Answer:
A = 1.54 x 10⁻⁵ m² = 15.4 mm²
Explanation:
The resistance of a wire can be given by the following formula:
where,
A = smallest cross-sectional area = ?
ρ = resistivity of copper = 1.54 x 10⁻⁸ Ωm
= resistance per unit length of wire = 0.001 Ω/m
Therefore,
<u>A = 1.54 x 10⁻⁵ m² = 15.4 mm²</u>
Answer:
96.21 ft/s
Explanation:
To solve this, you only need to use one expression which is:
Vf² = Vo² + 2gh
g = 9.8 m/s²
However, this exercise is talking in feet, so convert the gravity to feet first:
g = 9.8 * 3.28 = 32.15 ft/s²
Vo is zero, because it's a free fall and in free fall the innitial speed is always zero. With this, let's calculate the speed at 2 seconds, with a height of 64 ft, and then with the 256 ft:
V1 = √2*32.15*64
V1 = 64.15 ft/s
V2 = √2*32.15*256
V2 = 128.3 ft/s
So the average rate is:
V = 128.3 + 64.15 / 2
V = 96.22 ft/s
The answer is metabolic heat.
<span>Organisms from the higher trophic levels consume organisms from the lower trophic level and during that process, energy is lost as metabolic heat. Primary producers (plants) contain the greatest amount of energy originally from the sunlight. The next trophic level belongs to primary consumers that consume primary producers. During consumption, energy is lost. Similarly, secondary consumers eat primary consumers and energy is lost again. The highest trophic level is tertiary consumers that contains the least amount of energy.</span>
Answer/Explanation:
Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculation of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. We can just use the kinematic equations. Fortunately, we are given these values. So, we calculate as follows:
acceleration = v - v0 / t
acceleration = (80 mph - 50 mph) ( 1 h / 3600) / 5 s
acceleration = 1.67 x 10^-3 m / s^2
Answer:
V = 575.6 Volts
Explanation:
As we know that surface area of the equi-potential surface is given as
so we will say
Now the potential due to a point charge is given as
