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.
Answer:
t = 25 seconds
Explanation:
Given that,
Distance, d = 115 m
Initial speed, u = 4.2 m/s
Final speed, v = 5 m/s
We need to find the time taken in increasing the speed.
We know that,
Acceleration, 
....(1)
The third equation of kinematics is as follows :
Hence, it will take 25 seconds to increase the speed.
The braking force is -400 N
Explanation:
We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:
where in this problem, we have:
F is the force applied by the brakes
is the time interval
m = 13,000 kg is the mass of the ferry
u = 2.0 m/s is the initial velocity
v = 0 is the final velocity
And solving for F, we find the force applied by the brakes:
where the negative sign indicates that the direction is backward.
Learn more about impulse:
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To solve this problem, we must imagine that Jim’s initial
position, the position of the rock, and Jim’s final position all connects to
form a triangle. Now we can imagine that the triangle is a right triangle with
the 90° angle on the initial position.
The angle of 30° is directly opposite to the length of his
total stride while the width of the river is the side adjacent to the angle.
Therefore can use the tan function to solve for the width of the river:
tan θ = opposite side / adjacent side
tan 30 = total stride distance / width of river
where total stride distance = 65 * 0.8 = 52 m
width of river = 52 m / tan 30
<span>width of river = 90.07 m</span>
Answer:
El peso del cartel es 397,97 N
Explanation:
La tensión dada en cada segmento del cable = 2000 N
El desplazamiento vertical del cable = 50 cm = 0,5 m
La distancia entre los polos = 10 m
La posición del letrero en el cable = En el medio = 5
El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °
El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero
El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N
El peso del signo = 397,97 N.
