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myrzilka [38]
3 years ago
15

A 44.6 kg ice skater is moving at 5.14 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole.

She then moves in a circle of radius 0.525 m around the pole. Find the force exerted by the rope on her arms. The acceleration of gravity is 9.8 m/s 2 . Answer in units of kN.
Physics
1 answer:
hodyreva [135]3 years ago
6 0
This question is asking the force on the rope. We can find this by using centripetal acceleration, then multiply by the ice skater's mass.

First we need to find the centripetal acceleration, which is a = v^2/r

We are given v and r, so we solve for a.

a = [(5.14m/s)^2]/(0.525m) = 50.323 m/s^2

Next multiply by the ice skater's mass (because F=ma) which is 44.6 kg

We then get F = (50.323m/s^2)(44.6kg) = 2244.407924N

Divide by 10^3 to get kN, we get:

2.244kN

Hope this helps!
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AP Physics I, shouldn't be too hard.
Nana76 [90]

Answer:

The correct option is;

D. The kinetic energy decreases by 3·m₀·v₀²

Explanation:

The given parameters are;

The mass of object X = m₀

The initial velocity of object X = v₀

The mass of object Y = 2·m₀

The initial velocity of object Y = -2·v₀

By conservation of linear momentum, we have;

The total initial momentum = The total final momentum

Therefore, we have;

The total initial momentum = m₀·v₀ - 2·m₀·2·v₀ = The total final momentum

∴ The total final momentum = -3·m₀·v₀

The total mass of the two object after sticking together = 2·m₀ + m₀ = 3·m₀

Therefore, the velocity of the two objects after collision = (The total final momentum)/(Total mass) = -3·m₀·v₀/(3·m₀) = -v₀

The kinetic energy = 1/2 × Mass × (Velocity)²

Therefore, the kinetic energy after collision = 1/2 × (3·m₀) × v₀² = 3·m₀·v₀²/2

The kinetic energy before collision = 1/2 × m₀ × v₀² + 1/2 × (2·m₀) × (2·v₀)² = (1/2 + 4) × (m₀·v₀²)

∴ The kinetic energy before collision =  9·(m₀·v₀²)/2

The change in kinetic energy = The kinetic energy after collision - The kinetic energy before collision = 3·m₀·v₀²/2 - 9·(m₀·v₀²)/2 = -3·m₀·v₀²

Therefore, the kinetic energy decreases by 3·m₀·v₀².

5 0
3 years ago
PLS HELP WITH STEPS I'LL MARK U BRAINLISTT
Viktor [21]
35 m because if you count the space he took back it’s equivalent to 15 meters and so 20 + 15 = 35
4 0
3 years ago
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In the future, an experimental spacecraft leaves space dock for a test flight. After flying in a very large circle at constant s
Fantom [35]

Answer:

2.58 x 10⁸ m/s

Explanation:

Time dilation fomula will be applicable here, which is given below.

t = \frac{T}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}

Where T is dilated time or time observed by clock in motion , t is stationary time , v is velocity of clock in motion and c is velocity of light .

c is 3 times 10⁸ ms⁻¹ , T is 7.24 h , t is 3.69 h. Put these values in the formula

7.24 = \frac{3.69}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}\\

\frac{v^2}{c^2}=0.744\\\\

v=2.58\times 10^8

5 0
3 years ago
If the magnetic field in a traveling EM wave has a peak value of 17.9nT, what is the peak value of the electric field strength?
Doss [256]

Answer:

5.37 N/C

Explanation:

Peak value of magnetic field, Bo = 17.9 nT = 17.9 x 10^-9 T

The electromagnetic wave is produced when an oscillating electric and magnetic field interacts each other perpendicularly.

The direction of propagation of electromagnetic wave is perpendicular to both electric and magnetic field.

the relation between the electric field and magnetic field amplitudes is given by

c = \frac{E_{0}}{B_{0}}

where, c be the velocity of light, Eo be the peak value of electric field strength, Bo is the peak value of magnetic field strength.

3\times 10^{8} = \frac{E_{0}}{17.9\times 10^{-9}}}

Eo = 5.37 N/C

4 0
3 years ago
Which statement is true ?
Kisachek [45]

Answer:

it depends only on the forces acting along the y axis I think

3 0
3 years ago
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