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wolverine [178]
3 years ago
11

You and your friend Peter are putting new shingles on a roof pitched at 20°. You're sitting on the very top of the roof when Pet

er, who is on the edge of the roof, 5.1 m away, asks you for the box of nails. Rather than carry the 2.2 kg box of nails down to Peter, you decide to give the box a push and have it slide down to him.
If the coefficient of kinetic friction between the box and the roof is 0.53, with what speed should you push the box to have it gently come to rest right at the edge of the roof?
Physics
1 answer:
dexar [7]3 years ago
6 0

Answer:

vi = 3.95 m/s

Explanation:

We can apply the Work-Energy Theorem as follows:

W = ΔE = Ef - Ei

W = - Ff*d

then

Ef - Ei = - Ff*d   <em> </em>

If

Ei = Ki + Ui = 0.5*m*vi² + m*g*hi = 0.5*m*vi² + m*g*hi = m*(0.5*vi² + g*hi)

hi = d*Sin 20º = 5.1 m * Sin 20º = 1.7443 m

Ef = Kf + Uf = 0 + 0 = 0

As we know,   vf = 0  ⇒ Kf = 0

Uf = 0  since hf = 0

we get

W = ΔE = Ef - Ei = 0 - m*(0.5*vi² + g*hi)  ⇒   W = - m*(0.5*vi² + g*hi)   <em> (I)</em>

<em />

If

W = - Ff*d = - μ*N*d = - μ*(m*g*Cos 20º)*d = -  μ*m*g*Cos 20º*d    <em>(II)</em>

<em />

we can say that

<em />

- m*(0.5*vi² + g*hi) = -  μ*m*g*Cos 20º*d  

⇒ vi = √(2*g*(μ*Cos 20º*d - hi))  

⇒ vi = √(2*(9.81 m/s2)*(0.53*Cos 20º*5.1m - 1.7443 m)) = 3.95 m/s

<em />

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Aleonysh [2.5K]

Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil

Answer:

6.169 μA

Explanation:

Formula for induced EMF is given by the equation;

EMF = M(di/dt). We are given;

di/dt = 2.5 A/s

M = 19μH = 19 × 10^(-6) H

Thus;

EMF = 19 × 10^(-6) × 2.5.

EMF = 47.5 × 10^(-6) V

Formula for current is;

i = EMF/R. R is resistance given as 7.7 ohms.

Thus; i = 47.5 × 10^(-6)/7.7

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3 years ago
an ice cube placed in microwave melts in five minutes and it takes 3.50 kj of energy to melt it. what is the power of the microw
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If the ice absorbed 350,000 joules in 5 minutes, then it absorbed energy
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3 years ago
Which statement is true for both types of transistors?
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For both NPN and PNP this is true:

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Read 2 more answers
Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int
neonofarm [45]

Answer:

The  value is  V_n  =  2.2498 \  m^3

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  V_n  =  3.6 \  L=  3.6 *10^{-3} \ m^3

   The  density of  nitrogen at gaseous form   is  \rho_n =  1.2929 \  kg/m^3  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         \rho _l = 808 \  kg/m^3

And this is mathematically represented as

      \rho_l  =  \frac{m}{V_l }

=>   m  =  \rho_l  *  V_l

Now the density of  gaseous nitrogen is

       \rho_n  =  \frac{m}{V_n }

=>   m  =  \rho_n  *  V_n

Given that the mass is constant

       \rho_n  *  V_n  =   \rho_l  *  V_l

        1.2929*  V_n  =   808  *  3.6*10^{-3}

=>   V_n  =  2.2498 \  m^3

       

3 0
3 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
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<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
3 years ago
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