic field lines are originating from both poles here as well as these lines are repelling each other.

So here both poles must have North pole

So correction here must be

C)-Changing the S to N

hoa [83]3 years ago

4 0

I believe the answer is c

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A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w

slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

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3 years ago

A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo

Alla [95]

Answer:

$v_{ox}= 19.6\ m/s$

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = $v_{ox}$

Initial vertical velocity= $v_{oy}$ (Since ball was thrown horizontally only)

Acceleration acting horizontally, $a_x$ = 0 m/s² [ Since no acceleration acts horizontally) ]

Vertical Acceleration, $a_y$ = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

$H= v_{oy}t+\frac{1}{2}a_yt^2$

$5= (0)t+\frac{1}{2}(9.8)t^2$

$t= \frac{10}{9.8}=1.02 s$

Then using Equations of motion for horizontal motion,

$R= v_{ox}t+\frac{1}{2}a_xt^2$

$20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2$

$v_{ox}= 19.6\ m/s$

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3 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

bogdanovich [222]

Answer:

magnitude of force on charge 2Q = $\frac{KQ^{2} }{I^{2} }$

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force

║F║ = $\sqrt{f_{x}^{2} + f_{y}^{2} }$ = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = $\frac{KQ^{2} }{I^{2} }$

The direction of the force on charge 2Q is calculated as

tan ∅ = $\frac{f_{y} }{f_{x} }$ = 1.8284

therefore ∅ = $tan^{-1}$ 1.8284

= 61⁰

3 0

3 years ago

A charged particle moves through a magnetic field. In which situation is the magnetic force zero?

maksim [4K]

Answer:

The answer is the option a.

Explanation:

We know that magnetic force (Fm) is defined as

Fm = q (v x B)

Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.

"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is

v x B = vBsin (thetha)

So,

Fm = qvBsin (thetha)

And, in case in which v and B are parallel vectors, thetha is zero, and,

sin (thetha)=sin (0) = 0

So, Fm=0

7 0

3 years ago

What is acceleration produced by a force of 12 newton exerted on an object of mass 3kg

jarptica [38.1K]

Answer:

a=F/m

a=12N/3kg (here newton can be written as kgm/s^2 so kg will be cancelled)

a=4m/s^2

Explanation:

3 0

3 years ago