Suppose you are navigating a spacecraft far from other objects. The mass of the spacecraft is 2.6 × 104 kg (about 26 tons). The rocket engines are shut off, and you're coasting along with a constant velocity of <0, 20, 0> km/s. As you pass the location <7, 4, 0> km you briefly fire side thruster rockets, so that your spacecraft experiences a net force of <5 × 105, 0, 0> N for 24.5 s. The ejected gases have a mass that is small compared to the mass of the spacecraft. You then continue coasting with the rocket engines turned off. Where are you an hour later? (Think about what approximations or simplifying assumptions you made in your analysis. Also think about the choice of system: what are the surroundings that exert external forces on your system?)\

Answers

Answer 1

56 will be the answer all you do is divide 104 to 26 then divide 2.6


Related Questions

An air-track glider attached to a spring oscillates between the 15.0 cm mark and the 68.0 cm mark on the track. The glider completes 7.00 oscillations in 31.0 s . What is the period of the oscillations?

Answers

Answer:

4.42 s

Explanation:

The frequency of the oscillation is given by the ratio between the number of complete oscillations and the time taken:

[tex]f=\frac{N}{t}[/tex]

where for this glider, we have

N = 7.00

t = 31.0 s

Substituting, we find

[tex]f=\frac{7.00}{31.0 s}=0.226 Hz[/tex]

Now we now that the period of oscillation is the reciprocal of the frequency:

[tex]T=\frac{1}{f}[/tex]

So, substituting f = 0.226 Hz, we find:

[tex]T=\frac{1}{0.226 Hz}=4.42 s[/tex]

Final answer:

The period of oscillations for the air-track glider is found by dividing the total time of 31.0 seconds by the number of oscillations, which is 7.00, resulting in a period of approximately 4.43 seconds.

Explanation:

The question involves finding the period of oscillations for an air-track glider attached to a spring. To calculate the period, we use the formula for the period T, which is T = total time / number of oscillations. Given that the glider completes 7.00 oscillations in 31.0 seconds, we can calculate the period by dividing the total time by the number of oscillations.

The calculation would be as follows: T = 31.0 s / 7.00 oscillations, which equals approximately 4.43 seconds per oscillation.

Why aren't homes wired in series? A. Because it's too expensive. B. Parallel circuits are too complex for home wiring. O C. Because series circuits too easily cause fires. D. If a home were wired in series, every light and appliance would have to be turned on in order for any light or appliance to work.

Answers

Answer:

D. If a home were wired in series, every light and appliance would have to be turned on in order for any light or appliance to work.

Explanation:

In a series circuit, all the appliances are connected on the same branch of the circuit, one after the other. This means that the current flowing throught them is the same. However, this means also that if one of the appliance is turned off (so, its switch is open), that appliance breaks the circuit, so the current can no longer flow through the other appliances either.

On the contrary, when the appliances are connected in parallel, they are connected in different branches, so if one of them is switched off, the other branches continue working unaffacted by it.

Answer: D.

Explanation:

a p e x

A bungee jumper who has a mass of 80.0 kg leaps off a very high platform. A crowd excitedly watches as the jumper free-falls, reaches the end of the bungee cord, then gets “yanked” up by the elastic cord, again and again. One observer measures the time between the low points for the jumper to be 9.5 s. Another observer realizes that simple harmonic motion can be used to describe the process because several of the subsequent bounces for the jumper require 9.5 s also. Finally, the jumper comes to rest a distance of 40.0 m below the jump point. Calculate (a) the effective spring constant for the elastic bungee cord and (b) its unstretched length.

Answers

Answer:

(a) 35.0 N/m

Explanation:

Do two objects (two large hunks of lead suspended from two strong strings) exert a gravitational force on each other?

Yes
No
Not enough information to determine

Answers

Answer:

Yes

Explanation:

Gravitational force is an attractive force exertedn between every object that has mass.

The magnitude of the gravitational force between two objects is given by:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r the distance between the objects

In this problem, the two objects are two large hunks of lead. Since these objects have mass, they exert an attractive, gravitational force on each other.

1. Compare and contrast the two kinds of waves.
2.Draw a wave, label the 4 parts, and provide a description of each.
3.Draw a standing wave and label the nodes and antinodes.

Answers

Answer:

Look at the diagrams for 2 and 3.

