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marin [14]
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?
Physics
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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A double insulated drill ____. A. has two ground wires to insure proper grounding B. generally has a plastic case with the elect
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Option B is correct.

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8 0
3 years ago
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
Tanzania [10]

Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

Mass of platem_2=400 g=\frac{400}{1000}=0.4 kg

a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

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

Substitute the values  

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

0.4v_2=0.075\times 5.6+0.075\times 3.4

v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

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

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0
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

Fraction of power lost = \dfrac{80}{529} = 0.151

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

P_\text{loss} = I^2R

where R is the resistance of the wire. From the question, the wire has a resistance of 0.050\Omega per km. Since resistance is proportional to length, the resistance of the wire is

R = 0.050\times40 = 2\Omega

Hence,

P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2

The fraction lost = \dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151

3 0
3 years ago
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