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3 years ago

7

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

the object makes an angle α0 with the horizontal. Neglect air resistance. If α0 is 90∘, so that the object is thrown straight up (but misses the roof on the way down), what is the speed v of the object just before it strikes the ground?

1 answer:

Vsevolod [243]3 years ago

6 0

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

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Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

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$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

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Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

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3 years ago

A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.

nikitadnepr [17]

Answer:

Current = 8696 A

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Explanation:

Electric power is given by

$P=IV$

where I is the current and V is the voltage.

$I=\dfrac{P}{V}$

Using values from the question,

$I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}$

The power loss is given by

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Hence,

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