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4 years ago

One part of a freely swinging magnet always points

Physics

2 answers:

Sladkaya [172]4 years ago

8 0

One part of a freely swinging magnet always points to the Earth's magnetic pole in the Northern Hemisphere.

Readme [11.4K]4 years ago

3 0

Answer: The correct answer is (d).

Explanation:

Magnet has two poles: South pole and North pole. The same poles of the magnets repel each other. The opposite poles of the magnets attract each other.

In the absence of other magnet, a freely swinging magnet points in North-South direction. The earth has a magnetic field. The magnetic field of the bar magnet is same as the magnetic field of the earth's magnet.

The position of the earth's magnetic poles are not fixed. The south pole of the earth's magnet lies in the geographic north as it attracts the north pole of the freely swinging magnet.

If one part of a freely suspended magnet always points to Earth's magnetic pole in the Northern Hemisphere.

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Please answer correctly<br>will give the brainliest<br>Urgent

EastWind [94]

Answer:

f (frequency) = V / y where V is the speed of sound and y the wavelength

f = 1500 m/s / 1.5 m = 1000 / sec

T (period) = 1 / f = .001 sec

Suppose you replace the horn by a drum then the period would be the time between the beats of the drum - now if the source is moving towards the observer then the distance between crests of the wave produced by the drum will be shortened by V * T because of the motion of the drum "towards" the observer, and since the wavelength is shorter the frequency heard by the observer will be higher, and the higher the speed of of the car the shorter the wavelength as seen by the observer and the higher the frequency.

Also, if the car is moving away from the observer then the distance between the crests of the wave emitted will be further apart, and the observer will hear a lower frequency.

3 0

3 years ago

Do you think that solids can undergo convection

Alex_Xolod [135]

<span>Convection means the movement happens inside a fluid by the cause of too much heat and lesser density to rise, and also the colder cause denser materials to sink.
Thus, Solid cannot undergo convection because the molecules inside the solid are closer and can hardly move unlike liquid.
So, when the temperature got hotter or colder, the molecules on solid will only vibrate and not convert.
Therefore, this concludes that the solid do not undergo convection rather conduction of heat.</span>

7 0

4 years ago

A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the s

egoroff_w [7]

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt = 10.3 KN

Therefore, Net force of the driver is 10.3 KN

No sit belt,

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

4 0

3 years ago

An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t

docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

$y_1(t)=h+u_1 t + \frac{1}{2}gt^2$

where:

h = 50 m is the initial height

$u_1=0$ is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

$g=-9.8 m/s^2$ is the acceleration of gravity

So, its vertical position can be rewritten as

$y_1(t)=50-4.9t^2$

The position of object 2 instead can be written as

$y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2$

where

$u_2 sin \theta$ is the initial vertical velocity, where

$u_2 = 50 m/s$ is the initial velocity

$\theta=30^{\circ}$ is the angle of projection

Substituting, we get:

$y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2$

The two objects collide when their vertical position is the same, so:

$y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2$

And solving for t, we find:

$50=25t\\t= 2 s$

Note that this means that the two object at t = 2 s have the same vertical position: however, this is not true for the horizontal position.

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

$y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m$