Answer:
22.9 minutes
Explanation:
Distance = Speed x Time
We want the time to go 16 miles at 42 miles/hour (mph).
16 miles = (42 mph)*T
T = 16 miles/(42 miles/hr)
Time = 0.381 hours
(60 minutes/hour)*(.381 hours)= 22.9 minutes
Find the coefficient of static friction between the ramp and object with the following
Horizontal length of ramp: 75cm
Vertical length of ramp: 19cm
mass of object: 128g
The coefficient of static friction between the ramp of 75 cm horizontal length and a vertical length of 19 cm and object with a mass of 128 g is 0.25
Since the ramp forms a right angled triangle,
tan θ = Opposite side / Adjacent side
Opposite side = Vertical length of ramp = 19 cm
Adjacent side = Horizontal length of ramp = 75 cm
tan θ = 19 / 75
θ = [tex]tan^{-1}[/tex] ( 0.25 )
θ = 14°
m = 128 g = 0.128 kg
g = 9.8 m / s²
W = m g
W = 0.128 * 9.8
W = 1.25 N
Since the object is in static condition,
∑ [tex]F_{y}[/tex] = 0
N - [tex]W_{y}[/tex] = 0
N - W cos θ = 0
N = 1.25 * cos 14°
N = 1.25 * 0.97
N = 1.21 N
F[tex]_{s}[/tex] = μ[tex]_{s}[/tex] N
F[tex]_{s}[/tex] = Static frictional force
μ[tex]_{s}[/tex] = Co-efficient of static friction
N = Normal force
Since the object is in static condition,
∑ [tex]F_{x}[/tex] = 0
F[tex]_{s}[/tex] - [tex]W_{x}[/tex] = 0
F[tex]_{s}[/tex] = W sin θ
F[tex]_{s}[/tex] = 1.25 * sin 14°
F[tex]_{s}[/tex] = 1.25 * 0.24
F[tex]_{s}[/tex] = 0.3 N
μ[tex]_{s}[/tex] = F[tex]_{s}[/tex] / N
μ[tex]_{s}[/tex] = 0.3 / 1.21
μ[tex]_{s}[/tex] = 0.25
Therefore, the coefficient of static friction between the ramp and object is 0.25
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What element has 4 electrons 3 protons and 6 neutrons
The element that has 4 electrons, 3 protons and 6 neutrons is lithium ion.
What is an element?An element is one of the simplest chemical substances that cannot be decomposed in a chemical reaction or by any chemical means and made up of atoms all having the same number of protons.
The atom of an element is made up of three subatomic particles namely;
ProtonElectronNeutronIn a neutral element, the number of protons and electrons are the same i.e. equal. According to this question, an element has 4 electrons, 3 protons and 6 neutrons.
The element in the periodic table with 3 protons is Lithium. This means that the element is lithium ion because it has gained one electron.
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An 85.0 kg man stands on a spring scale in an elevator when it begins to accelerate upwards. If the scale at that moment indicates that his "weight" is 1020 N, what is the acceleration of the elevator?
2.20 m/s^2 is the acceleration of the elevator.
Option c is correct answered .
What does the scale read when the elevator accelerates upward?If you stand on a scale in an elevator accelerating upward, you feel heavier because the elevator's floor presses harder on your feet, and the scale will show a higher reading than when the elevator is at rest. On the other hand, when the elevator accelerates downward, you feel lighter.
How do you find the acceleration of an elevator with weight?support force F = mass x acceleration + weight
For a mass m= kg, the elevator must support its weight = mg = Newtons to hold it up at rest. If the acceleration is a= m/s² then a net force= Newtons is required to accelerate the mass.
Fnet = m a
Fn + (-Fg) = m a
a = 1020 - (85)(9.8) /85
a = 2.20 m/s^2
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compare and contrast the energy of moving water and the energy of water that is stored.
I need to get it done
The main difference between the energy of moving water and the energy of stored water is that the energy of moving water has kinetic energy whereas the energy of water that is stored has potential energy.
What are kinetic and potential energy?Potential energy is defined as the stored energy in any object due to its position while on the other hand, while Kinetic energy is defined as the energy an object has due to its motion. The energy stored in the water molecules which is used for the movement of these molecules is called water potential. Stored water has potential energy but the kinetic energy is present only in the moving molecules of water.
So we can conclude that moving water has kinetic energy while on the other hand, the energy of water that is stored has potential energy.
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two satellites orbit the earth in stable orbits. satellite a is three times the mass of satellite b. satellite a orbits with a speed vvv at a distance rrr from the center of the earth. satellite b travels at a speed that is greater than vvv . at what distance from the center of the earth does the satellite b orbit?
Satellite B's radius will be less than Satellite A's radius.
m = Satellite Mass
v = Satellite Velocity
r = Orbital radius of the satellite
Relating the gravitational and centripetal forces
mv^2/ r= GmM/r^2
v^2/r=GM/r^2
v^2=GM/r
v = sqrtGM/r
As can be shown, the mass of the satellite has no bearing on the relationship between velocity and radius.
This implies that the radius must shrink for v to increase
Here, vb>va
Therefore, satellite b's radius will be smaller than satellite a's radius.
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I need help with question 7, I just need the answer you don’t have to explain.
Explanation
Step 1
free body diagram
so, for m1
m1= 10 kg
so,
[tex]\begin{gathered} \sum ^{forces}_{\text{ y}}=T_1-mg=ma \\ T_1=m_1(a+g)\Rightarrow equation(1) \\ T_1=10(a+g) \end{gathered}[/tex]and for m2
[tex]\begin{gathered} \sum ^{forces}_{\text{ x}}=m_2g\sin (37)-T_1=m_2a \\ \text{solve for T}_1 \\ T_1=m_2g\sin (37)-m_2a \\ \text{replace} \\ T_1=(3.6\text{ kg)(9.8 }\frac{m}{s^2})\sin 37-3.6a \\ T_1=21.23\text{ -}3.6a\Rightarrow eq(2) \end{gathered}[/tex]Step 2
now, replace in equaiotn (1) and solve for a
[tex]\begin{gathered} T_1=m_1(a+g)\Rightarrow equation(1) \\ 21.23\text{ -}3.6a=10(a+g) \\ 21.23\text{ -}3.6a=10a+10g \\ -7.2a=10a-98.1 \\ -17.2a=-98.1 \\ a=-\frac{98.1}{-17.2} \\ a=5.703\text{ }\frac{m}{s^2} \end{gathered}[/tex]finally, replace in equation (2) to find Tension
[tex]\begin{gathered} T_1=21.23\text{ -}3.6a \\ T_1=21.23\text{ -}3.6(5.703) \\ T_2=3.891\text{N} \end{gathered}[/tex]I hope this helps you
7 The distance-time graph for a motorcyclist riding
off from rest is shown in Figure 1.2.18.
a Describe the motion.
b
Calculate how far the motorbike moves in
30 seconds.
c Calculate the speed.
Answer:
a) Uniform rectilinear motion
b) 600 m
c) 20 m/s
Explanation:
a) average velocity is described as the change of position divided by the time of travel, so v= ∆x/∆t. This is a linear function (y = mx+ q) which is the same structure of the equation needed to calculate average velocity (x2= x1 + vt). If you do the ratio between a distance and the corresponding time you'll see that the average velocity is the same, which is typical of this kind of motion.
b) At the the time of 30 seconds the corresponding distance on the y axis is 600 m.
c) Again, v=∆x/∆t and we can rewrite is as v=x2-x1/t. If you take the same data we've used before you can write v= 600m-0/ 30s= 20 m/s. The average velocity is 20 m/s.
a window-mounted air conditioner removes 3.5 kj from the inside of a home using 1.75 kj work input. how much energy is released outside and what is its coefficient of performance?
The amount of heat released outside is 3.5 kJ and the coefficient of performance is 2.
What is efficiency of a machine?The efficiency of the a machine or any device is the measure of effectiveness of the device. Efficiency describes how a machine converts input energy to output energy without wasting much of the input energy.
Mathematically, efficiency of a machine is given as;
E = output energy / input energy x 100%
The coefficient of performance (COP) of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required.
The energy released outside = 3.5 kJ
COP = 3.5 kJ / 1.75 kJ
COP = 2
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24) A 0.042-kg pet lab mouse sits on a 0.35 kg air-track cart. The cart is at rest, as is a second cart with a mass of 0.25 kg. The lab mouse now jumps to the second cart. After the jump, the 0.35-kg cart has a speed of u1=0.88m/s. What is the speed v2 of the mouse and the 0.25-kg cart? |
The mass of the mouse, m₁=0.042 kg
The mass of the 1st cart, m₂=0.35 kg
The mass of the second cart, m₃=0.25 kg
The speed of the 1st car after the jump, u₁=0.88 m/s
From the law of conservation of momentum, the momentum of a system should always remain the same.
Before the mouse jumps from one cart to another, all the objects were at rest. Therefore the total momentum of the system was zero.
Thus after the mouse jumps, the total momentum of the system should be equal to zero.
Thus the momentum of the second cart and the mouse will be equal in magnitude to the momentum of the first cart but it will be in the opposite direction.
Thus,
[tex]m_2u_1=(m_1+m_3)v_2[/tex]On substituting the known values,
[tex]\begin{gathered} 0.35\times0.88=(0.042+0.25)v_2 \\ \Rightarrow v_2=\frac{0.35\times0.88}{(0.042+0.25)} \\ =1.05\text{ m/s} \end{gathered}[/tex]Thus the speed of the mouse and the second cart after the mouse jumps is 1.05 m/s
now assume that the crate is sliding along the floor with constant velocity. what is the magnitude fp of the force you need to exert on the crate to make it continue to slide along the floor with constant velocity?
The magnitude[tex]F_{P}[/tex] of the force you need to exert on the crate to make it continue to slide along the floor with constant velocity is 437.5N.
How friction force can be calculated?The Normal Force (in a horizontal plane, it is comparable to the mass by gravity) and the Kinetic Coefficient of Friction must be multiplied in order to obtain the Friction Force, which must be used in order to solve this problem.
Given that the sentence does not specify the value of the coefficient of friction, I will ascribe a value of 0.5 between the crate and the ground. If a different value has been specified for this coefficient, the below equation will simply need to be changed to reflect it.
[tex]F_{P}[/tex] = µK N
µK = Friction coefficient
Normal Force is N.
Replacing,
[tex]F_{P}[/tex] = 0.5 × 875
[tex]F_{P}[/tex] = 437.5N.
Consequently, the crate is subject to a friction force of 437.5N.
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Complete question is "Suppose you have to move a heavy crate of weight 875 {\rm N} by sliding it along a horizontal concrete floor. You push the crate to the right with a horizontal force of magnitude 300 {\rm N}, but friction prevents the crate from sliding. Now assume that the crate is sliding along the floor with constant velocity. What is the magnitude F_p of the force you need to exert on the crate to make it continue to slide along the floor with constant velocity?"
Slim Jim, continually maintaining his svelte body, lifts a 70 kg barbell 1.4m above the grounda) How much energy did the barbell have when it was on the ground?b) How much energy does it have after being lifted 1.4m? What kind of energy does it have afterbeing lifted? Where did it come from?c) How much work did Jim do to the lift the object?dIf he lifted it in 1.5s how much power did he use?
Given,
The mass of the barbell, m=70 kg
The height to which Slim Jim lifts the barbell, h=1.4 m
a)
When the barbell was on the ground, it will have zero kinetic energy as it has no velocity. And if assume the height of ground as zero meters, then its potential energy is also zero.
Thus when the barbell was on the ground, its energy was zero joules.
b)
The energy of the barbell when it is at a height of h is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=70\times9.8\times1.4 \\ =960.4\text{ J} \end{gathered}[/tex]Thus the energy that the barbell has after being lifted 1.4 m is 960.4 J
This energy is the energy stored in the barbell due to its position. Thus the energy stored in the barbell is the potential energy.
Slim Jim has to do some work to lift the barbell to the given height. This work done will be stored in the barbell in the form of potential energy. That is, the energy of the barbell is supplied to it from Slim Jim through the work.
c)
All the work done by Jim will be stored in the barbell in the form of potential energy. Thus, the work done by Jim is equal to the potential energy of the barbel.
Therefore, the work done by Jim is 960.4 J
d)
Given,
The time interval, t=1.5 s
The power is given by,
[tex]P=\frac{W}{t}[/tex]Where W is the work done by Jim.
