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r-ruslan [8.4K]
2 years ago
13

In this dark-colored forest, you started with 50% light moths and 50% dark moths. At the end of your simulation, was there a hig

her percentage of light or dark moths?

Physics
1 answer:
krek1111 [17]2 years ago
5 0
Dark moths because they use their color to blend in with the trees to hide from birds
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Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice w
abruzzese [7]

Answer:

3.85*10^8m

Explanation:

We know that

Speed v = distance/time t

So distance = v x t

So = 3*10^8 x 2.56s

= distatance = 7.7*10^8m

However this is double the distance from the earth to the moon, so to get distance from earth to the moon we divide by 2

7.7*10^8/2= 3.85*10^8m

8 0
3 years ago
During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.
neonofarm [45]

Answer:

work is =7590joules

power = 23watts

4 0
3 years ago
Read 2 more answers
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
3 years ago
A charged particle is located in an electric field where the magnitude of the electric field strength is 2.0x10^3 newtons per co
Alexxandr [17]

Answer:

20.7

Explanation:

:0 because basis  of the daily occured

6 0
2 years ago
What will be the final velocity of a rock if we drop it off of a bridge and it strikes the ground 2.8s later (ignoring air resis
mars1129 [50]
Formula for final velocity: Vf= vi+(a*t)
Vi- initial velocity, a=acceleration, t-time

Vf=vi+(at)
Vf= 0+(9.8m/s*2.8s)
Vf= 27.44 m/s

The acceleration of the Earth when dropping something would be 9.8 m/s

Here is an reference that can help you answer problems like these.
Hope this helps and good luck :)

5 0
3 years ago
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