1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>
Galileo Galilei was the first scientist to perform experiments in order to test his ideas. He was also the first astronomer to systematically observe the skies with a telescope.
:)
Complete Question
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,
A :
a force repels the glass out of the capacitor.
B :
a force attracts the glass into the capacitor.
C :
no force acts on the glass.
D :
a net charge appears on the glass.
E :
the glass makes the plates repel each other.
Answer:
The correct option is B
Explanation:
Generally when the glass dielectric is slowly inserted between the plated,
The positive plate of the capacitor will induce a negative charge on the glass while the negative plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass
Answer:
Ae/A* = 1.115
Explanation:
Let the reservoir pressure be 
Let the exit pressure be 
Ratio of reservoir pressure and exit pressure
= 3.182
For the above value of pressure ratio
Obtain the area ratio from the isentropic flow table
Ae/A* = 1.115
The value of pressure ratio is Ae/A* = 1.115
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
