The velocity of the apple right before impact with the ground is approximately 26.2005 m/s, while its initial velocity was about 25.8235 m/s.
So, the final velocity (Vf) equals 26.2005 m/s,
and the initial velocity (Vi) equals 25.8235 m/s.
This difference in velocity arises because the apple was thrown from a starting height of 1 meter.
Answer:
Explanation:
Given,
Voltage of the primary coil (V_p) = 30 kV-rms
Voltage of the secondary coils (V_s) = 345 kV-rms
number of turns in the primary coil (n_p) = 80 turns
number of turns in the secondary coil (n_s) =?
the ratio of turns between primary and secondary coils
The number of turns in the secondary coil is equal to
1) C 2) E. Upon reviewing, the explanatory text reveals that identical satellites X and Y, with mass m, orbit a planet of mass M. Satellite X is at an orbital radius of 3R, while satellite Y is at 4R. The gravitational potential energy of the X system, in terms of Kx, is determined as (A) -2Kx, (B) -Kx, (C) -Kx/2, (D) Kx/2, or (E) 2Kx. Moreover, upon moving satellite X to the same orbit as satellite Y through external work, the work done in relation to Kx is evaluated as (A) -Kx/2, (B) -Kx/4, (C) 0, (D) +Kx/4, or (E) +Kx/2. A detailed analysis of each question follows.
The mass is m=4.03 kg. According to the provided data, we have r = 1.2 m, I = 24 kg·m², N₁ = 40 rpm converting to ω₁= 4.1 rad/s. For N₂ = 32 rpm, ω₂ = 3.3 rad/s. Let's denote the mass of the clay as m kg. The initial angular momentum L₁ is calculated as L₁ = Iω₁, while the final angular momentum L₂ is expressed as L₂= I₂ ω₂ with I₂ indicating the moment of inertia after adding the clay, computed as I₂ = I + m r². Since no external torque is introduced, angular momentum will be conserved, leading to L₁ = L₂. Rearranging gives us Iω₁ = I₂ω₂, or Iω₁ =( I + m r²)ω₂. By inserting the known values of I, r, and ω into the equation, we solve for m, yielding m=4.03 kg.
Given data:
bullet mass (m) = 25 g = 0.025 kg,
gun mass (M) = 0.9 kg,
bullet speed (v) = 230 m/s,
gun speed (V) =?
From the information, it is evident that momentum is conserved. Following the "law of conservation of momentum", the overall momentum pre- and post-collision is equal.
For this instance, the momentum prior to the collision (bullet + gun) is zero.
Thus, after the bullet is fired from the gun, the momentum must remain zero.
Mathematically,
M × V + m × v = 0
where,
M × V = momentum of the gun
m × v = momentum of the bullet
(0. 9 × V) + (0.025 × 230) = 0
0.9 V = -5.75
V = -5.75/0.9
= -6.39 m/s
The gun recoils at a speed of 6.39 m/s
