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3 years ago

11

A spinning ice skater will slow down if she extends her arms away from her body. Which of the following statements explains this

phenomenon?

1 answer:

klasskru [66]3 years ago

6 0

Drag from her armas would slow her down if she was spinning at a fast speed

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The unstretched rope is 20 meters. After getting dunked a few times the 80 kg jumper comes to rest above the water with the rope

Kamila [148]

Answer:

Therefore maximum stretch is y2 = 32.36 m

Explanation:

In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium

$F_{e}$ - W = 0

k Δx = mg

k = mg / Δx

k = 80 9.8 / (30-20)

k = 78.4 N / m

now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m

starting point. When will you jump

Em₀ = U = mg y

final point. Just when the rope starts to stretch

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

mg y = ½ m v²

v = √ 2g y

v = √ (2 9.8 20)

v = 19.8 m / s

now all kinetic energy is transformed into elastic energy

starting point

Em₀ = K = ½ m v²

final point

Em_{f} = $K_{e}$ + U = ½ k y² + m g y

Emo = Em_{f}

½ m v² = ½ k y² + mgy

k y² + 2 m g y - m v² = 0

we substitute the values and solve the quadratic equation

78.4 y² + 2 80 9.8 y - 80 19.8² = 0

78.4 y² + 1568 y - 31363.2 = 0

y² + 20 y - 400 = 0

y = [- 20 ±√ (20 2 +4 400)] / 2

y = [-20 ± 44.72] / 2

the solutions are

y₁ = 12.36 m

y₂ = 32.36 m

These solutions correspond to the maximum stretch and its rebound.

Therefore maximum stretch is y2 = 32.36 m

7 0

3 years ago

An originally stationary car with a mass of 1500 kg which velocity of 15 m/s, five second after starting. What is the car’s acce

gladu [14]

15=0+a*5
a=3m/s^2

F=1500*3=4500N

4 0

3 years ago

The neck is a _______.

valkas [14]

Answer:

muscle

Explanation:

if not correct provide more information next time

4 0

3 years ago

Read 2 more answers

One star has a mass slightly more than the mass of the sun. Which of these graphs would best model and predict the lifetime of t

Oksanka [162]

Answer:

b

Explanation:

6 0

4 years ago

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(c)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 50 cm = 0.50 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:

$E=\dfrac{kQ}{r^2}$

Plug in the given values and solve for 'E'. This gives,

$E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

8 0

4 years ago