The car starts from a room with a constant acceleration of 5 ms-2. What path will it pass in 6 seconds and what speed will it re
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Answer:
The pressure difference is 25.8kPa while the force exerted on the window by air pressure is 3.02 kN
Explanation:
Using the Bernoulli's equation
Here
- Pin-Pout is the pressure difference which is to be calculated.
- ρ is the density of air whose value is 1.29 kg/m^3
- vin is the velocity of air inside which is 0 m/s
- vout is the velocity of air outside which is 200 m/s
Substituting values in the above equation yields
So the pressure difference is 25.8kPa.
As force is given by
Here
- ΔP is the pressure difference calculated above
- A is the area of the window given as
Now force is
So the force exerted on the window by air pressure is 3.02 kN
Answer:
q = 2.066* 10⁻¹³ C.
n = 1,291,250 electrons.
Explanation:
1)
- If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:
- where Fg is the gravitational attraction, that can be written as follows according Newton's Universal Law of Gravitation:
- Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming we can treat both spheres as point charges), as follows:
- since m₁ = m₂ = 0.0024 kg, and r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:
- Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:
- Since both charges are the same, the charge on each sphere is just the square root of (5):
- Q = 2.066* 10⁻¹³ C.
2)
- Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
- Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
Force on a moving charge is given by formula
here we know that this force will be maximum when velocity is perpendicular to magnetic field
here we know that
now we have