Answer:
175 N/m
Explanation:
Given:
Force = F= 14.0 N
Distance = x = 8.00 cm = 0.08 m
To find:
spring constant
Solution:
spring constant is calculated by using Hooke's law:
k = F/x
Putting the values in above formula:
k = 14.0 / 0.08
k = 175 N/m
The answer is 7 (its like the only question thats easy so far lol)
Answer:
vₓ = 0.566 m / s, W_total = 9.1 J
Explanation:
This exercise is a parabolic type movement, for the x axis where there is no acceleration
x = v t
vₓ = x / t
vₓ = 0.34 / 0.6
vₓ = 0.566 m / s
the work done is
X axis
In this axis there is no acceleration, therefore the sum of the forces is zero and since the work is the force times the distance, we conclude that the lock in this axis is zero.
W₁ = 0
Y axis
in this axis the force that exists is the force of gravity, that is, the weight of the body
W₂ = Fy y
W₂ = mg and
W₂ = m 9.8 0.70
W₂ = m 9.1
the work is a scalar for which we have to add the quantities obtained
W_total = W₁ + W₂
W_total = 0 + 9.1 m
W_total = 9.1 m
In order to complete the calculation, the mass of the body is needed if we assume that the mass is m = 1
W_total = 9.1 J
Whenever the motion of an object <em><u>changes</u></em> . . . speeding up, or slowing down,
or changing direction . . . that change is called "<em><u>acceleration</u></em>". Acceleration is
produced by force on the object.
If there is <em><u>no force</u></em> on the object, then there is no acceleration. That means that
its motion <em><u>doesn't change</u></em>. The object remains in constant, uniform motion ...
moving with steady speed, in a straight line.
No force is necessary to <em><u>keep</u></em> an object moving, only to <em><u>change</u></em> its motion.
Answer:
V=14.9 m/s
Explanation:
In order to solve this problem, we are going to use the formulas of parabolic motion.
The velocity X-component of the ball is given by:
The motion on the X axis is a constant velocity motion so:
The whole trajectory of the ball takes 1.48 seconds
We know that:
Knowing the X and Y components of the velocity, we can calculate its magnitude by:
