Answer:
The position of the particle is -2.34 m.
Explanation:
Hi there!
The equation of position of a particle moving in a straight line with constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the particle at a time t:
x0 = initial position.
v0 = initial velocity.
t = time
a = acceleration
We have the following information:
x0 = 0.270 m
v0 = 0.140 m/s
a = -0.320 m/s²
t = 4.50 s (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).
Then, we have all the needed data to calculate the position of the particle:
x = x0 + v0 · t + 1/2 · a · t²
x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²
x = -2.34 m
The position of the particle is -2.34 m.
Answer:
35.7 m
Explanation:
Let


We have to find the distance between Joe's and Karl'e tent.


Substitute the values then we get




Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.
By triangle addition of vector






Hence, the distance between Joe's and Karl's tent=35.7 m
Answer: Choose the normal force acting between the object and the ground. Let's assume a normal force of 250 N.
Determine the friction coefficient.
Multiply these values by each other: 250 N * 0.13 = 32.5 N .
You just found the force of friction!
Explanation:
Answer:
huh? do you need help on math?
Explanation:
what do you mean?