Explanation:

1. There are two ways to categorize waves.

Direction of particles of the wave:

If you need to differentiate them based on direction of particles of the waves then you have either longitudinal or transverse.

Particles of the medium of longitudinal waves move parallel to the direction or movement of the wave.  On the other hand, transverse waves are waves where the particles of the medium it travels through move perpendicular to the motion of the wave.  

Ability to transmit energy through a medium or vacuum

You have the mechanical wave and the electromagenetic wave (em wave).

The main difference between these two is that mechanical waves travel through a medium. Basically, they need the molecules in the medium, which collide or bump into each other to pass on the energy. An example would be sound waves.

Electromagnetic waves differ because they do not need a medium. They can travel through a vacuum. Like light waves.

2.

Crest - It is the displacement of a wave in the upwards direction. In short it is the peak or the highest point of a wave.  

Trough - It is the opposite of the crest, so it is the displacement of a wave going downwards. To put it shortly, it is the depth or lowest point of a wave.

If you will get the distance between the crest and trough, you will see that it is twice the measure of the amplitude, which you will be defined later on.

Wavelength - is the distance between two crests or two troughs of two consecutive waves. It is measured in meters and goes with the direction of the wave.

Amplitude - height or depth of the crest or trough from the rest position. It is also measured in meters. It is defined as the displacement of the wave from the rest position or point.

Look at image B, to see the different parts.

3.

Standing waves are waves that vibrate vertically and have the same frequency and amplitude.

Nodes are points in the wave where the amplitude is equal to zero or at their resting point. Antinodes are points in the wave where the amplitudes are at their maximum.

Look at image C.

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×104 V/m . (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

Answers

(a) 3000 V

For two parallel conducting plates, the potential difference between the plates is given by:

[tex]\Delta V=Ed[/tex]

where

E is the magnitude of the electric field

d is the separation between the plates

Here we have:

[tex]E=7.50\cdot 10^4 V/m[/tex] is the electric field

d = 4.00 cm = 0.04 m is the distance between the plates

Substituting,

[tex]\Delta V=(7.50\cdot 10^4 V/m)(0.04 m)=3000 V[/tex]

(b) 750 V

The potential difference between the two plates A and B is

[tex]\Delta V = V_B - V_A = 3000 V[/tex]

Let's take plate A as the plate at 0 volts:

[tex]V_A = 0 V[/tex]

The potential increases linearly going from plate A (0 V) to plate B (3000 V).

So, if the potential difference between A and B, separated by 4 cm, is 3000 V, then the potential difference between A and a point located at 1 cm from A is given by the proportion:

[tex]3000 V : 4 cm = V(1 cm) : 1 cm[/tex]

and solving for V(1 cm) we find:

[tex]V(1 cm)=\frac{(3000 V)(1 cm)}{4 cm}=750 V[/tex]

Potential difference is the difference in electrical potential. when the lowest potential is taken to be at zero volts then the potential 1.00 cm from that plate is 750 V.

What is the potential difference?

A potential difference is defined as the difference in the electrical potential between two points.

Given to us

Distance between the plate, d = 4 cm = 0.04 m

Electric field, E = 7.50×104 V/m .

A.) We know that for two parallel conducting plates, the potential difference between the plate is given as,

[tex]\triangle V = Ed[/tex]

E = Magnitude of the electric field

d = distance between the plates

Substitute the values,

[tex]E = 7.5 \times 10^4 \times 0.04\\\\E = 3000\rm\ V[/tex]

B.) We already know the potential difference between the two plates, therefore,

The potential difference between the two plates,

[tex]\triangle V= V_A - V_B[/tex]

Since, one of the plates is having zero potential, therefore,

[tex]\triangle V= V_A = 3000\rm\ V[/tex]

We know that the potential difference between the two plates is 300 which are 4 cm apart, therefore,

[tex]\dfrac{\triangle V}{d} = \dfrac{V}{1}\\\\V = 750 \rm\ V[/tex]

Hence, when the lowest potential is taken to be at zero volts then the potential 1.00 cm from that plate is 750 V.

