Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:

Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution

Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
Answer:

Explanation:
Mass of block=10 kg
Applied horizontal force =F=20 N
Friction force=f=10 N
We have to find the acceleration of block.
Net force=Applied horizontal force-friction force

Where F= Horizontal force
f=Friction force
m=Mass of object
a=Acceleration of object


Hence, the acceleration of the block=
<h2>
Answer:g=9.79
,A object of mass

at the surface of earth experiences a force

</h2>
Explanation:
Let
be the mass of earth.
Let
be the radius of earth.
Let
be the universal gravitational constant.
Given,




Let
be the acceleration due to gravity.
Then,


A object of mass
at the surface of earth experiences a force 