Two vectors A and B are directed such that there is an angle θ between them. show answer Incorrect Answer Which of the following

A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40° and fly ano

serg [3582]

Answer:

(a). The distance traveled is 7.06 km towards the west.

(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.

Based on the diagram,

(a). To determine the distance

We will apply the distance formula

$AD=AB+BD$

Insert the values into the equation

$D= 4+4.0\cos40^{\circ}$

$D=7.06\ km$

The resultant distance is 7.06 km westward.

(b). We will find the total displacement's magnitude

Using the displacement formula

$AC=\sqrt{(CD)^2+(AD)^2}$

Insert the values into the equation

$AC=\sqrt{(4.0\sin40)^2+(7.06)^2}$

$AC=7.51$

The total displacement magnitude is 7.51 km.

In conclusion, (a). The traveled distance is 7.06 km towards the west.

(b). The magnitude of their total displacement is 7.51 km.

7 0

3 months ago

A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the

kicyunya [3294]

Answer:

Induced EMF is 2 x 10⁻³ volts

Explanation:

B = strength of the magnetic field aligning with the loop's axis = 1 T

$\frac{dA}{dt}$ = area change rate of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = the angle formed by the magnetic field and area vector = 0

E = the induced EMF across the loop

EMF can be calculated using the formula

E = B $\frac{dA}{dt}$

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0

2 months ago

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

kicyunya [3294]

Response:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Reasoning:

The laser power P = 1 mW = 1 x 10^-3 W

duration t = 250 ms = 250 x 10^-3 s

Taking a beam diameter of 2 mm = 2 x 10^-3 m

therefore

the beam's cross-sectional area A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) The intensity I = P/A

where P refers to the laser's power

and A represents the beam's cross-sectional area

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) The total energy delivered E =Pt

where P is the beam's power

and t is the exposure duration

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field can be computed as

E = $\sqrt{2I/ce_{0} }$

where I signifies the beam's intensity

and E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

6 0

2 months ago