Answer:
That is human_scale thinking and understanding
Explanation..
If you are going to consider stars(sun's), then you have to thik on a more cosmological scale stars are prodigiously immense which means they can burn 4 million tons of their fuel per second and not even realize it.
Consider this_if the life of our sun is of the order of 8 billion years, then it takes two billion years to bum only a quit of its fuel.
The universe is a very large place and if you are going to understand it, you have to think large.
A gas is compressed at constant pressure of 0.8atm from 9litre to 2 litre in the process, 400J of a energy leves the gas by heat.
a. what is work done on the gas?
b, What is the change in its internal energy?
According to the give statement:
a)Work done on the gas is 567 J.
b)The change in its internal energy is 167J.
What is an example of internal energy?The temperature and condition of a substance are examples of internal energy. For instance, the internal energy of water is influenced by its temperature as well as whether it is in a solid, liquid, or gas state. Due to its condition, liquid water has greater stored energy than copper metal at the same temperatures.
Briefing:The work done on the gas is evaluated using formulas:
W = -p * ΔV
Only the large volumes which is signified by large data Substituting the known values we obtain
W= 0.8 * 1.01325 * 10⁵ Pa * -1 (-7) * 10⁻³ m³
W = 567 J
Change in Internal energy is given by
ΔU=W - Q
=567−400)
=167J
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Calculate the mass of the Sun using Venus.
Venus is 1.08 x 1011 meters from the Sun and has a period of 224.7 days.
Show your computations.
The mass of the Sun is 1.98 * 10³⁰ kg.
What is the mass of the Sun?The mass of the Sun is calculated from the data given about Venus as follows:
radius of Venus from the Sun = 1.08 x 10¹¹ meters
Period of rotation of Venus around the Sun, T = 224.7 days.
T in seconds = 224.7 * 24 * 3600 = 1.94 * 10⁷ s
Force of gravity = Centripetal force
GMm/r₂ = mv²/r
M = mass of Sun
m = mss of Venus
M = rv²/G
G = 6.67 * 10⁻¹¹ m³kg⁻¹s⁻²
But v = 2πr/T
v = 2 * 3.14 * 1.08 x 10¹¹ / 1.94 * 10⁷
v = 3.496 * 10⁴ m/s
Hence;
M = 1.08 x 10¹¹ * (3.496 * 10⁴)² / 6.67 * 10⁻¹¹
M = 1.98 * 10³⁰ kg
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A hummingbird flies forward and backward. Its motion is shown on the above graph of horizontal Position and Time.
What is the total distance the hummingbird travels from t=0 sec to t=12 sec?
A.4m
b.0m
c.1m
d.2m
Answer:
what is the speed of the bird
how do zou plan your day
Answer: i usually use the calendar on my phone to set my day or i make plans with other people.
D-A6 kg cart moving at 2 m/s hits another car that has a mass of 3 kg. What is the velocity of the second car after impact?(Use the conservation of momentum equation)
The conservation of momentum states that the total intiial momentum is equal to the total final momentum. In this cae this will mean that:
[tex]p_{1i}+p_{2i}=p_{1f}+p_{2f}[/tex]module 2 question 16
(a) Calculate the tension in a vertical strand of spiderweb if a spider of mass 7.00 ✕ 10-5 kg hangs motionless on it.
N
(b) Calculate the tension in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.13. The strand sags at an angle of 13.0° below the horizontal.
N
Compare this with the tension in the vertical strand (find their ratio).
