Answer:
Final speed of car = 12 m/s
Explanation:
We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.
a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s
v = ?
u = 0 m/s
a = 4.0 m/s²
t = 5 s
v = u + at = 0 + 4 x 5 = 20 m/s
b) Then maintains that velocity for 10 s
v = ?
u = 20 m/s
a = 0 m/s²
t = 10 s
v = u + at = 20 + 0 x 10 = 20 m/s
c) Then decelerates at the rate of 2.0 m/s² for 4.0 s
v = ?
u = 20 m/s
a = -2.0 m/s²
t = 4 s
v = u + at = 20 + -2 x 4 = 12 m/s
Final speed of car = 12 m/s
Answer:
The speed of the cart after 8 seconds of Low fan speed is 72.0 cm/s
The speed of the cart after 3 seconds of Medium fan speed is 36.0 cm/s
The speed of the cart after 6 seconds of High fan speed is 96.0 cm/s
Explanation:
took the test on edgenuity
Answer is 6.84 approx
reason:-
(2.78^2+6.25^2)^1/2=6.84 approx
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J
