(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)
</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
Answer:
33.33 %
Explanation:
given,
error last year = 63
error this year = 42
percent decrease in the error = ?
to find the percentage difference in the error formula used is
= 
= 
= 
= 33.33 %
Percentage decrease in the number of error is equal to 33.33%.
Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
The force of F=10 N produces an extension of

on the string, so the spring constant is equal to

Then the string is stretched by

. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J