Answer:
μ = 0.408
Explanation:
given,
speed of the automobile (u)= 20 m/s
distance = 50 m
final velocity (v) = 0 m/s
kinetic friction = ?
we know that,
v² = u² + 2 a s
0 = 20² + 2 × a × 50
a = 4 m/s²
We know
F = ma = μN
ma = μ mg
a = μ g
μ = 0.408
hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408
<span>Mechanical association learning used by an actor to memorize his lines</span>
<span>1.5 minutes per rotation.
The formula for centripetal force is
A = v^2/r
where
A = acceleration
v = velocity
r = radius
So let's substitute the known values and solve for v. So
F = v^2/r
0.98 m/s^2 = v^2/200 m
196 m^2/s^2 = v^2
14 m/s = v
So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin.
Its circumference is 2*pi*r, so
C = 2 * 3.14159 * 200 m
C = 1256.636 m
And we need a velocity of 14 m/s, so
1256.636 m / 14 m/s = 89.75971429 s
Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.</span>
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
efficiency= [useful energy transferred ÷ total energy supply]×100%
So, [5500÷10000]×100%=0.55×100
=55%