(tension in horizontal strand / tension in vertical strand)
a ) The tension in a vertical strand = 68.6 * [tex]10^{-5}[/tex] N
b ) The tension in a horizontal strand = 149.13 * [tex]10^{-5}[/tex] N
c ) Tension in horizontal strand / Tension in vertical strand = 2.17
a ) The tension in a vertical strand,
m = 7 * [tex]10^{-5}[/tex] kg
g = 9.8 m / s²
Since the strand is vertical and the spider is motionless,
∑ [tex]F_{y}[/tex] = 0
T - mg = 0
T = 7 * [tex]10^{-5}[/tex] * 9.8
T = 68.6 * [tex]10^{-5}[/tex] N
b ) The tension in a horizontal strand,
θ = 13.0°
The strands left and right have same magnitude and makes the same angle with the horizontal. Resolving,
sin θ = [tex]T_{y}[/tex] / T
[tex]T_{y}[/tex] = T sin 13.0°
∑ [tex]F_{y}[/tex] = 0
[tex]T_{y}[/tex] + [tex]T_{y}[/tex] - mg = 0
T sin 13.0° + T sin 13.0° = 7 * [tex]10^{-5}[/tex] * 9.8
2 T sin 13.0° = 68.6 * [tex]10^{-5}[/tex]
T = 68.6 * [tex]10^{-5}[/tex] / 0.46
T = 149.13 * [tex]10^{-5}[/tex] N
c ) Ratio of strands,
Tension in horizontal strand / Tension in vertical strand = 149.13 * [tex]10^{-5}[/tex] / 68.6 * [tex]10^{-5}[/tex]
Tension in horizontal strand / Tension in vertical strand = 2.17
Therefore,
a ) The tension in a vertical strand = 68.6 * [tex]10^{-5}[/tex] N
b ) The tension in a horizontal strand = 149.13 * [tex]10^{-5}[/tex] N
c ) Tension in horizontal strand / Tension in vertical strand = 2.17
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A Student Bends Her Knees 15 Cm And Jumps Up 80 Cm Off The Ground. Find The A.) The Velocity With Which The Student Leaves ground
The initial velocity of the student as he leaves the ground is 3.57 m/s.
What is the velocity of the student as he leaves the ground?The velocity of the student as he leaves the ground can be calculated using the following kinematic equation for upward motion.
v² = u² - 2gh
where;
v is the final velocity of the studentu is the initial velocity of the studenth is the maximum height jumped by the student = 80 cm - 15 cm = 65 cm = 0.65 mAt the maximum height, the final velocity of the student = 0
0 = u² - 2gh
u² = 2gh
u = √2gh
u = √(2 x 9.8 x 0.65)
u = 3.57 m/s
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What helped Northwest Coast tribes develop complex societies? Choose three correct answers.
having to travel to search for resources
being able to plant and harvest crops
having an abundant supply of fish
being able to save and store food
having a wealth of resources
The development of complex societies depends of the Northwest Coast tribes on;
having a wealth of resourceshaving an abundant supply of fishbeing able to plant and harvest cropsWhat are complex societies?The term society just has to do with people that are found to live around a give geographical area. This is the way the people tend to live along with the civilization that they may tend to have actually created over a long period of time as we can see.
Having said thus, the people of the are the people that there originally found around the west of which the original native Americans also formed a huge part of. These were the people that tend to be found around the Americas before the onset of modern civilization as we can se.
The factors that helped Northwest Coast tribes develop complex societies are found in the section above. These factors saw to the development of the region.
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What is the mass of an object that needs a force of 5,500 N to accelerate it at a
rate of 5 m/s/s?
THE MASS OF AN OBJECT IS 1100
A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height of 3 cm above the top of the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 20 cm. What is the oscillation frequency?
The oscillation frequency of the spring is equal to 1.23 s⁻¹.
What is the gravitational potential energy?Gravitational potential energy can be described as the energy contained by the object because of its displacement of some height above the surface.
Given the height of the block before dropping, h = 3cm = 0.03 m
The amplitude of oscillation, A = 20 cm = 0.2m
The expression for the oscillation frequency is:
[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]
From the equation of motion: v²= u² + 2gh
v = √2gh
Because of the equilibrium between the block and the spring, the spring force is equal to the weight of the block.
