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Nikolay [14]
2 years ago
13

Which of the following is described here? "Point your skis straight down the fall line with your skis parallel and about a foot

apart. Lean forward and bend at the knees, ankles, and hips."
a. linked turn
b. christie
c. torsion
d. downhill schussing
Physics
2 answers:
MariettaO [177]2 years ago
4 0

Answer:

Downhill schussing

Explanation:

Can I have brainliest please im trying to level up

elena55 [62]2 years ago
3 0

answer d. down hill schussing

Explanation:

I took the quiz

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Can a machine like a lever, produce a greater force than what you put into it? Can it increase the energy that you put into it?
morpeh [17]

Answer:

yes

Explanation:

the force is multiplied by the levers length of the handle

7 0
2 years ago
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A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, its t
Lina20 [59]
E_{total} = KE + PE \\ KE = 500 J \\ PE = 1500 J \\ E_{total} = 500J + 1500J \\ E_{total} = 2000J
4 0
3 years ago
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You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

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So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0
2 years ago
Jak wpłynie na czas oczekiwania na pierwsze ogrzanie wody (pytanie 1) zwiększenie mocy
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Answer:

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Explanation:

10x100=

5 0
3 years ago
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