Answer:
– 2.5 m/s²
Explanation:
We have,
• Initial velocity, u = 180 km/h = 50 m/s
• Final velocity, v = 0 m/s (it stops)
• Time taken, t = 20 seconds
We have to find acceleration, a.
a = (v ― u)/t
a = (0 – 50)/20 m/s²
a = –50/20 m/s²
a = – 5/2 m/s²
a = – 2.5 m/s² (Velocity is decreasing) [Answer]
Answer:
= ( ρ_fluid g A) y
Explanation:
This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force
for the first part, let's write Newton's equilibrium equation
B₀ - W = 0
B₀ = W
ρ_fluid g V_fluid = W
the volume of the fluid is the area of the cube times the height it is submerged
V_fluid = A y
For the second part, the body introduces a quantity and below this equilibrium point, the equation is
B - W = m a
ρ_fluid g A (y₀ + y) - W = m a
ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a
ρ_fluid g A y + (B₀-W) = ma
the part in parentheses is zero since it is the force when it is in equilibrium
ρ_fluid g A y = m a
this equation the net force is
= ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position
Answer:
Explanation:
capacitance of sphere 2 will be 4.5 times sphere 1
a ) when spheres are in contact they will have same potential finally . So
V_1 / V_2 = 1
b )
Charge will be distributed in the ratio of their capacity
charge on sphere1 = q x 1 / ( 1 + 4.5 )
= q / 5.5
fraction = 1 / 5.5
c ) charge on sphere 2
= q x 4.5 / 5.5
fraction = 4.5 / 5.5
d ) surface charge density of sphere 1
= q /( 5.5 x A ) where A is surface area
surface charge density of sphere 2
= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area
= q /( 5.5 x 4.5 A )
q_1/q_2 = 4.5
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
Answer:
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
The box is initially at rest, so . Solving for :