The acceleration of the wagon along the ground is 3.6 m/s².
To solve the problem above, we need to use the formula of acceleration as related to force and mass.
Acceleration: This can be defined as the rate of change of velocity.
⇒ Formula:
- Fcos∅ = ma................. Equation 1
⇒ Where:
- F = Force
- ∅ = angle above the horizontal
- m = mass of the wagon
- a = acceleration of the wagon
⇒ make a the subject of equation 1
- a = Fcos∅/m..................... Equation 2
From the question,
⇒ Given:
⇒ Substitute these values into equation 2
- a = 44(cos35°)/10
- a = 44(0.8191)/10
- a = 3.6 m/s²
Hence, The acceleration of the wagon along the ground is 3.6 m/s²
Learn more about acceleration here: brainly.com/question/9408577
Answer:
(A) The correct answer is option (B) three halves that of the old unit.
(B) The answer is three fourth that of old unit
Explanation:
from the relation;
(A) Fromthe expression;
K = Qd/AΔT
Anew = 3/2 A(old)
(B)
K¹ = 2K(old), so we get
A(old) = A(old)/2
Combining with part A, we have;
Anew = 3/2 *A(old)/2
= 3/4A(old)
The answer is three fourth that of old unit
Answer:
g = 5 m/s square
Explanation:
Weight(W), Mass(m), Gravity(g)
W = mg
1,000N = 200g
g = 1000/200
g = 5 m/s square
The isobars in the conventional series that will be needed
to complete the pressure analysis between the lowest and highest values on this
map are: 1008, 1012, 1016, 1020.
To add, an isobar is <span>a line on a map connecting points having the
same atmospheric pressure at a given time or on average over a given period.</span>
