Answer:
In these reactions the products are higher in energy than the reactants. ... This barrier is due to the fact that to make CO2 and H2O we have to break 4 carbon-hydrogen bonds and some ...
Explanation:
Answer:
a) F = 1.26 10⁵ N, b) F = 2.44 10³ N, c) F_net = 1.82 10³ N directed vertically upwards
Explanation:
For this exercise we must use the relationship between momentum and momentum
I = Δp
F t = p_f -p₀
a) It asks to find the force
as the man stops the final velocity is zero
F = 0 - p₀ / t
the speed is directed downwards which is why it is negative, therefore the result is positive
F = m v₀ / t
F = 63.5 7.89 / 3.99 10⁻³
F = 1.26 10⁵ N
b) in this case flex the knees giving a time of t = 0.205 s
F = 63.5 7.89 / 0.205
F = 2.44 10³ N
c) The net force is
F_net = Sum F
F_net = F - W
F_net = F - mg
let's calculate
F_net = 2.44 10³ - 63.5 9.8
F_net = 1.82 10³ N
since it is positive it is directed vertically upwards
Answer:
a = 1.41 m/s²
Explanation:
Given that
mass ,m= 41 kg
F₁ = 65 N , θ = 59°
F₂ = 35 N ,θ = 32°
The component of Force F₁
F₁x= F₁cos59° i
F₁x= 65 x cos59° i = 33.47 i
F₁y= - F₁ sin 59° j
F₁y= - 65 x sin 59° j = - 55.71 j
The component of Force F₂
F₂x= F₂ sin 32° i
F₂x= 35 x sin 32° i = 18.54 i
F₂y= F₂ cos 32° j
F₂y= 35 x cos 32° j = 29.68 j
The total force F
F= 33.47 i + 18.54 i - 55.71 j + 29.68 j
F= 52.01 i - 26.03 j
The magnitude of the force F

F=58.16 N
We know that
F= m a
a= Acceleration
m=mass
58.16 = 41 x a
a = 1.41 m/s²
Answer:
The thickness of the foil is 0.017 mm.
Explanation:
Given that,
Weight = 7.5 oz = 212.6175 gm
Density = 2.70 g/cm³
Area of aluminium = 50 ft² = 46451.52 cm²
We need to calculate the thickness of the foil
Using formula of density



Where, A = area
t = thickness
m = mass
Put the value into the formula



Hence, The thickness of the foil is 0.017 mm.
The total kinetic energy of the system after collision is 223.5J
In elastic collision kinetic energy and momentum are conserved.
According to the question
mass of object a = 25kg
mass of object b = 25 kg
initial velocity of a u1 = 5.98 m/s
initial velocity of b u2 = 0
so from momentum conservation-
mau1 + mbu2 = (m1+m2)v
25kg × 5.98m/s + 25×0 = (25+25)v
v = 2.99 m/s
Now the total kinetic energy after the collision will be:
final kinetic energy = 1/2 (m1+m2) v²
= 1/2 (25+25)× (2.99)²
= 223.5 J
Thus, total kinetic energy of the system after collision is 223.5J
Learn more about elastic collision here:
brainly.com/question/1808045
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