Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
<h3>
Magnitude of electric field </h3>
The magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
- E is the electric field
- m is mass of the particle
- g is acceleration due to gravity
- q is charge of the particle
<h3>For an electron</h3>
E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
<h3>For proton</h3>
E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
Learn more about electric field here: brainly.com/question/14372859
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Answer:
The value of the centripetal forces are same.
Explanation:
Given:
The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.
The expression for centripetal force is given by,

where,
is the mass of the object,
is the velocity of the object and
is the radius of the path.
The value of the centripetal force depends on the mass of the object, not on its weight.
As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.
The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

Where,
Depth of glass
Refraction index of water
Refraction index of glass
Refraction index of air
Depth of water
I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to



Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm