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Natali [406]
2 years ago
9

A bird flies 2km to the south, then 0.2km to the north.What is the displacement of the bird ?

Physics
2 answers:
Lunna [17]2 years ago
6 0

Explanation:

It is given that,

A bird flies 2 km to the south, then 0.2 km to the north. The displacement of the bird is equal to the shortest path covered by it.

Initially, the bird is moving in south direction and then it move to north. Displacement can be calculated using difference between initial position and final position.

d = 2.2 km - 0.2 km = 1.8 km

Hence, the displacement of the bird is 1.8 km

vivado [14]2 years ago
3 0
It would be D because it is flying back to where it started 2-.2=1.8
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2 years ago
A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to
dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence,i=30^{\circ}

Angle of refraction,r=19.24^{\circ}

a.Refractive index of air,n_1=1

We know that

n_2sinr=n_1sini

n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517

b.Wavelength of red light in vacuum,\lambda=632.8nm=632.8\times 10^{-9} m

1nm=10^{-9} m

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\lambda'=\frac{632.8}{1.517}=417nm

c.Frequency does not change .It remains same in vacuum and solution.

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Where c=3\times 10^8 m/s

Frequency,\nu=4.74\times 10^{14}Hz

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5 0
3 years ago
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen
Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

a=\dfrac{v^2}{R}

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

a'=\dfrac{v^2}{R'}

a'=\dfrac{v^2}{R/4}

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So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0
3 years ago
What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe
Eduardwww [97]

Answer:

Final temperature, T_f=21.85^{\circ}

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, T_i=41^{\circ}C

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, c=0.235\ J/g\ C

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{-Q}{mc}+T_i

T_f=\dfrac{-18}{4\times 0.235}+41

T_f=21.85^{\circ}

So, the final temperature of silver is 21.85 degrees Celsius.

5 0
3 years ago
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