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4 years ago

Magnetic field lines exit out of the . Magnetic field lines enter into the . Magnetic field lines travel around a bar magnet in

Physics

2 answers:

Inessa05 [86]4 years ago

8 0

Answer:

Magnetic field lines exit out of the

✔ north pole

Magnetic field lines enter into the

✔ south pole

Magnetic field lines travel around a bar magnet in

✔ a closed loop

Explanation:

Airida [17]4 years ago

3 0

Answer:

Magnetic field lines exit out of the North pole . Magnetic field lines enter into the South pole. Magnetic field lines travel around a bar magnet in closed loops.

Explanation:

Magnetic field lines shows the direction of a magnetic force and how it acts, it gives the direction of the magnetic field at that point in time.

For a bar magnetic, the magnetic field lines runs from the north pole to the south pole, i.e. it exits the north pole and enters into the south pole. This magnetic field lines also go through the magnet forming closed loops without ends.

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After 20 seconds, a 200-kg object increases its velocity from 15 m/s to 40 m/s. Determine the impulse applied to the object. (Al

sleet_krkn [62]

Answer:

Imp_{1-2}=5000[kg*m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of momentum, which is defined as the product of mass by Velocity.

It must be defined that the impulse after the force is applied is equal to the momentum before the impulse applied on the body.

ΣPbefore = ΣPafter

P = momentum = m*v [kg*m/s]

In this way, we will construct the following equation.

$(m_{1}*v_{1})+ Imp_{1-2}=(m_{1}*v_{2})$

where:

m₁ = mass of the object = 200 [kg]

v₁ = velocity of the object before the impulse = 15 [m/s]

v₂ = velocity of the object after the impulse = 40 [m/s]

Now replacing:

$(200*15) + Imp_{1-2} = (200*40)\\Imp_{1-2}=5000[kg*m/s]$

7 0

3 years ago

A horizontal force of 200 n is applied to move a 55-kg cart (initially at rest) across a 10-m level surface. what is the final k

Vesnalui [34]

<span>Kinetic energy is the energy that is possessed by an object that is moving. It is calculated by one-half the product of the mass and the square of the velocity of the object. We need to determine the velocity of the cart that is moving.
</span>
First, we use Newton's Second Law of motion;
<span>Force = ma
200 = 55a
a = 3.64 m/s^2

Then, from the kinematic equation we calculate the velocity;
v^2 = v0^2 + 2ax

where v is the final velocity, v0 is the initial veocity (zero since it initially start at zero), a is the acceleration ( 3.64 m/s^2) and x is the distance traveled.

v^2 = 0^2 + 2 (3.64) (10)
v^2 = 72.73 m^2 / s^2
v = 8.53 m / s

KE = mv^2 / 2
KE = 1/2 (55) (8.53^2)
KE= 2000 J</span>

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3 years ago

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r