[tex]kx = mg\\k =\frac{mg}{x}[/tex]
From the law of the conservation f energy, find the value of 'x' displacement:
[tex]m\frac{v^2}{2} + mgx = \frac{mg}{2x}(x+a )^2-mgA[/tex]
[tex]x^2 +2hx-A^2 =0[/tex]
[tex]x^2 + 2\times x \times 0.03- (0.2)^2 =0\\x^2 +0.0.6x -0.04 =0\\x = 0.172 \;m[/tex]
The oscillation frequency is equal to:
[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]
[tex]f = \frac{1}{2\pi }\sqrt{\frac{g}{x} }[/tex]
[tex]f = \frac{1}{2\pi }\sqrt{\frac{9.81}{0.172} }[/tex]
f = 1.20 s⁻¹
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A hot air balloon with a mass of 371 kg is flying 615 m above the ground. How much gravitational potential energy does the balloon have?
The formula for determining the gravitational potential energy of the balloon is expressed as
Energy = mgh
where
m is the mass of the ballon
g is the accelaration due to gravity and the value is 9.8m/s^2
h is the height of the balloon above the ground
From the information given,
m = 371
h = 615
Thus,
= 371 x 89.8 x615 = 22316017 NKJoulesJoules8298.651
module 2 question 14
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 12,500 kg. The thrust of its engines is 31,000 N. (Assume that the gravitational acceleration on the Moon is 1.67 m/s2.)
(a) Calculate its magnitude of acceleration in a vertical takeoff from the Moon.
m/s2
(b) Could it lift off from Earth? If not, why not?
No, the thrust of the module's engines is less than its weight on Earth.
Yes, the thrust of the module's engines is greater than its weight on Earth.
No, the thrust of the module's engines is equal to its weight on Earth.
Yes, the thrust of the module's engines is equal to its weight on Earth.
If it could, calculate the magnitude of its acceleration. (If not, enter NONE.)
m/s2
a ) The acceleration in a vertical takeoff from the Moon = 0.81 m / s²
b ) No, it could not lift off from Earth. Because, the thrust of the module's engines is less than its weight on Earth.
a ) Vertical takeoff from the Moon,
F = 31000 N
m = 12500 kg
[tex]g_{m}[/tex] = 1.67 m / s²
W = m g
[tex]W_{m}[/tex] = 12500 * 1.67
[tex]W_{m}[/tex] = 20875 N
∑ [tex]F_{y}[/tex] = m [tex]a_{y}[/tex]
F - [tex]W_{m}[/tex] = m [tex]a_{y}[/tex]
31000 - 20875 = 12500 [tex]a_{y}[/tex]
[tex]a_{y}[/tex] = 0.81 m / s²
b ) Lift off from the Earth,
Force required for a 12500 kg object to lift off from Earth,
F = m g
F = 12500 * 9.8
F = 122500 N
Since the force required to lift off from Earth is much higher than the actual force, the module cannot lift off from Earth
Therefore,
a ) The acceleration in a vertical takeoff from the Moon = 0.81 m / s²
b ) No, it could not lift off from Earth. Because, the thrust of the module's engines is less than its weight on Earth.
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What is a 65 kg
k
g
rider's apparent weight as the ride is coming to rest?
The weight of the rider will remain the same that is 65 kg after the ride is finished because the weight does not depend on speed.
What is Weight?It gauges how much gravity is pulling on a body.
Weight being a force, the SI unit of weight is the Newton, which is also the same as the SI unit of force (N). When we look at how weight is expressed, we can see that it depends on both mass and the acceleration caused by gravity; while the mass may not change from one location to another, the acceleration caused by gravity does.
According to the question, the weight will remain same because weight depends on mass and gravity but not the speed. As per the definition of weight it is the product of mass and gravity, it means for different plants the weight will be different but for the same planet the weight will be the same everywhere.
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What kinematics equation would i use for this ? Btw one of those are the answers
Given that the time taken to reach the ground is t = 6 s
The final velocity is v = 0m/s
We have to find the height of the monument, h.
The initial velocity can be calculated by the formula
[tex]u=gt[/tex]Here, g is the acceleration is due to gravity whose value is 9.81 m/s^2
Substituting the values, the initial velocity will be
[tex]\begin{gathered} u=9.81\times6 \\ =58.86\text{ m/s} \end{gathered}[/tex]The formula to find height is
[tex]h=\frac{u^2}{2g}[/tex]Substituting the values, the height will be
[tex]\begin{gathered} h=\frac{(58.86)^2}{2\times9.81} \\ =176.58 \\ \approx177\text{ m} \end{gathered}[/tex]Thus, the correct option is 177 m.
