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3 years ago

A hiker starts at point P and walks 2.0 km due east and then walks at an angle of 30 degrees north of east for 1.4 km.

Physics

1 answer:

olchik [2.2K]3 years ago

8 0

Answer:

The magnitude of the hiker’s displacement is 2.96 km

Explanation:

Let the initial displacement of the hiker, = x = 2km

the final displacement of the hiker, = y = 1.4 km

The resultant of the two vectors, According to Pythagorean theorem is the vector sum of the two vectors.

R' = x' + y'

Check the image uploaded for solution;

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4 years ago

Read 2 more answers

Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed

Ivanshal [37]

Answer:

Part a)

$f_w = f_g = 4.57 \times 10^{14} Hz$

Part b)

$\lambda_w = 492 nm$

$\lambda_g = 437.3 nm$

Part c)

$v_w = 2.25 \times 10^8 m/s$

$v_g = 2.0 \times 10^8 m/s$

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

$f = \frac{v}{\lambda}$

$f = \frac{3 \times 10^8}{656 \times 10^{-9}}$

$f = 4.57 \times 10^{14} Hz$

Part b)

As we know that the refractive index of water is given as

$\mu_w = 4/3$

so the wavelength in the water medium is given as

$\lambda_w = \frac{\lambda}{\mu_w}$

$\lambda_w = \frac{656 nm}{4/3}$

$\lambda_w = 492 nm$

Similarly the refractive index of glass is given as

$\mu_w = 3/2$

so the wavelength in the glass medium is given as

$\lambda_g = \frac{\lambda}{\mu_g}$

$\lambda_g = \frac{656 nm}{3/2}$

$\lambda_g = 437.3 nm$

Part c)

Speed of the wave in water is given as

$v_w = \frac{c}{\mu_w}$

$v_w = \frac{3 \times 10^8}{4/3}$

$v_w = 2.25 \times 10^8 m/s$

Speed of the wave in glass is given as

$v_g = \frac{c}{\mu_g}$

$v_g = \frac{3 \times 10^8}{3/2}$

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Answer:

B. d(low)=4d(high)

Explanation:

Frequency of a string can be written as;

f = v/2L

Where;

v = sound velocity

L = string length

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f = v/2L = (1/2L)√(T/u) ......1

Where;

m= mass,

u = linear density of string,

T = tension

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4f(low) = f(high)

Then,

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