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Helga [31]
3 years ago
14

A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops

directly downward 1.4m to hover in front of the flower. What is the hummingbirds total displacement

Physics
2 answers:
polet [3.4K]3 years ago
7 0

Answer:

1.84m, 49.4⁰ South west

Explanation:

<u>Step 1:</u> construct a right angled triangle from the birds displacement

⇒From the diagram uploaded; we calculate the resultant/total displacement (R) of the bird using Pythagoras theorem.

R² = 1.2² + 1.4²

R² = 3.4

R = √3.4

R = 1.84 m

<u>Step 2:</u> calculate the direction of the bird's displacement

From vector diagram head-to-tail rule, we determine the cardinal direction;

From the diagram; tan(\alpha) = \frac{1.4}{1.2}

tan(\alpha) = 1.1667

\alpha =tan^{-1} 1.1667

\alpha = 49.4^o

Therefore the hummingbirds total displacement is 1.84m, 49.4⁰ South west

Anit [1.1K]3 years ago
4 0

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

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A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

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