Answer:
a) A = 500 N/s²
b) I = 5812.50 N-s
c) Δv = 2.7034 m/s
Explanation:
Given info
m = 2150 kg
F(t) = At²
F(1.25 s) = 781.25 N
a) A = ?
We use the equation
F(t) = At² ⇒ 781.25 N = A*(1.25 s)²
⇒ A = F(t) / t² = (781.25 N) / (1.25 s)²
⇒ A = 500 N/s²
b) I = ? if 2.00 s ≤ t ≤ 3.50 s
we apply the equation
I = ∫F(t) dt = ∫At² dt = A ∫t² dt = (500/3)*t³ + C
Since the limits of integration are 2 and 3.5, we obtain
I = (500/3)*((3.5)³-(2)³) = 5812.50 N-s
c) Δv = ?
we can apply the equation
I = m*Δv ⇒ Δv = I / m
⇒ Δv = 5812.50 N-s / 2150 kg
⇒ Δv = 2.7034 m/s
Answer:
Many objects are designed specifically to store elastic potential energy, for example: The coil spring of a wind-up clock. An archer's stretched bow. A bent diving board, just before a divers jump
Explanation:
Answer:
The Average velocity is 750m.
Explanation:
1miles is equal to 3km and 3km is equal to 3000m so a.v = 3000/4= 750m.
Answer:
v = 2.18m/s
Explanation:
In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.
Then you have:
(1)
M: mass Betty = 40kg
m: mass of the dog = 15kg
v1o: initial speed of Betty = 3.0m/s
v2o: initial speed of the dog = 0 m/s
v: speed of both Betty and her dog = ?
You solve the equation (1) for v:
The speed fo both Betty and her dog is 2.18m/s
