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1 year ago

If a spear is thrown at a fish swimming in a lake, it will often miss the fish completely. Why does this happen?

Physics

2 answers:

givi [52]1 year ago

5 0

Answer:

C. Refraction of light between the air and water causes the fish to appear in a different location

Explanation:

When you something in the water you see that the objects appear shallower or the real position and the apparent position differ. This happens due to refraction of light. Refraction is a phenomenon of light where it bends when it moves from one medium to another.

In this scenario when the observer looks at the fish, the position of the fish seen to the observer will be different than its actual position because of which the spear will miss the fish most often.

andreev551 [17]1 year ago

3 0

C. refraction of light between the air and water causes the fish to appear in a different place

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For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a

jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

$R=\frac{L}{A\sigma }$

$R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }$

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

$i = \frac{V}{R}$

$i = \frac{0.060}{0.0178}$

i = 3.4 A

(c)

Current density is given as

$J = \frac{i}{A}$

$J = \frac{3.4}{5.3\times10^{-6}}$

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

$E = \frac{J}{\sigma }$

$E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}$

E = 9.01 x 10⁻³ V/m

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1 year ago

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear

Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

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1 year ago

3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta

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Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

Heat pump can be find using the equation

β = TH / TH - TC

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / 288.15 k - ( 253.15 K )

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TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / 288.15 k - ( 278.15 K )

β = 28.815 K

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1 year ago

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Answer:

time taken is 20 μs

Explanation:

given data

temperature = 20°C = 293 K

radius = 1 cm

atomic mass of air = 29 u

to find out

how long it will take for air to refill

solution

we find here rms velocity of air particle that is

$\frac{1}{2}mv^2 = \frac{3}{2}RT$

here m is mass and t is temperature and v is speed and R is ideal gas constant i.e. 8.3145 (kg·m²/s²)/K·mol

v = $\sqrt{\frac{3RT}{M} }$ ............................1

v = $\sqrt{\frac{3(8.314)293}{29*10{-3}kg} }$

v = 501.99 m/s

so now for cover 1 cm

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time take = $\frac{r}{v}$

time taken = $\frac{1*10^{-2}m}{501.99}$

time taken = 19.92 × $10^{6}$ s = 20μs

so time taken is 20 μs

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Explanation subtract

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