Answer:
The magnitude of the electric force on a protein with this charge is
Explanation:
Given that,
Electric field = 1500 N/C
Charge = 30 e
We need to calculate the magnitude of the electric force on a protein with this charge
Using formula of electrostatic force
Where, F = force
E = electric field
q = charge
Put the value into the formula
Hence, The magnitude of the electric force on a protein with this charge is
Answer:
More extreme weather.
Explanation:
The Conveyor Belt of tides functions on a local and global level to spread out the cold and hot temperature differences on the planet. It is a delicate but important process that is easily disrupted, which causes it to slow down. And when it slows down, all those temperature differences will become more concentrated, causing colder places to be colder and hotter places to be hotter, ultimately leading to more extreme weather events as these cold and hot spots collide more violently than before.
Here's a picture I found on it:
Answer:
a. The ratio of their resistance is 2783:64
b. The ratio of their energy is 4:23
c. The charge on the first bulb is 5.75 C
The charge on the second bulb is C
Explanation:
The voltage on one of the electric bulbs, V₁ = 40 volts
The power rating of the bulb, P₁ = 230 w
The voltage on the other electric bulbs, V₂ = 110 volts
The power rating of the bulb, P₂ = 40 w
a. The power is given by the formula, P = I·V = V²/R
Therefore, R = V²/P
For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96
The resistance of the second bulb, R₂ = 110²/40
The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64
∴ The ratio of their resistance, R₂:R₁ = 2783:64
b. The energy of a bulb, E = t × P
Where;
t = The time in which the bulb is powered on
∴ The energy of the first bulb, E₁ = 230 w × t
The energy of the second bulb, E₂ = 40 w × t
The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23
∴ The ratio of their energy, E₂:E₁ = 4:23
c. The charge on a bulb, 'Q', is given by the formula, Q = I × t
Where;
I = The current flowing through the bulb
From P = I·V, we get;
I = P/V
For the first bulb, the current, I = 230 w/40 V = 5.75 amperes
The charge on the first bulb per second (t = 1) is therefore;
Q₁ = 5.75 A × 1 s = 5.75 C
The charge on the first bulb, Q₁ = 5.75 C
Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = C
The charge on the second bulb, Q₂ = C.
d. The question has left out parts