Answer:
Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.
Explanation:
Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.
Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.
The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.
Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].
The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].
The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:
[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].
The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:
[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:
[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:
[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].
[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.
[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].
[tex]9995\, \Delta v = 25000[/tex].
[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].
Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.
A racing car on the straight accelerates from 100 km/h to 316 km/h in three seconds.
What is its acceleration?
40m/s2
30m/s2
20m/s2
72m/s2
Answer:
[tex]20m/s^2[/tex]
Explanation:
Solution is attached. I apologize if it is a little messy.
Two objects are a distance of 1.7 x 103 meters apart. One object has a mass of 3 x 107 kg and the other has a mass of 6 x 108. Determine the gravitational force between the objects.
Answer: You need to use Newton's law for the equation --->
Explanation: G × M × m / separation. Thats how youll get your answer !!
1. How far does a train travel in the fourth second if it starts from rest with a uniform
acceleration of 2.0 m/s2?
Answer:
Explanation:
d = ½(2.0)4² - ½(2.0)3² = 7 m
F=mass x what does this equal?
Answer:
Force = mass × acceleration.
According to the table what was the hikers total displacement the graph has 4km 6km 4km 6km
Answer:
0
Explanation:
0 is the answer
A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3 meters.
What is the SPE?
22.5 J
Explanation:
Given:
x = 3 m
[tex]k = 5\:\text{N/m}[/tex]
The spring potential energy [tex]PE_s[/tex] is
[tex]PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2[/tex]
[tex]\:\:\:\:\:\:\:=22.5\:\text{J}[/tex]
Which of the following can cause an object to accelerate?
Select one:
a. Force
b. Inertia
c. Mass
d. Kinetic Energy
I think force can accelerate
Explanation:
i think force of an object
a coconut falls from the top of a tree and takes 3.5 seconds to reach the ground. How tall is the tree?
Hello!
To solve, we can begin by using the kinematic equation:
[tex]d = v_it + \frac{1}{2}at^2[/tex]
Where:
vi = initial velocity (m/s)
t = time (s)
a = acceleration (in this case, due to gravity. g = 9.8 m/s²)
Since the object falls from rest, the initial velocity is 0 m/s.
[tex]d = \frac{1}{2}at^2[/tex]
Plug in the given values:
[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]
Objects 1 and 2 attract each other with a gravitational force of 34 units. If the distance separating objects 1 and 2 is changed to one-third the original value, then the new gravitational force will be ____ units.
Answer:
F12 = G M1 M2 / R12^2
F12' = G M1 M2 / R12'^2
F12' / F12 = R12'^2 / R12^2 = (1/3)^2
F12' = 1/9 F12
The new force is 1/9 the of the old force
Convert 6 picoseconds into seconds.
Answer:
6e-12
Explanation:
divide the time value by 1e+12
What causes an astigmatism?
A. damaged lens
B. retina not focusing the image
C. cornea being wavy or not spherical
D. sclera not refracting light properly
Answer:
c) cornea being wavy or not spherical
PLSSS HELP IT'S VERY IMPORTANT FOR ME WILL MAKE AS BRAINLIEST
Fill in the blank
Valence electrons are______
"The answer is not electrons in the outer shells that are not filled!!"
Answer:
negatively charged particle
valence electron is a electron of an atom, located in the outermost shell (valence shell ) of the atom, that can be transferred to or shared with another atom.
Why is it important for the community to take action
Answer:
Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself
Make one comparison between the moral condition of the world at the time of the Flood with our day. Only One Short explanation.
The moral condition of the world today appears to be worse than it was in the antediluvian era.
The biblical account of the flood records that the world delved into apostasy in the days of Noah so much so that God regretted the fact that he created man. Some of the evils of the antediluvian world include; sodomy, drunkenness, lewdness and several forms of immorality.
We can see that these vices that led to the destruction of the world due to moral bankruptcy in the antediluvian era is still very much prevalent in our world today. The moral condition of the world today appears to be worse than it was in the antediluvian era.
Learn more: https://brainly.com/question/8646601
2. Two equipotential lines are separated by a distance of 2.17 cm. If the potential values of the lines are 5.9 volts and 8.6 volts, what would the strength of the electric field between the lines be
Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?
Answer:
d = -33.1 m and Vf = -25.5 m/s
Explanation:
Given:a = -9.8 m
t = 2.6 s
Vᵢ = 0 m/s
To Find:d = ?
Vf = ?
Now,
d = Vᵢ × t + 0.5 × a × t²
d = (0 m/s) × (2.60 s) + 0.5 × (-9.8 m/s²) × (2.60 s)²
d = -33.1 m (- indicates direction)
Vf = Vᵢ + a × t
Vf = 0 + (-9.8 m/s²) × (2.60 s)
Vf = -25.5 m/s (- indicates direction)
Thus, d = -33.1 m and Vf = -25.5 m/s
-TheUnknownScientist 72
in rotational motion the angular frequency of all the particles is
Answer:
L=mvr
α=0
v=ωr
L=mωr
2
2
1
mv
2
=k
L=Iω
k=
2
1
Iω
2
ω
1
=2ω
k
1
=1/2k
∴I
1
=
8
1
I
L
1
=
8
1
I×2ω
=
4
1
Iω
=L/4
Physical motion that occurs when an item rotates or spins on an axis is known as “rotatory motion,” sometimes known as “rotational motion” or “circular motion.”
What angular frequency of particles in rotational motion?Angular displacement per unit of time is measured by angular frequency, also referred to as radial or circular frequency (). Therefore, it has degrees (or radians) per second as its units. 1 Hz = 6.28 rad/sec, thus. 1 radian equals 57.3° because 2 radians equal 360°.
The frequency of rotation is the rate at which an item rotates around an axis, sometimes referred to as rotational speed or rate of rotation.
