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oksano4ka [1.4K]
1 year ago
11

How high would a skater need to start on a previous incline to make it up and around a loop that is 6.1meters high?

Physics
1 answer:
OlgaM077 [116]1 year ago
6 0

The skater need to start 7.525 meter high.

Let,

The skater need to start on a previous incline at height = h.

Mass of the skater = m

Given, height of the loop = 6.1 meter.

Radius of the loop; r = 6.1/2 meter = 3.05 meter.

Let to make it up and around a loop that is 6.1meters high, the man required minimum velocity v on the highest point of the loop.

So, centripetal force at highest point = weight of the man

⇒ mv²/r = mg

⇒ v = √(gr)

Then, potential energy of the skater at height h is = mgh.

And, minimum energy  of the skater at the highest point of the loop is = potential energy + minimum kinetic energy.

= mg(2r) + 1/2 mv²

= 2mgr + 1/2 mgr

= 5/2 mgr

According to conservation of energy,

potential energy of the skater at height h = minimum energy  of the skater at the highest point of the loop

⇒ mgh = 5/2 mgr

⇒ h = 5/2 r =( 5/2 )× 3.05 meter = 7.525 meter.

Hence, required height is 7.525 meter.

Learn more about energy here:

brainly.com/question/1932868

#SPJ1

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Answer:

The unit you should use for work done and energy is the joule (J) which is indeed the same as the newton metre (N m).

There is another physical quantity which is the product of force and distance and that is torque or moment of a force.

The unit you should use for torque is the newton metre (Nm) and not the joule.

Naming the units of work done and torque differently helps to emphasis the fact that work done and torque refer to two different physical quantities although the definitions of both quantities have the product of force and distance in them.

work done=force→⋅displacement→ and torque→=force→×displacement→

Hope I helped

4 0
3 years ago
6, P 14 are consecutive terms in an AP<br>find the value of P.​
xenn [34]

In an arithmetic progression, consecutive terms differ by the same value.

So, we have

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The equation solves to

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And in fact, if you start with

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6 0
3 years ago
A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the
REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0
3 years ago
A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev
Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

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v= 2πrf   where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

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Answer:

Range, R = MV²/2QE

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The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

5 0
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