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lorasvet [3.4K]
3 years ago
15

A worker on the roof of a house drops his 0.58 kg hammer, which slides down the roof at constant speed of 6.69 m/s. The roof mak

es an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. what is the the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground?
Physics
1 answer:
shusha [124]3 years ago
3 0

Answer:

17.3 m

Explanation:

Given that,

Mass of a hammer is 0.58 kg

Velocity with which the hammer slides is 6.69 m/s at constant speed.

The roof makes an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. We need to find the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground. Firstly, we will find the time taken by the hammer when it reaches ground in vertical direction.

y=y_0+v_ot +\dfrac{1}{2}gt^2

Putting all the values,

0=18.2+6.69t-\dfrac{1}{2}\times 9.8t^2\\\\-4.9t^2+6.69t+18.2=0\\\\t=\dfrac{-6.69+\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}, \dfrac{-6.69-\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}\\\\t=-1.36\ s\ \text{and}\ t=2.72\ s

Neglecting negative value,

To find horizontal distance, multiply 2.72 s with the horizontal component of velocity.

d=2.72\times 6.69\times \cos(18)\\\\d=17.3\ m

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In an economy, the demand for labor is given by the equation W = 15 - (1/200) L and the supply of labor is given by the equation
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Answer:

the equilibrium wage rate is 10  and the equilibrium quantity of labor is 1000 workers

Explanation:

The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

W_{Demand} = W_{Supply}

15 - (1/200) L = 5 + (1/200) L

15 - 5 =  (1/200) L +  (1/200) L

10 = (2/200) L

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Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:

W = 15 - (1/200) L

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Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha
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Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

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Explanation:

From the question we are told that

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           The charge carried by B is 0 C

            The charge carried by C is 0 C

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When C and B are brought close and then separated the charge carried by  C and B  is mathematically evaluated as    

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Looking at these calculation we can see that the charge carried by B is

        \frac{q}{4} C

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