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Kazeer [188]
3 years ago
10

Hi ) how to do this??

Physics
1 answer:
pav-90 [236]3 years ago
4 0

Answer:

X = 12N

Explanation:

Anticlockwise moment = Clockwise moment

Taking the moment at the pivot,

Anticlockwise moment = X×2.0m

Clockwise moment = (2.0N × 4.0m) + (4.0N × 4.0m)

Hence, 2X = (8 + 16)

2X = 24

X = 12N.

I HOPE IT'S OK.

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Answer:

but not to

D). float in the air

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3 years ago
Why are the loops on a roller coaster tear-drop shaped instead of circular? It would be simpler to design & build tracks wit
VLD [36.1K]
The centripetal force acting on the rider is much greater if the roller coaster has a circular loop, rather than oval. This is because the change in direction is much sharper throughout the loop, causing the rider to experience a much more intense G-Force throughout the loop. 
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5 0
3 years ago
A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its
lana [24]

At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.

We have a circular loop of radius ' r ' carrying current ' i '.

We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.

<h3>What is the formula to calculate the Magnetic field intensity due to a current carrying circular loop at a point on its axis?</h3>

The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

B(0) = \frac{\mu_{o} i}{2R}

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -

\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}

2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }

4R^{6} = (r^{2} +R^{2} )^{3}

r^{2} =0.587R^{2}

r = 0.766R

Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.

To solve more questions on magnetic field intensity, visit the link below-

brainly.com/question/15553675

#SPJ4

6 0
2 years ago
a loaded sack of total mass is 1000 gramme falls down from the floor of a lorry 200cm high, calculate the workdone by the gravit
Helen [10]

Answer:

W = 20 J

Explanation:

Given that,

The mass of a loaded sack, m = 1000 g = 1 kg

It falls down from the floor of a lorry 200 cm high, h = 2 m

We need to find the work done by the gravity. The work done by an object under the action of gravity is given by :

W = mgh

Substitute all the values,

W = 1 × 10 × 2

= 20 J

Hence, the required work done by gravity is equal to 20 J.

8 0
3 years ago
NET FORCE ( PLEASE ANSWER ASAP )
slava [35]

First picture (black background):  50 Newtons UP

Second picture (white background):  30 Newtons RIGHT

3 0
4 years ago
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