Answer:
Explanation:
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.
Hence, his average velocity is 300m/150s=2ms^−1
(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.
Therefore, his average speed for this journey is 400m210s=1.9ms−1.
For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.
Hence, his average velocity for this case is 200m/210s=0.95ms^−1
1 volt = 1 joule per coulomb.
Current doesn't actually pass 'through' a battery.
But if it did, then each coulomb would gain or lose 6 joules in traversing 6 volts, depending on its sign, and whether it climbed or fell.
Answer:
the square of the average atomic velocity.
Explanation:
From the formulas for kinetic energy and temperature for a monoatomic gas, which has three translational degrees of freedom, the relationship between root mean square velocity and temperature is as follows:
(1)
Where is the root mean square velocity, M is the molar mass of the gas, R is the universal constant of the ideal gases and T is the temperature.
The root mean square velocity is a measure of the velocity of the particles in a gas. It is defined as the square root of the mean square velocity of the gas molecules:
(2)
substituting 2 in 1, we find the relationship between mean square speed and temperature:
Answer:
6.20×10⁴ V/m
Explanation:
The magnitude of electric field is:
E = √(Eₓ² + Eᵧ²)
where Eₓ = ∂φ/∂x and Eᵧ = ∂φ/∂y.
φ = 1.11 (x² + y²)^-½ − 429x
Eₓ = -0.555 (x² + y²)^-(³/₂) (2x) − 429
Eᵧ = -0.555 (x² + y²)^-(³/₂) (2y)
Evaluating at (0.003, 0.003):
Eₓ = -44034 V/m
Eᵧ = -43605 V/m
The magnitude is:
E = 61971 V/m
Rounded to three significant figures, the strength of the electric field is 6.20×10⁴ V/m.