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vlada-n [284]
3 years ago
14

Strength of electromagnets

Physics
1 answer:
Triss [41]3 years ago
6 0

Lines of Force around an Electromagnet. ... The magnetic field strength of an electromagnet is therefore determined by the ampere turns of the coil with the more turns of wire in the coil the greater will be the strength of the magnetic field.

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Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t
elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play .  But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t  + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only  horizontal velocity initially , Vy = 0

h = 1/2 gt²

t = \sqrt{\frac{2h}{g} }

Horizontal distance covered  = t x Vx

= \sqrt{\frac{2h}{g} } \times  V_x

From this expression also

Horizontal distance covered is proportional to \sqrt{h} .

7 0
3 years ago
What is the scientific name given to potato​
Burka [1]
The answer: Solanum tuberosum
8 0
3 years ago
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In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas
Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

3 0
3 years ago
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what is the answer for the path that one body in space takes as it revolves around another body is called a .......... a orbit,a
salantis [7]
The answer is orbit, we are orbiting the sun as the moon orbits us
6 0
2 years ago
a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur
Doss [256]
Refer to the diagram shown below.

Still-water speed  = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°

Answer:  10.2 m/s at 21.5° downstream.

7 0
3 years ago
Read 2 more answers
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