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atroni [7]
3 years ago
7

A migrating bird flew across a lake at an average speed of 18 meters per second. Was the distance that the bird flew across the

lake less than 21 kilometers? (1 kilometer = 1,000 meters) (1) It took the bird less than 20 minutes to fly across the lake. (2) It took the bird less than 0.4 hour to fly across the lake.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

It cannot be said with 100% guarantee.

Explanation:

Since bird is travelling with 18 meters per second, that means that it covers a distance of 18 meters in 1 second.

for part 1 we convert 20 minutes into seconds, thus, 20 * 60 = 1200 seconds.

Distance covered in 1200 seconds will be:

S = v * t (S is distance , v is speed, t is time)

18 * 1200 = 21600 meters = 21.6 Kilometers

If the bird took less than 20 minutes than distance covered would be less than 21.6 Kilometer which could be greater than or lesser than 21 Kilometers. On average we can say that the distance bird travelled will be less than 21 Kilometer. But it is not definite.

For part B convert 0.4 hours into seconds to find the distance it travelled.

0.4 * 60 * 60 = 1440 seconds

Distance covered will be

18 * 1440 = 25920 meters = 25.9 Kilo meters.

Again, if the bird takes less than 0.4 hours to cover the lake, the lake will be less than 25.9 kilometers long. This means that it could be greater than or lesser than 21 kilometers. It cannot be definitive

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Arisa [49]

Answer:

31.6\:\mathrm{m/s}

Explanation:

The elastic potential energy of a spring is given by Us=\frac{1}{2}kx^2, where k is the spring constant of the spring and x is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by KE=\frac{1}{2}mv^2, where m is the mass of the object and v is the object's velocity.

Thus, we have:

Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2

Substituting given values, we get:

\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}

4 0
2 years ago
According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energ
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Answer:

c.   V = k Q1 * Q2 / R1      potential energy of Q1 and Q2 separated by R

V2 / V1 = (R1 / R2) = 1/4

V2 = V1 / 4

8 0
3 years ago
A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t
Readme [11.4K]

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis  ( tan^-1 (13.5 / 16.4)

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3 years ago
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
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Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

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E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0
2 years ago
The engineer of a train traveling at 30 m/sec sees a cow on the tracks. He applies the brakes and causes the train to accelerate
sineoko [7]

Answer:The train travels 105 meters after applying the brakes

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8 0
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