Answer:
Explanation:
Refer to the attachment.
What is the weight in newtons of lily’s convertible, which has a mass of 1800 kg?
1 answer:
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The question is incomplete. The complete question is :
The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.
A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.
Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?
Body 3 + Engine 1
Body 2 + Engine 2
Body 1 + Engine 2
Body 1 + Engine 1
Solution :
Given :
Body Mass (gram) Engine Force (newtons)
1 500 1 25
2 1500 2 20
3 750 3 30
The body 1 has a mass of 500 gram which is equal to 0.5 kg
And engine 2 has a force of 20 newtons.
We know that according to Newton's laws of motion,
Force = mass x acceleration
20 = 0.5 x acceleration
Acceleration
Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of at the start of the launch.
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s