On substituting the known values,
[tex]\begin{gathered} P=\frac{960.4}{1.5} \\ =640.27\text{ W} \end{gathered}[/tex]Thus the power used by Jim is 640.27 W
A 0. 20-kg baseball is struck with a force of 100 n from a 0. 94-kg baseball bat. What will be the acceleration of the ball and the bat, in that order, while the bat and ball are in contact?.
The acceleration of the baseball is 500 m/s².
A force is an influence that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull.
The mass of the baseball, m = 0.2 kg
The force acting on the baseball is 100 N.
By Newton's second law of motion,
F = ma
Substituting the values in the above equation,
100 N = 0.2 kg × a
a = 100 / 0.2
a = 500 m/s²
The total mass when the baseball and baseball bat are in contact is:
m = 0.2 + 0.94 = 1.14 kg
Then the acceleration will be:
F = ma
100 = 1.14 × a
a = 87.7 m/s²
The acceleration of the ball after the contact is 500 m/s².
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Determine normal boiling point of chloroform if its heat of vaporization is 31. 4 KJ/mol and it has a vapor pressure of 190 mmHg at 25 °C
Answer:
The answer is 61.45 °C
Explanation:
Clausius-Clapeyron equation:
P2= Normal pressure
T2= Normal boiling point
ln(p2/p1)=−(ΔHv/R)*(1/T2 − 1/T1)
ln(760/190) = –(31400/8.314)*[(1/T2) – (1/298)]
T2 = 334.6 K = 61.45 °C
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Which Combination of graphs best describes free-fall motion? [Neglect air resistance.]
The combination of the graphs that shows a free fall are A and C.
What is the description of free fall?We know that motion has to do with the change of position with time. When an object is falling freely, the velocity of the object would remain constant because the velocity or the speed does not change.
Recall that when we throw objects down from a height, the objects would fall at the same rate irrespective of the masses of the objects. This can be understood to mean that the acceleration that acts on the objects is the same and does not change.
However, the distance that is covered by the object does change with tome and this can be seen from the graph as we have in the question. We can now see that the graph that has a flat speed time graph and a straight distance time graph can describe on object that is undergoing free fall.
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To which category does galaxy #2 belong? Why does it belong in this category?
Galaxy #2 belongs to spiral. This is because it has a bulge at the centre, a disk as well as the curved arms. Contains both Young and Old stars - contains gas and dust - contains more red giants and red supergiant's than disk - some regular rotation about centre.
What Is Dark Matter?Dark matter is a form of matter thought to account for approximately 85% of the matter in the world. Dark matter consists of particles that do not absorb, reflect or emit light, so they cannot be detected by observing electromagnetic radiation. Dark matter is material that cannot be directly seen. We know dark matter exists because it affects objects we can directly observe.
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a very thin horizontal, 2.00-m long, 5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. at the east end of the beam, a 200-n force pushes downward. at the west end of the beam, a 200-n force pushes upward. what is the angular acceleration of the beam?
The angular acceleration of the beam of length 2 meters and mass 5 kg is 240 radian/s².
The length of the rod is 2 meters, the mass of the rod is 5kg, a force of 200N is applied downward in the east direction while a force of 200N applied upward in west direction.
We know,
That the torque on a body is,
T = Fr
T is torque,
F is the force,
r is the perpendicular distance from the axis of rotation.
So, assuming the centre of the beam as the axis of rotation, the torque on the body is,
T = 200(1)+200(1)
T = 400N-m
We also know,
That torque on the body is,
T = IA
Where I is moment of inertia of the beam,
A is the angular acceleration of the body,
We know, moment of inertia of the beam assuming the centre of the rod as axis of rotation is,
I = MR²/12
M is the mass of the body,
R is the length of the rod,
So,
I = 5(2)²/12
I = 20/12 kg-m²
So, now we can write,
T = 20/12A
As both are value of torque, so,
20/12A = 400
A = 240 radian/s²
The angular acceleration is 240 radian/s².
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In 2 - 3 sentences, explain the difference in Kinetic energy and Potential energy.
The main difference between Kinetic energy and Potential energy is that kinetic energy refers to movement while potential energy refers to storage.
