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garik1379 [7]
3 years ago
14

. Calculate the magnetic force on a current carrying conductor.

Physics
1 answer:
Phantasy [73]3 years ago
6 0

Answer:

gggggggggggggggggggggggggggggggggg

Explanation:

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A block with a mass of 6.0 kg is
Umnica [9.8K]

(a) The normal force on the block is 50.92 N.

(b) The horizontal force on block keeping it in equilibrium is 29.4 N.

The given parameters;

  • <em>mass of the block, m = 6 kg</em>
  • <em>angle of inclination, θ = 30.0°</em>

The normal force on the block is calculated as follows;

F_n = W \times cos(\theta)

where;

  • <em>W is the weight of the block</em>

F_n = mg \times cos(\theta)\\\\F_n = 6 \times 9.8 \times cos(30)\\\\F_n = 50 .92 \ N

The horizontal force on block keeping it in equilibrium is calculated as follows;

F- F_x = 0\\\\F-mgsin\theta= 0\\\\F = mgsin\theta\\\\F = 6 \times 9.8 \times sin(30)\\\\F = 29.4 \ N

Learn more here:brainly.com/question/24860178

5 0
2 years ago
Water vapor condensing to liquid water is what type of process?
Natalka [10]

Answer:

Endothermic reaction

Explanation:

The heat from the object is being released or taken from an outer object causing it to turn back into a water also known as precipitation.

8 0
4 years ago
A lens is used to make an image of magnification -1.50. The image is 30.0 cm from the lens. What is the object’s distance (unit=
photoshop1234 [79]

Answer:

The object distance is 20cm

Explanation:

Given

Magnification = -1.5

Image distance = 30 cm.

Required

Object Distance

We can calculate the object's distance using magnification formula;

M = -V/U

Where M = Magnification = -1.50

V = Image Distance = 30cm

U = Object Distance

Substitute the above parameters in the formula above.

-1.50 = -30/U

Multiply both sides by -1

-1.50 * -1 = -30/U * -1

1.50 = 30/U

Multiply both sides by U

1.50 * U = 30/U * U

1.50U = 30

Divide through by 1.50

1.50U/1.50 = 30/1.50

U = 30/1.50

U = 20cm

Recall that U represented the object distance.

Hence, the object distance is 20cm

5 0
4 years ago
9 The diagram shows a uniform beam PQ. The length of the beam is 3.0 m and its weight is 50 N. The beam is supported on a pivot
tankabanditka [31]

equilibrium = 1/1

50 N/x = 1/1

x = 1/1 × 50 N

x = 50 N (B)

#LearnWithEXO

6 0
3 years ago
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su
aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0
3 years ago
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