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:
4miles/hour
Explanation:
the solution for this question requires that the quantities are converted to the appropriate units as required by the question.
Rate in miles per hour = distance in miles / time in hour
to convert 12 minutes to hours; recall that 60 minutes make 1 hour
12 minutes to hour = 12/60 = 0.2hr
to convert 4224 feet to miles; recall 5280 feet is equivalent to 1 mile
4224 feet to miles = 4224/5280 = 0.8 miles
∴ rate = 0.8 / 0.2
rate = 4 miles per hour
the constant rate in miles per hour = 4 miles/hour
Answer: he did travel 15 meters.
Explanation:
We have the data:
Acceleration = a = 1.2 m/s^2
Time lapes = 3 seconds
Initial speed = 3.2 m/s.
Then we start writing the acceleration:
a(t) = 1.2 m/s^2
now for the velocity, we integrate over time:
v(t) = (1.2 m/s^2)*t + v0
with v0 = 3.2 m/s
v(t) = (1.2 m/s^2)*t + 3.2 m/s
For the position, we integrate again.
p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0
Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m
Then the displacement at t = 3s will be equal to p(3s).
p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m
Answer:
It's constant everywhere in its trajectory.
Explanation:
the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.
The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.
This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.
