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anyanavicka [17]
3 years ago
14

A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s.

Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop
Physics
2 answers:
GREYUIT [131]3 years ago
6 0

Answer:

The velocity is  v = 8.85 m/s

Explanation:

From the question we are told that

    The mass of the skier is m_s =  60 \ kg

      The initial speed is u =  4.0 \ m/s

       The height is  h =  10 \ m

According to the law of energy conservation

     PE_t  + KE_t  =  KE_b + PE_b

Where PE_t is the potential energy at the top which is mathematically evaluated as

       PE_t =  mg h

substituting values

       PE_t  =  60 * 4*9.8

      PE_t  =  2352 \ J

And  KE_t is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill

And  KE_b is the kinetic energy at the bottom of the hill which is mathematically represented as

         KE_b  =  0.5 * m * v^2

  substituting  values

         KE_b  =  0.5 * 60 * v^2

=>     KE_b  = 30 v^2

Where v is the velocity at the bottom

   And PE_b is the potential  energy at the bottom which equal to zero due to the fact that height  is zero at the bottom of the hill

So  

        30 v^2 =  2352

=>      v^2 =   \frac{2352}{30}

=>       v =  \sqrt{ \frac{2352}{30}}

        v = 8.85 m/s

         

Romashka [77]3 years ago
5 0

Answer:

The Skier's velocity at the bottom of the hill will be 18m/s

Explanation:

This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.

That is

mgh = 0.5 mv^{2}

solving for velocity, we will have

v= \sqrt{2gh}

hence her velocity will be

v=\sqrt{2 \times 9.81 \times 10}=14.00m/s

This is the velocity she gains from the slope.

Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s

The Skier's velocity at the bottom of the hill will be 18m/s

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Answer:

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Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

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Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

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3 years ago
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A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Over [174]
B) law of conservation of momentum

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In this case total momentum before impact:

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After Impact:

10kg*0m/s + 5kg*10m/s = 50 kg m/s

You can see the momentum before and after impact is same as 50 kg m/s  

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Two long, straight wires both carry current to the right, are parallel, and are 25 cm apart. Wire one carries a current of 2.0 A
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Answer:

x = 7.14 meters

Explanation:

It is given that,

Current in wire 1, I_1=2\ A

Current in wire 2, I_2=5\ A  

Distance between parallel wires, r = 25 cm

Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :

B=\dfrac{\mu_oI}{2\pi r}

Let the distance is x from wire 1. So,

\dfrac{I_1}{r}=\dfrac{I_2}{(0.25-r)}

\dfrac{2}{r}=\dfrac{5}{(25-r)}

x=\dfrac{50}{7}\ m

x = 7.14 meters

So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.

6 0
3 years ago
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
boyakko [2]

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

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Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

V = V_{0}(1-e^{\frac{-t}{RC} }  )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} })  \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }

Taking natural log on both sides,

e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }  \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })

t = -RC ln (1-\frac{V}{V_{0} })        (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

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Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

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 = 15×10⁻⁹s

= 15 ns

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3 years ago
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