However, the linear velocity also depends on how far the particles are from the axis of rotation. Because every particle distances are different, so too will the linear velocities for each particle.
Therefore, All the particles in a rigid body's rotational motion move in the same direction in a given interval. Therefore, all the particles' angular velocities will be the same.
Learn more about angular frequency here:
https://brainly.com/question/29107224
#SPJ2
When a 25000-kgkg fighter airplane lands on the deck of the aircraft carrier, the carrier sinks 0.30 cmcm deeper into the water.
What is the meaning of eddy currents in electromagnetic series?
Answer:
currents which circulate in conductors like swirling eddies in a stream in electromagnetic series....
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.44 m/s2. Determine the orbital period of the satellite.
Explanation:
The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as
[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]
where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.
Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write
[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]
[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]
where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is
[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]
[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]
Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,
[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]
or
[tex]4\pi^2r = gT^2[/tex]
Solving for T, we get
[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]
We can further simplify the above expression into
[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]
Plugging in the values for r and g, we get
[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]
What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1000 m/sm/s
Explanation:
Since the neutron is only moving at 1000 m/s, we are going to ignore the relativistic effects on its mass and energy. The mass of a neutron in [tex]m_n = 1.67×10^{-27}\:\text{kg}[/tex] so its kinetic energy KE is
[tex]KE = \frac{1}{2}m_nv^2[/tex]
[tex]\:\:\:\:\:\:\:\:= \frac{1}{2}(1.67×10^{-27}\:\text{kg})(10^3\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:\:= 8.35×10^{-22}\:\text{J}[/tex]
A photon's energy E is defined as
[tex]E = h\nu[/tex]
where [tex]\nu[/tex] is the photon's frequency and h is the Planck's constant. Solving for the frequency, we get
[tex]\nu = \dfrac{E}{h} = \dfrac{8.35×10^{-22}\:\text{J}}{6.63×10^{-34}\:\text{J-s}}[/tex]
[tex]\:\:\:\:\: = 1.26×10^{12}\:\text{Hz}[/tex]
which is right around the infrared radiation range.
A metal wire of length 1.2 m and cross-sectional area 2.0 x 10 raised to power-7 m 2 is stretched by a force of 50 N. assuming the force constant of the metal is 6000 Nm-1. Calculate the tensile stress
L=1m
=2mm²=(2/1000²)=2(10^-6)m
y=4x10¹¹N/m²
∆l=2mm=2/10.00=0.002m
=(4x10¹¹x2x10^-6x2x10^-3)÷1m
=16x10¹¹-⁶-³
=16x10¹¹-⁹
=16x10²
=1600N
where a=cross sectional area=2x10^-6m
C=2mm= 2x10^-3m
L=1m
hope it helps
Nikolai is using a hand-operated grain mill to grind wheatberries into flour. The mill is operated by spinning a fly-wheel with radiusR= 23 cm, which has a handle attachedto the outer edge. After grinding for a few minutes at a con-stant angular speedωi, he lets go of the handle and allows themechanism to come to rest as it undergoes constant angularacceleration. This happens over the course oft= 0.50 s, andthe flywheel undergoes a quarter of a rotation during this time.What is the linear tangential accelerationaof the handle as itcomes to rest? For the limits check, investigate what happenstoaas the time required to stop the flywheel becomes small(t→0).
Answer:
Explanation:
α = Δω/t = (0 - ωi)/0.50 = -2ωi rad/s²
ωf² = ωi² + 2αθ
θ = (ωf² - ωi²) / 2α
2π/4 = (0² - ωi²) / (2(-2ωi))
2π/4 = ωi / 4
ωi = 2π rad/s
α = -2(2π) = -4π rad/s²
a = rα = 0.23(-4π) = 0.92π m/s² ≈ -2.89 m/s²
as the time to stop approaches zero, acceleration goes toward infinity.
MCQ
A body of mass 5kg is pushed for distance x with accleration a. Then workdone against static friction is
1.ma*X cosB
2.ma*X sinB
3.zero
4.ma/X
Answer:
ma*XsinB
option 2 is correct
What do alcohol, drugs, and tobacco all have in common?
All have some medicinal value.
All are harmful to the body.
All are depressants.
All are stimulants.
Answer:
all are harmful to the body
PLEASE HELP I DONT GET THISS
Answer:
I feel like its the second one but I'm not completely sure..
Explanation:
The average normal body temperature measured in the mouth is 310 K. What would Celsius thermometers read for this temperature?
54.1°C
23.8°C
36.9°C
42.7°C
________
= 310 - 273
= 37°C
Actually,310 Kelvin is same with 37°C, and as you see, there is no 37°C
So, The Nearest Number To 37°C is 36,9°C
Answer:
36.9
Explanation:
Plato
A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg of the wateris in the liquid state. What is the pressure in the vessel, and what is the volume of the vessel
In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).
Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:
[tex]x=\frac{m_{steam}}{m_{total}}[/tex]
Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:
[tex]x=\frac{2kg}{10kg} =0.20[/tex]
Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:
[tex]v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg[/tex]
This means that the volume of the container will be:
[tex]V=10kg*0.4727m^3/kg\\\\V=4.73m^3[/tex]
Learn more:
https://brainly.com/question/23339302https://brainly.com/question/15081976At the molecular level, as the kinetic energy increases, what happens to the temperature?
decreases
increases
stays the same
Answer: increases
Explanation:
Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.
What is the birth rate of a population of 3000 chipmunks if 200 chipmunks are born each year?
o 0.022 births per chipmunk per year.
0.067 births per chipmunk per year.
O 0.82 births per chipmunk per year.
o 15 births per chipmunk per year.
Answer:
0.067 births per chipmunk per year
Explanation:
I took the test