Answer:D
Explanation: i found the answer in an answer key!
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
Answer:
Potential energy
Explanation:
Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.
Answer:
The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.
Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis
