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3 years ago

how do the retinas of the eyes of night hunting animals differ from the retinas of animals that hunt during the day time

2 answers:

5 0

Hello,

Night-hunting animals have more rod cells than day-hunting animals. This is most likely because night-hunting animals have to be able to see better than day-hunting animals.

maksim [4K]3 years ago

4 0

Night hunting animals have more rod cells in their retinas which allows them to see better in the dark.

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

2 years ago

The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an

Troyanec [42]

Answer:

$3400$ Hz

Explanation:

We know that

1 cm = 0.01 m

$L$ = Length of the human ear canal = 2.5 cm = 0.025 m

$V$ = Speed of sound = 340 ms⁻¹

$f$ = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

$f = \frac{(2n - 1)V}{4L}$

for first resonant frequency, we have n = 1

Inserting the values

$f = \frac{(2(1) - 1) 340}{4(0.025)}$

$f = \frac{340}{4(0.025)}$

$f = 3400$ Hz

4 0

3 years ago

How does the gravitational force between two objects change if the distance

Anestetic [448]

Answer:

The Gravitational Force is reduced 4 times

Explanation:

The equation of Gravitational force follows:

F = (G*m1*m2)/r^2

Assume that G*m1*m2 = 1 and r = 1:

F = 1/1^2 = 1 N

Multiply the radius by 2

F = 1/2^2 = 1/4 N

So doubling the distance reduces the force 4 times.

7 0

3 years ago

A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the

zysi [14]

As per the question, the mass of meteorite [ m]= 50 kg

The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

$Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2$

$=\frac{1}{2}50kg*[1000\ m/s]^2$

$=\frac{1}{2}50* 10^{6}\ J$

$=25*10^6\ J$

Here, J stands for Joule which is the S.I unit of energy.

$Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J$

4 0

3 years ago

Read 2 more answers

Two cars are moving with velocities 70km/hr and 50km/hr in east and west direction respectively.

viva [34]

Answer:

70 + 50 = 120 km/hr

Explanation:

The driver of either car would see the other car approaching or departing at 120 kph

5 0

2 years ago