Question

1. - / 1 points details my notes ask your teache a satellite is in orbit about earth. its orbital radius is 7.9×10⁷ m. the mass of the satellite is 2523 kg and the mass of earth is 5.974×10²⁴ kg. determine the orbital speed of the satellite in mi/s. 1 mi/s = 1609 m/s. mi/s submit answer 7. - / 1 points details my notes ask your teache a closed system consists of 4 objects. the table below shows how much energy each object had to start. after 298 s have passed, object a has 49 j, object b has 109 j, and object c has 32 j. how much energy does object d have? j submit answer 8. - / 3 points details my notes ask your teache a 8.9 kg bird is flying 47 m above the ground at a speed of 3.6 m/s. calculate the ke, pe, and momentum of the bird. ke = j pe = j p = kg·m/s submit answer 9. - / 2 points details my notes ask your teache a closed system consists of 2 objects. initially, object a has a momentum of 24 kg·m/s north and object b has a momentum of 61 kg·m/s south. the two objects have a head - on collision. afterwards, object a is observed to have a momentum of 41 kg·m/s south. after the collision, what is object bs momentum? kg·m/s east west up down north south left right no direction because object b has stopped

Explanation:

Problem 7

Step1: Calculate the total initial energy

The total initial energy of the system is the sum of the initial energies of all objects.
$$174 + 197+192 + 192=755\space J$$

Step2: Calculate the sum of the known final energies

The sum of the known final energies of objects A, B, and C is \(49+109 + 32\).
$$49+109+32 = 190\space J$$

Step3: Find the energy of object D

Since energy is conserved in a closed system (total initial energy = total final energy), the energy of object D, \(E_D\), is \(755-190\).
$$E_D=755 - 190=565\space J$$

Kinetic Energy (KE)

The formula for kinetic energy is \(KE=\frac{1}{2}mv^{2}\), where \(m = 8.9\space kg\) and \(v=3.6\space m/s\)
$$KE=\frac{1}{2}\times8.9\times(3.6)^{2}=\frac{1}{2}\times8.9\times12.96 = 57.984\space J$$

Potential Energy (PE)

The formula for gravitational potential energy is \(PE = mgh\), where \(m = 8.9\space kg\), \(g = 9.8\space m/s^{2}\), and \(h = 47\space m\)
$$PE=8.9\times9.8\times47=8.9\times460.6=4099.34\space J$$

Momentum (\(p\))

The formula for momentum is \(p=mv\), where \(m = 8.9\space kg\) and \(v = 3.6\space m/s\)
$$p=8.9\times3.6 = 32.04\space kg\cdot m/s$$

Step1: Define the direction

Let north be the positive direction and south be the negative direction. So, \(p_{A_i}=24\space kg\cdot m/s\), \(p_{B_i}=- 61\space kg\cdot m/s\), and \(p_{A_f}=-41\space kg\cdot m/s\)

Step2: Apply the law of conservation of momentum

The law of conservation of momentum states that \(p_{i}=p_{f}\), or \(p_{A_i}+p_{B_i}=p_{A_f}+p_{B_f}\)
We want to solve for \(p_{B_f}\), so \(p_{B_f}=p_{A_i}+p_{B_i}-p_{A_f}\)
Substitute the values: \(p_{B_f}=24+( - 61)-( - 41)\)
$$p_{B_f}=24-61 + 41=4\space kg\cdot m/s$$
Since \(p_{B_f}=4>0\), the direction is north.

Answer:

\(565\)

Problem 8