What are Kinetic energy and Potential energy?Kinetic energy is a type of energy in motion or energy in movement such as a turbine (mechanical energy), while Potential energy refers to the energy that is stored to be used when required (e.g. chemical bonds of foods).
Therefore, with this data, we can see that Kinetic energy is used as movement, while Potential energy is stored to be used in the future.
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the chemical makeup of a star's surface is obtained by ... a. examining rocks recovered from the moon. b. examining rocks here on earth. c. examining material collecting on solar panels here on earth.
option c is the correct answer
What is Spectroscopy ?
It is a common method used by astronomers to determine the composition of stars, planets, and other objects. Today, this process uses instruments with gratings that distribute the light from the object by wavelength. This spread of light is called the spectrum.
The science of spectroscopy is very difficult. From the spectral lines, astronomers can not only determine the element, but also the temperature and density of this element within the star. The spectral lines also tell us about each of the star's magnetic fields. The width of the line tells us how fast the matter is moving.
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suppose you were hired to build a dam. what features would you look for in a site? be sure to consider the impact on living things as well as the physical characteristics of the site.
I need to get it done right now
Answer: The wall and base
Explanation:
A freight train moves due north with a speed of 8 m/s. The mass of the train is 6 x 106 kg. How fast would a 1500 kg automobile have to be moving due north to have the same momentum as the train
The freight train has a momentum of mv = (4.5 × 105 kg)(1.4 m/s) = 6.3 ×105 kg-m/s.
An automobile with a mass of 1800 kg would need to go at a speed of 6.3 × 105 kg-m/s ÷ 1800 kg = 350 m/s in order to have a momentum of the same magnitude. (Roughly 800 mph)
A moving body's amount of motion is referred to as momentum. It has units of kg m/s and is theoretically represented as p = m * v.
The momentum equation connects the total force exerted on a fluid element to its acceleration or rate of change of momentum and is a formulation of Newton's Second Law. You're certainly familiar with the formula F = ma, which links applied force and acceleration in solid mechanics analysis.
Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force applied on a particle is equal to the time rate of change of momentum.
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a 2 900-kg truck moving at 11.0 m/s strikes a car waiting at a traffic light, hooking bumpers. the two continue to move together at 7.00 m/s. what was the mass of the struck car?
The mass of the car that was stuck is 1657 Kg.
What is the mass of the struck car?We know that we have to use the laws of the conservation of the linear momentum here. It is known that the momentum before collision must be equal to the total momentum that we have after collision.
In this case, we have;
m1u1 + m2u2 = m1v1 + m2v2
m1 = mass of the truck
m2 = mass of the car
u1 and v1 = the initial and final velocities of the truck
u2 and v2 = initial and final velocities of the car
Again we are told that the two continue to move together hence v1 + v2 = v
m1u1 + m2u2 = (m1 + m2) v
(2900 * 11.0) + 0 = (2900 + m2) 7
31900 = 20300 + 7m2
31900 - 20300 = 7m2
m2 = 31900 - 20300 /7
m2 = 1657 Kg
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why are the rocky inner planets different than the gaseous outer planets? heat from the growing sun allowed rock to mainly form close in gases come from further away from the center of the solar system the rocky inner planets got rocks from the sun later solar wind pushing material away had more effect on metals gas has more momentum and flies further away naturally
The correct answer is option A.
The rocky inner planets are different than the gaseous outer planets because heat from the growing sun allowed rock to mainly form close in.
In our solar system, the inner planets are rocky while the outer planets are in a gaseous state.
Rocky planets are also called terrestrial planets as their outer surface is solid and composed of rocks and metal.
They have large thick gaseous atmospheres which are abundant in gases such as nitrogen, carbon dioxide, and oxygen.
While the outer planets are giant balls of gases that orbit around the sun.
Unlike terrestrial planets, these gaseous planets do not have thick atmospheres and are 2x larger in size than the rocky inner planets.
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the sweet spot on a baseball bat is 55 cm from the axis of rotation during the swing of the bat. if the sweet spot on the bat moves 36 m/s, what is the angular velocity of the bat (in rad/s)?
The Angular velocity is 54.55 rad/s
What is angular velocity?
