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SCORPION-xisa [38]
3 years ago
13

An eagle that has a mass of 4.5 kg is spotted 5 m above a lake. It flies up to its nest in a tree at a height of 25 m. How much

gravitational potential energy did the eagle gain
Physics
2 answers:
krek1111 [17]3 years ago
8 0
The eagle gained 882 J of gravitational potential energy.
musickatia [10]3 years ago
8 0

Answer:

The eagle gained 882 J of  gravitational potential energy.

Explanation:

Given that,

mass of an eagle = 4.5 kg

Initial height h_{i} = 5\ m

Final height  h _{f} = 25\ m

We need to calculate the gravitational potential energy gain

The gravitational potential energy is defined as,

E = mgh

Change the gravitational potential energy is

\Delta E= mgh_{f}-mgh_{i}

Put the value in to the formula

\Delta E = 4.5\times9,8\times(25-5)

\Delta E = 4.5\times9,8\times20

\Delta E=882\ J

Hence, The eagle gained 882 J of  gravitational potential energy.

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An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displace
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Answer:

Explanation:

Displacement is the shortest distance possible between initial and final position.

As the athlete is in circular motion , displacement is zero. [ THis is because , the initial and final position is the same ]

Distance covered in 1 round = Circumference of the circle

Circumference of the circle = 2πr

Diameter = 200 m

Radius = 100 m

Distance covered in 1 round[ 40 sec ] = 2 × 22/7 × 100 = 628.57 m

Distance covered in 1 sec = 628.57 ÷ 40 = 15.71 m

2min 20 sec = 140 sec

Distance covered in 140 sec = 15.71 × 140 = 2199.4 m

For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.

At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.

Hence, the distance covered is 2200 m and the displacement is 200 m.

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4 years ago
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An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV
ra1l [238]

Answer:

t = 5.48 × 10⁻³ s

Explanation:

Given:

ΔV = ΔVmax × sin(2πft)

frequency, f = 16.9Hz

thus,

ΔV = ΔVmax × sin(2π×16.9×ft)

Now,

Let 'R' be the resistance

Also according to the ohms law

i = V/R

where,

i = current

V = voltage

hence,

i=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}

also, given at time 't' the current in the circuit is 55.0% of the peak current

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i=\frac{55}{100}\times \frac{\Delta V_{max}}{R}=0.55\times \frac{\Delta V_{max}}{R}

thus,

0.55\times \frac{\Delta V_{max}}{R}=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}

or

0.55=sin(2\pi \times 16.9\times t)}

or

0.5823=(2\pi \times 16.9\times t)}

or

t = 5.48 × 10⁻³ s (Answer)

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