Answer:
-3.5m/s²
Explanation:
Given
Initial Velocity, u = 30m/s
Final Velocity, v = 23m/s
Time, t = 2.0s
Required
Determine the magnitude of acceleration
This is calculated using first equation of motion.
v = u + at
Substitute values for v, u and t
23 = 30 + a * 2.0
23 = 30 + 2.0a
23 - 30 = 2.0a
-7 =2.0a
Solve for a
a = -7 ÷ 2.0
a = -3.5m/s²
Electricidade is electricity in English
Explanation:
We will calculate the gravitational potential energy as follows.
=
= 1164000 J
or, = 1164 kJ (as 1 kJ = 1000 J)
Now, we will calculate the change in potential energy as follows.
=
=
= -873000 J
or, = -873 kJ
Thus, we can conclude that change in gravitational potential energy is -873 kJ.
Answer:
I'm pretty sure it's the third one where velocity goes from positive to negative
Explanation:
the positive velocity is before the object hits the ground and the negative is after
Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle