The distance travelled is 10 m
The velocity gained at the end of the time is 2 m/s
<h3>Motion</h3>
From the question, we are to determine distance travelled and the velocity gained
From one of the equations of motion for <u>linear motion</u>, we have that
S = ut + 1/2at²
Where S is the distance
u is the initial velocity
t is the time taken
and a is the acceleration
First, we will calculate the acceleration
Using the formula,
F = ma
Where F is the force
m is the mass
and a is the acceleration
∴ a = F/m
Where F is the force
and a is the acceleration
From the given information,
F = 50 N
m = 250 kg
Putting the parameters into the equation,
a = 50/250
a = 0.2 m/s²
Thus,
From the information,
u = 0 m/s (Since the object was initially at rest)
t = 10 s
S = 0(t) + 1/2(0.2)(10)²
S = 10 m
Hence, the distance travelled is 10 m
For the velocity
Using the formula,
v = u + at
Where v is the velocity
v = 0 + 0.2×10
v = 2 m/s
Hence, the velocity gained at the end of the time is 2 m/s
Learn more on Motion here: brainly.com/question/10962624
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Answer:
-45
Step-by-step explanation:
In order to keep the equation right, both sides have to balance each other. When you only perform an operation ( I'm assuming you're talking about division or multiplication) on one side, it messes up the balance because now one side is larger than the other.
The answer is Letter D - 3.42
Size - Mean
10 - 3.51
20 - 3.42
30 - 3.39
40 - 3.34
To solve for the mean, add up all the given values then divide it by how many numbers there are. In this case, we will add the mean of the samples then divide the result by the number of the given samples.
= (3.51 + 3.42 + 3.39 + 3.34) / 4
= 13.66 / 4
= 3.415 or 3.42
I assume that sequence is 3,4,6,9,13,18,24
delta y=1,2,3,4,5,6
delta delta y=1,1,1,1
So this sequence has a constant acceleration, which is the property of a parabola or quadratic equation of the form ax^2+bx+c=y. Using three data points to create a system of equations we can solve for the three variables.
9a+3b+c=6, 4a+2b+c=4, a+b+c=3, getting differences
5a+b=2, 3a+b=1, and getting differences again
2a=1, a=1/2, making 5a+b=2 become:
2.5+b=2, b=-1/2, making a+b+c=3 become:
1/2-1/2+c=3, c=3
So the quadratic equation that produces that sequence is:
y=x^2/2-x/2+3 or more neatly
y=(x^2-x+6)/2, so the 8th term is:
y(8)=(8^2-8+6)/2
y(8)=(64-8+6)/2
y(8)=62/2
y(8)=31
