If you write down the formula for friction, you will get an answer.
Ff = u * N Where N is a push down force that an object experiences.
u (mu) is a constant and has no units
It may not be accelerating and still experience friction. A is not correct.
Color and Density will not affect the frictional force. B is not so.
Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.
The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.
Answer
given,
given,
small cube side = 10 cm
larger cube side = 12 cm
density of steel = 7 g/cm³
density of aluminium = 2.7 g/cm³
density of the water (ρ₁)= 1 g/cm³
Cube A and B made of steel
buoyant force of Cube A
B₁ = ρ₁ V g = 1 x 10 x 10 x 10 x g= 1000 g
for cube B
B₂ = ρ₁ V g = 1 x 12 x 12 x 12 x g= 1728 g
buoyant force of Cube C
B₃ = ρ₁ V g = 1 x 10 x 10 x 10 x g= 1000 g
for cube D
B₄ = ρ₁ V g = 1 x 12 x 12 x 12 x g= 1728 g
buoyant force acting on the cube depends on the density of the fluid
hence,
B₂ = B₄ > B₁ = B₃
Answer:
(a) 1.093 rad/s^2
(b) 4.375 rad/s
(c) 8.744 rad/s
(d) 67.845 rad
Explanation:
initial angular velocity, ωo = 0
time, t = 8s
angular displacement, θ = 35 rad
(a) Let α be the angular acceleration.
Use second equation of motion for rotational motion
By substituting the values
35 = 0 + 0.5 x α x 8 x 8
α = 1.093 rad/s^2
(b) The average angular velocity is defined as the ratio of total angular displacement to the total time taken .
Average angular velocity = 35 / 8 = 4.375 rad/s
(c) Let ω be the instantaneous angular velocity at t = 8 s
Use first equation of motion for rotational motion
ω = ωo + αt
ω = 0 + 1.093 x 8 = 8.744 rad/s
(d) Let in next 5 seconds the angular displacement is θ.
By substituting the values
θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5
θ = 67.845 rad
Answer:
0.2m
The solution is in the picture
Answer with Explanation:
The direction of the electric field line at any point gives us the direction of the electric force that will act on a positive charge if placed at the point. We know that if we place a charge in an electric field it will experience a force, as we know that force is a vector quantity hence it requires both magnitude and direction for it's complete description. The direction of this electric force that acts on a charge is given by the direction of the electric field in the space. In case the charge is negatively charged electric force will act on it in the direction opposite to the direction of electric field at the point.