The rate at which an angle between two bodies changes is known as angular velocity. Something changes when an object rotates or revolves around an axis. This displacement is depicted in the image by the angle formed by a line on just one body and a line on the other. Technical terms like degrees, 360-degree revolutions, as well as revolutions per minute have been frequently used to describe angles or angular displacements (rpm). Angles and angular velocities are typically expressed in radians as well as radians per second, respectively, in mathematics and physics.
These conversion factors link these measurements together:
A radian equals 1/180 (or 0.0175) of a degree, and a radian per second = 1/30 (or 0.105) of a radian.
Angular velocity W= U/r
Given V= 30m/s ; r=0.55mW=3/0.55 = 54.55 rad/s
Hence the answer is 54.55 rad/s
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two boxes of the same mass are connected by a string of negligible mass and pulled across a level tabletop with negligible friction by another horizontal string. the force applied by the leading string is f. m m t f a how does the magnitude of the force t exerted by the string connecting the boxes compare to the magnitude f?
The importance of the pressure t exerted via the string connecting the bins evaluate to the value f is T< f.
A force is a power that can trade the movement of an object. A force can encourage an item with mass to exchange its tempo, i.e., to boost up. stress can also be described intuitively as a push or a pull. A strain has both importance and path, making it a vector quantity.
considering both masses together
The acceleration of the system = a
a = f/m+m = f/2m
Now, considering one mass towards left m
T = ma
T = (f/2m)
m = f/2
T = f/2
T<f
Therefore, T must have a magnitude less than f.
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ake a 40-pf capacitor and charge it to a potential difference of 500 v. then disconnect it from the battery and connect its terminals to those of an uncharged 10-pf capacitor. what are: (a) the
If we rearrange the circuit so that the inductor and capacitor were connected in series,the impedance would be Decreased and the resonant frequency would be unchanged.
What is capacitor ?
Frequency is the rate at which something occurs over a particular period of time or in a given sample.
Sol-An LC circuit is made up of an inductor (L) and a capacitor (C). At the resonance condition of the LC circuit, the inductive reactance XL. becomes equal to the capacitive reactance Xc.
It is defined as-
XL= 2πfL
Xc= 1/2πfC
So when reactances are equal we have-
XL=Xc
2πfL= 1/2πfC
f^2=1/4π^2LC
f=1/2π√LC
The resonance condition is the same for both the parallel and series LC circuit, so the resonance frequency will not change.
The impedance of a parallel LC circuit is:
Z(w)= jL w^2-w^2•/w
In a series LC circuit, when w
, the impedance becomes equal to 0.
Therefore the impedance of a series LC circuit will decrease and the resonant frequency will remain the same.
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A rocket is launched from the surface of the earth with a speed of 9.0x103 m/s. What is the maximum altitude reached by the rocket? (MEarth=5.98x1024 kg, REarth=6.37x106 m)
From the Law of conservation of energy, we know that the sum of the kinetic and potential energy of the rocket is the same at the surface of the Earth and at the maximum altitude. Nevertheless, the kinetic energy of the rocket when it is at the maximum altitude is 0:
[tex]\begin{gathered} K_1+U_1=K_2+U_2 \\ K_2=0 \\ \Rightarrow K_1+U_1=U_2 \end{gathered}[/tex]The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]On the other hand, the gravitational potential energy for big changes in altitude (comparable to the radius of the Earth) is given by the expression:
[tex]U=-\frac{GMm}{r}[/tex]Where M is the mass of the Earth, m is the mass of the rocket, r is the distance from the center of the Earth to the rocket and G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\cdot\frac{m^2}{\operatorname{kg}}[/tex]At the beggining of the movement, the value of r corresponds to the radius of the Earth:
[tex]U_1=-\frac{GMm}{R_E}[/tex]At the end of the movement, the value of r corresponds to the radius of the Earth plus the maximum altitude h:
[tex]U_2=-\frac{GMm}{R_E+h_{}}[/tex]Substitute the expressions for U_1, K_1 and U_2 and simplify the equation by eliminating the factor m:
[tex]\begin{gathered} \frac{1}{2}mv^2-\frac{GMm}{R_E}=-\frac{GMm}{R_E+h} \\ \Rightarrow\frac{1}{2}v^2-\frac{GM}{R_E}=-\frac{GM}{R_E+h} \end{gathered}[/tex]Isolate the term GM/(R_E+h):
[tex]\Rightarrow\frac{GM}{R_E+h}=\frac{GM}{R_E_{}}-\frac{1}{2}v^2[/tex]Divide both sides by the factor GM:
[tex]\Rightarrow\frac{1}{R_E+h}=\frac{1}{R_E}-\frac{v^2}{2GM}[/tex]Take the reciprocal to both sides of the equation:
[tex]\Rightarrow R_E+h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}[/tex]Isolate h:
[tex]h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}-R_E[/tex]Substitute the values of each variable: R_E=6.37x10^6m, M=5.98x10^24kg, G=6.67x10^-11 N*m^2/kg^2, and v=9.0x10^-3 m/s:
[tex]\begin{gathered} h=\frac{1}{\frac{1}{6.37\times10^6m}-\frac{(9.0\times10^3\cdot\frac{m}{s})^2}{2(6.67\times10^{-11}N\cdot\frac{m^2}{kg^2})(5.98\times10^{24}kg)}}-6.37\times10^6m \\ =18.03\times10^6m-6.37\times10^6m \\ =11.7\times10^6m \end{gathered}[/tex]Therefore, the maximum altitude reached by a rocket with an initial speed of 9.0x10^3m is:
[tex]11.7\times10^6m[/tex]mass of bullet 0.25kg, velocity of bullet 230kg, mass of gun 9kg. What is the recoil speed of the gun
The recoil speed of the gun is 6.39m/s
What is recoil speed of the gun ?
After firing the bullet, the bullet moves forward with a large momentum and the gun moves backward with an equal momentum. Let m and M be the masses of bullet and gun, respectively. If v and V are the velocities of the bullet and gun, respectively after the firing then:-
0= mv+ MV
Therefore V=-mv/M
The negative sign indicates that the gun moves in a direction to that of
the bullet.
m1v1 = m2v2
m1 = mass of bullet
v1 = velocity of bullet
m2 = mass of gun
v2 = velocity of gun
m1v1=m2v2
0.25*230=9*v2
v2=0.25*230/9 = 6.39m/s
The recoil speed of the gun is 6.39m/s
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Please help me with this, it's a Kinematics Equation 2 problem!
The distance covered by the object is given as,
[tex]d=ut+\frac{1}{2}at^2[/tex]Plug in the known values,
[tex]\begin{gathered} d=(0\text{ m/s)(9 s)+}\frac{1}{2}(2m/s^2)(9s)^2 \\ =0\text{ m+}81\text{ m} \\ =81\text{ m} \end{gathered}[/tex]Thus, the distance covered by the object is 81 m.
a group of students are provided with three objects all of the same mass and radius. the objects include a solid cylinder, a thin hoop (or cylindrical shell), and a solid sphere. the students are asked to predict which will get to the bottom of a ramp first if all three are released together from the same distance up the ramp. which prediction is correct if the objects are listed in order from fastest to slowest?
Sphere, cylinder, hoop is the correct prediction. Moment of inertia is defined as the quantity represented by a body that resists an angular acceleration that is the sum of the product of each particle's mass times the square of its distance from the axis of rotation.
Ball moment of inertia Is = 2/5 M r²
Cylinder Ic = ½ M r²
Tire Ih = M r²
Substitute each to calculate time
Ball
ts = (d / √2gy) √ ( 1 + 2/5 Mr² / mr²)
ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)
ts = (d / √ 2gy) 1.1
Cylinder
tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)
tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5
tc = (d / √ 2gy ) 1.2
tires
th = (d / √2gy) √ (1 + mr² / mr²)
th = (d / √2gy) √(1 + 1) = (d / √ 2gy ) √ 2
th = ( d / √ 2gy) 1.41
The result is that the bullet hits the fastest and the body with the most tires. So the correct answer is
ts < . tc < th
ball, cylinder tire
Learn more about inertia here-
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a. What is pressure? Write down its Sl unit
Answer:
Pascal (Pa)
Explanation:
the si for pressure is the pascal pa equal to one newton per sqaure meter (N/m^2, or kgxm-^1xs-^2)
[tex] \boxed {\huge \: \sf \red{~~~Answer~~~} \: }[/tex]
Pressure is defined as the physical force exerted on an object. SI unit of pressure is Pascal (pa).[tex] \\ \\ [/tex]
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