Answer:
The acceleration of the object is 9.3 m/s²
Explanation:
For a straight movement with constant acceleration, this equation for the position applies:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:
x₁ = x0 + v0 t + 1/2 a (1 s)²
x₂ = x0 + v0 t + 1/2 a (2 s)²
x₂ - x₁ = 14 m
we know that the object starts from rest, so v0 = 0
substracting both equations of position we will get:
x₂ - x₁ = 14
x0 + v0 t + 1/2 a (2 s)² - (x0 + v0 t + 1/2 a (1 s)²) = 14 m
x0 + v0 t + 2 a s² - x0 -v0 t - 1/2 a s² = 14 m
2 a s² - 1/2 a s² = 14 m
3/2 a s² = 14 m
a = 14 m / (3/2 s²) = <u>9.3 m/s² </u>
Answer:
Scientific models have two basic types. FALSE.
Hoped I helped
Answer:
Pressure = 24000 N/m²
Explanation:
Given the following data;
Force = 120N
Area = 0.5cm² to meters = 0.5/100 = 0.005 m²
To find the pressure;
Pressure= force/area
Pressure= 120/0.005
Pressure = 24000 N/m²
Therefore, the pressure on the archer's fingers is 24000 Newton per meters square.
Answer:
The speed of the ball B is 6.4 m/s. The direction is 50 degrees counterclockwise.
Explanation:
Assuming the collision is elastic, use the conservation of momentum to solve this problem. The conservation law implies that:
(the total momentum of the two balls is the same before (index 0) and after (index 1) the collision). Since B is stationary and A and B have the same mass, this simplifies to:
and allows us to determine the velocity of ball B after the collision:
The above involves vectors. Your problem suggests to use the component method, which I am assuming means solving the above equation separately along the x and y axes. Define x to align with the original line of motion of the ball A before the collision, and y to be perpendicular to x, pointing up:
We just need to compute the x- and y-components of the known velocity of the ball A. Drs. Sine and Cosine come to help here.
so
The speed of the ball B is 
. The direction (angle from horizontal) is 
, i.e., 50 degrees counterclockwise.