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An atom emits a photon when one of its electrons:

collides with another of its electrons
exchanges quantum states with another of its electrons
undergoes a transition to a quantum state of lower energy
undergoes a transition to a quantum state of higher energy

Answers

Answer:

undergoes a transition to a quantum state of lower energy

Explanation:

When electrons in an atom move to another quantum state, they emit/absorb a photon according to the following:

- If the electron is moving to a higher energy state, it absorbs a photon (because it needs energy to move to a higher energy level, so it must absorb the energy of the photon)

- if the electron is moving to a lower energy state, it emits a photon (because it releases the excess energy)

In particular, the energy of the absorbed/emitted photon is exactly equal to the difference in energy between the two levels of the electron transition:

[tex]E=|E_1 - E_2|[/tex]

An atom emits a photon when one of its electrons undergoes a transition to a quantum state of lower energy. Hence, the correct option is 3.

This happens when an electron in an excited energy state returns to a lower energy state or the ground state. The energy difference between these states is released as a photon, which corresponds to the energy of the gap between the two levels.

To summarize, the emission of a photon occurs due to:

Transition of an electron from a higher energy level to a lower energy level.The release of excess energy in the form of a photon.

In contrast, if an electron moves from a lower energy level to a higher one, it must absorb a photon, not emit it.

Which of the following is emitted from the nucleus?

alpha particle
beta particle
gamma Ray
photoelectron

Answers

Answer:

gamma Ray

Explanation:

Stable or unstable nucleus will have high energy in form of nuclear energy stored in it.

When this energy exist in the unstable state of the nucleus then in order to attain the stability the nucleus will will go to lower energy state.

This energy change is of higher order due to which energy released is of large order and high frequency electromagnetic waves.

so it is given as

[tex]E_2 - E_1 = h\nu[/tex]

since this energy is of higher range so it is given as gamma rays.

so correct answer will be

Gamma Rays

Answer:

Gamma Ray

Explanation:

Gamma Ray is produced furring nuclear decay which occurs in the nucleus of the atom.

Other radiations are produced from outside the nucleus.

Gamma Ray has the shortest wavelength and the highest penetrating power and ionises matter.

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball when it reaches its highest point? (b) What is its velocity 1 s before it reaches its highest point? (c) What is the change in its velocity during this 1-s interval? (d) What is its velocity 1 s after it reaches its highest point? (e) What is the change in velocity during this 1-s interval? (f) What is the change in velocity during the 2-s interval? (Careful!) (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity

Answers

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

[tex]K=\frac{1}{2}mv^2[/tex]

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

[tex]U=mgh[/tex]

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

[tex]a=\frac{v-u}{t}[/tex]

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

[tex]u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s[/tex]

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

[tex]\Delta v = v -u[/tex]

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

[tex]\Delta v = 0 - (+9.8 m/s)=-9.8 m/s[/tex]

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

[tex]a=\frac{v-u}{t}[/tex]

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

[tex]v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s[/tex]

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

[tex]\Delta v = v -u[/tex]

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

[tex]\Delta v = -9.8 m/s - 0=-9.8 m/s[/tex]

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

[tex]\Delta v = v -u[/tex]

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

[tex]\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s[/tex]

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

[tex]F=mg[/tex]

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

[tex]mg = ma[/tex]

which means that the acceleration is

[tex]a= g = -9.8 m/s^2[/tex]

and the negative sign means it points downward.

Final answer:

The ball's velocity changes in intervals of 9.8 m/s before reaching, at, and after its highest point due to gravity. The acceleration remains constant at -9.8 m/s² at all time intervals and even when the velocity is zero at the highest point.

Explanation:

(a) When the ball reaches its highest point, its velocity is 0 m/s. It momentarily stops before reversing its direction and falling down.

(b) The ball's velocity 1 second before it reaches its highest point is determined by the acceleration due to gravity, which on Earth is approximately -9.8 m/s². Thus, the ball's velocity at this time would be 9.8 m/s.

(c) The change in velocity during this 1-s interval is 9.8 m/s, calculated by the difference in velocity between the highest point and 1 second before.

(d) The ball's velocity 1 second after it reaches its highest point would likewise be -9.8 m/s (as the ball is now moving in the opposite direction).

(e) The change in velocity during this 1-s interval is the same as in part (c), 9.8 m/s.

(f) The change in velocity during the total 2-s interval is again 9.8 m/s. This is because the ball's velocity changes at the constant rate of 9.8 m/s each second due to the constant acceleration of gravity.

(g) The acceleration of the ball at any of these time intervals and at the moment the ball has zero velocity is constant and equal to the acceleration due to gravity, -9.8 m/s².

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