Answer:
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Explanation:
Given:
Radius of sphere (r) = 12 cm = 0.12 m
Distance from the electric field R = 24 cm = 0.24 m
Magnitude (E) = 640 N/C
Find:
Charge density on the sphere
Computation:
Charge on the sphere (q) = (1/K)ER² (K = 9 × 10⁹)
Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²
Charge on the sphere (q) = 4 × 10⁻⁹ C
Charge density on the sphere = q / [4πr²]
Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]
Charge density on the sphere = [4 × 10⁻⁹] / [0.18]
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Answer:
The speed of galaxy relative to the earth is
Explanation:
Given :
Observed frequency
Actual frequency
For finding the speed of galaxy relative to the earth.
From the doppler effect of light,
Where , relative speed
Therefore, the speed of galaxy relative to the earth is
From the basic "heat lost by hot object=heat gained by colder object" principle, we have
m1c1ΔT1=m2c2ΔT2
where m1= 1kg
m2=4kg
c1=900J/kg k
c2=4200J/kg k
With this information at hand we have
m1c1(90-T)=m2c2(T-25)
after substituting the given values we can find that
T=28.3^{0}c
Answer:
Explanation:
In our previous articles, we observed the theoretical formulas of Ohm’s law, its calculations in the lab report, and experiment. Today you’ll learn the verification of theory vs experimental results on a 1 kΩ resistor.
The theoretical results are obtained from the formula of Ohm’s law: V = IR. The experimental verification is provided for a metal film 1 kΩ (±0.05%). We have used a high-quality resistor with negligible tolerance value so as to reduce the tolerance error.
The little problem in our calculations arises due to improper handling of multimeter probes. You can learn the complete method to perform the Ohm’s experiment here and can calculate the current values by using the Ohm’s law calculator.
Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1.
2.
Since both cars will travel the same distance x, we can equal both formulas and solve for t:
We simplify the fraction present and rearrange for our formula so that it equals 0:
In the very last step we factored a common factor t. There is two possible solutions to the equation at and:
What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at (when the SUV passed the police car) and (when the police car catches up to the SUV)