X^2+y^2+z^2=A^2
But here XY and Z are all equal so
3X^2=A^2
X=A/(sqrt(3))
Each component is the value of a divided by the square root of three. This way if you square then and add them up it equals a squared
Answer:
Centripetal acceleration of the car is 17.4 m/s²
Explanation:
It is given that,
Radius of the track, r = 57 m
Speed of car, v = 31.5 m/s
We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :



So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.
Answer:
a) W_total = 8240 J
, b) W₁ / W₂ = 1.1
Explanation:
In this exercise you are asked to calculate the work that is defined by
W = F. dy
As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.
W = F dy = F Δy
let's apply this formula to our case
a) Let's use Newton's second law to calculate the force in the first y = 5 m
F - W = m a
W = mg
F = m (a + g)
F = 80 (1 + 9.8)
F = 864 N
The work of this force we will call it W1
We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)
F₂ - W = 0
F₂ = W
F₂ = 80 9.8
F₂ = 784 N
The work of this fura we will call them W2
The total work is
W_total = W₁ + W₂
W_total = (F + F₂) y
W_total = (864 + 784) 5
W_total = 8240 J
b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use
W₁ / W₂ = F y / F₂ y
W₁ / W₂ = 864/784
W₁ / W₂ = 1.1
Answer:

Explanation:
The equation that relates heat Q with the temperature change
of a substance of mass <em>m </em>and specific heat <em>c </em>is
.
We want to calculate the final temperature <em>T, </em>so we have:

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add
to our temperature in
to have it in
as it must be):

Answer:
Speed of 1.83 m/s and 6.83 m/s
Explanation:
From the principle of conservation of momentum
where m is the mass,
is the initial speed before impact,
and
are velocity of the impacting object after collision and velocity after impact of the originally constant object
Therefore
After collision, kinetic energy doubles hence
Substituting 5 m/s for
then
Also, it’s known that
hence
Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then
Substituting,
Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s