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3 years ago

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A 85.4 kg ice skater is moving at 5.97 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 0.895 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.

1 answer:

Nimfa-mama [501]3 years ago

5 0

Answer:

$F_c=35.81kN$

Explanation:

When the ice skater grabs the loose end of the rope, he experiences a centipretal force, where the speed he had before grabbing the rope is the tangential speed, this force is given by:

$F_c=ma_c\\F_c=m\frac{(v_T)^2}{r}\\F_c=85.4kg\frac{(5.97\frac{m}{s})^2}{0.085m}\\F_c=35.81kN$

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A car in motion has kinetic energy. What happens to the kinetic energy when the car brakes to a stop? The kinetic energy is tran

Pie

Answer:

<em>When a moving car brakes to a stop the </em><em>kinetic energy of the car is converted to heat energy. </em>

Explanation:

A moving car has kinetic energy.

It is given by the equation $k=\frac{1}{2} mv^2$

Where m denotes mass of the car and v denote sits velocity. When the brakes are applied the velocity becomes zero and the car doesn’t possess kinetic energy anymore.

According to law of conservation of energy can neither be created nor be destroyed but can only be transformed from one form to another. On coming to a stop, the kinetic energy of the car gets converted to heat. The friction between the tyre and the road heats up the tyre.

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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

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