Question

a satellite is in orbit about earth. its orbital radius is ( 7.9\times10^{7} ) m. the mass of the satellite is 2523 kg and the mass of earth is ( 5.974\times10^{24} ) kg. determine the orbital speed of the satellite in mi/s. ( 1 ) mi/s ( = 1609 ) m/s. mi/s

1. - / 1 points 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

1. - / 3 points 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. ( mathrm{ke}= ) j ( mathrm{pe}= ) j ( p= ) kg - m/s
2. - / 2 points 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 \(E_{initial}\) is the sum of the initial energies of all objects.

$$E_{initial}=174 + 197+192 + 192$$

$$E_{initial}=755\space J$$

Step2: Calculate the sum of the energies of objects A, B and C after 298 s

Let \(E_{A}\), \(E_{B}\) and \(E_{C}\) be the energies of objects A, B and C after 298 s.

$$E_{A}+E_{B}+E_{C}=49 + 109+32$$

$$E_{A}+E_{B}+E_{C}=190\space J$$

Step3: Calculate the energy of object D

By the law of conservation of energy in a closed system (\(E_{initial}=E_{final}\)), if \(E_{D}\) is the energy of object D after 298 s, then \(E_{D}=E_{initial}-(E_{A}+E_{B}+E_{C})\)

$$E_{D}=755 - 190$$

$$E_{D}=565\space J$$

Step1: Calculate the kinetic energy (KE)

The formula for kinetic energy is \(KE=\frac{1}{2}mv^{2}\), where \(m = 8.9\space kg\) and \(v = 3.6\space m/s\)

$$KE=\frac{1}{2}\times8.9\times(3.6)^{2}$$

$$KE = 57.456\space J$$

Step2: Calculate the 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 \(h = 47\space m\)

$$PE=8.9\times9.8\times47$$

$$PE=4100.74\space J$$

Step3: Calculate the momentum (p)

The formula for momentum is \(p = mv\), where \(m = 8.9\space kg\) and \(v = 3.6\space m/s\)

$$p=8.9\times3.6$$

$$p = 32.04\space kg\cdot m/s$$

Step1: Define the initial and final momenta

Let the north - direction be positive. Initial momentum of object A, \(p_{A1}=24\space kg\cdot m/s\), initial momentum of object B, \(p_{B1}=- 61\space kg\cdot m/s\) (south - direction is negative). Final momentum of object A, \(p_{A2}=-41\space kg\cdot m/s\)

Step2: Apply the law of conservation of momentum (\(p_{1}=p_{2}\))

\(p_{A1}+p_{B1}=p_{A2}+p_{B2}\)
\(p_{B2}=p_{A1}+p_{B1}-p_{A2}\)
Substitute the values:

$$p_{B2}=24+( - 61)-(-41)$$

$$p_{B2}=24 - 61 + 41$$

$$p_{B2}=4\space kg\cdot m/s$$

Since \(p_{B2}>0\), the direction is north.

Answer:

\(565\)

Problem 8