Answer:
Explanation:
Given
Airplane is flying with horizontally with a constant momentum during time interval 
Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant
(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction
To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as
The potential energy, 
The volume, 
The potential energy per unit volume is defined as the energy density.



The energy density related with electric field is given by

Here, the permitivity of the free space is

Therefore, rerranging to find the electric field strength we have,



Therefore the electric field is 2.21V/m
Answer:
pressure= force/area
A solid resting on a horizontal surfaceexerts a normal contactforce equals to its weight. The pressure of the solid on the surface depends on the area of contact. (b) the area of contact between the two surfaces. The greater the force or the smaller the area the greater the pressure.
Answer:
Elements in a period have wide range of chemical properties.
Explanation:
- Period is the row of chemical elements's arrangement in the periodic table. It is numbered from 1 to 7.
- In period as we move towards right the atomic number of element constantly increases along with the change in chemical properties.
- As we move from left towards right in period, the valence electron number regularly increases in every element.
- This valence electron differs the reactivity of the elements.
M = mass of aluminium = 1.11 kg
= specific heat of aluminium = 900
= initial temperature of aluminium = 78.3 c
m = mass of water = 0.210 kg
= specific heat of water = 4186
= initial temperature of water = 15 c
T = final equilibrium temperature = ?
using conservation of heat
Heat lost by aluminium = heat gained by water
M
(
- T) = m
(T -
)
(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)
T = 48.7 c