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Daniel [21]
3 years ago
10

Which has more gravitaatoinal potental energy: a bird on the ground or the same bird in the tree? why?

Physics
1 answer:
julsineya [31]3 years ago
7 0
The same bird on the tree has more gravitational potential energy. This is because it is at a higher distance from the ground as it is on the tree, than when it is on the ground.

Considering also the formula for Gravitational Potential Energy GPE = mgh

For the bird on the ground, h =0,  therefore GPE = m*9.8*0 = 0

For that on the tree = mgh = m*9.8*h

Of course the one on the tree has a value greater than zero.  
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[A] Write an expression for the equivalent resistance of three resistors connected in parallel.[ no derivation needed]
anzhelika [568]

Answer:

1) R1 + ((R2 × R3)/(R2 + R3))

2) 0.5 A

3) 3.6 V

Explanation:

1) We can see that resistors R2 and R3 are in parallel.

Formula for sum of parallel resistors; 1/Rt = 1/R2 + 1/R3

Making Rt the subject gives;

Rt = (R2 × R3)/(R2 + R3)

Now, Resistor R1 is in series with this sum of R2 and R3. Thus;

Total resistance of circuit = R1 + ((R2 × R3)/(R2 + R3))

2) R_total = R1 + ((R2 × R3)/(R2 + R3))

We are given;

R1 = 7.2 Ω

R2 = 8 Ω

R3 = 12 Ω

R_total = 7.2 + ((8 × 12)/(8 + 12))

R_total = 7.2 + 4.8

R_total = 12 Ω

Formula for current is;

I = V/R

I = 6/12

I = 0.5 A

3) since current through the circuit is 0.5 and R1 is 7.2 Ω.

Thus, potential difference through R1 is;

V = IR = 0.5 × 7.2 = 3.6 V

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3 years ago
Layer of the atmosphere where jets fly through and contains ozone layer is called ________
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Answer:

The stratosphere

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a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec
Inessa [10]

Answer: 49.5 m

Explanation:

The speed of sound s is given by a relation between the distance d and the time t:

s=\frac{d}{t} (1)

Where:

s=330 m/s is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

d=412.5 m-x since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance x towards the cliff

t=\frac{2.2 s}{2}=1.1 s Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate d and then find x:

d=st (2)

412.5 m-x=(330 m/s)(1.1 s) (3)

Finding x:

x=49.5 m This is the distance at which the hunter is from the cliff.

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