I need help on homework
Given:
Let's determine the preferred units used when measuring the velocity of the center of mass.
The center of mass can be said to be the mean position of matter in a given system.
This is the point in a body where the whole mass of the body is concentrated upon.
The SI unit for center of mass is meters (m).
When measuring the center of mass, the preferred unit is meters per second.
herefiore, when measuring the velocity of the center of mass, the preferred unit is meters/second.
• NSWER:
A. meters/second
module 2 question 7
An owl is carrying a mouse to the chicks in its nest. It is 4.00 m west and 12.0 m above the center of the 30 cm diameter nest and is flying east at 4.50 m/s at an angle 27° below the horizontal when it accidentally drops the mouse. Will it fall into the nest? Find out by solving for the horizontal position of the mouse (measured from the point of release) when it has fallen the 12.0 m.
m (from the point of release)
Answer:
the mouse will overshoot and will not land in the nest.
A parallel plate capacitor is connected to a battery that maintains a constant potential difference. How will the magnitude of the charge on the plates change if the separation between the plates is doubled?It will be cut in half.It will not change.It will double.It will quadruple.
A parallel plate capacitor is connected to a battery that maintains a constant potential difference.
As we know,
[tex]\begin{gathered} Q=CV \\ Q=\frac{A\epsilon_0V}{d} \end{gathered}[/tex]Here, Q is the charge which is inversely proportional to the separation between the plates.
[tex]Q\propto\frac{1}{d}[/tex]That means if the separation between the plates increases the magnitude of the charge on the plates decreases.
Thus, if the separation between the plates is doubled the magnitude of the charge will be cut in half.
PLEASE HURRY!
Give a short description of how our early solar system formed. Be sure to correctly address the following terms: solar nebulae, accretion disk, protostar, planetesimals, star, and planets. (3 to 5 sentences)
Subject is Earth and Space science btw not physics they just didnt have an option to put it under :,)
Solar system formation began approximately 4.5 billion years ago, when gravity pulled a cloud of dust and gas together to form our solar system.
The planets, moons, asteroids and the whole lot else in the solar device shaped from the small fraction of material in the region that wasn't included within the younger sun.
The nebular hypothesis is the most extensively regularly occurring version within the field of cosmogony to explain the formation and evolution of the sun gadget. The protoplanetary disk is an accretion disk that feeds the significant superstar.
Protostar are the stars are thought to shape internal giant clouds of cold molecular hydrogen massive molecular clouds kind of 300,000 instances the mass of the solar and 20 parsecs in diameter.
Beneath positive circumstances the disk, that could now be called protoplanetary, may additionally deliver beginning to a planetary device. Large planet core formation is notion to proceed more or less alongside the lines of the terrestrial planet formation.
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module 1 question 4
(a) How many significant figures are in the numbers 99.0 and 100.0?
99.0 :
100.0 :
(b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (Give your answer to 3 significant figures.)
99.0 :
100.0 :
a) there are 2 and 3 significant figures in the numbers 99.0 and 100.0
b)for the uncertainity of 1 in each number, the percentage uncertainity for in each is 1 an 1.01
Here the problem we are dealing with is related to significant numbers which are the number of digits in a value, frequently an estimation, that contributes to the degree of accuracy of the value. We begin counting significant figures at the primary non-zero digit. in the first case, we are given two numbers 99 and 100, where in 99 we have 2 significant numbers and in 100, there are 3 significant numbers. For the second case, the formula for calculating the uncertainity percentage is percentage = uncertainty /value x 100
So,the uncertainty is 1, so the certaintiy percentage for three significant numbers are which is 100 , the uncertainty percentage = 1/100 *100 = 1
For 99 , the uncertainty percentage = 1/99* 100 = 1.01
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A 63.1 kg weight-watcher wishes to climb a mountain to work off the equivalent of a large piece of chocolate cake rated at 797 (food) Calories. How high must the person climb? The acceleration due to gravity is 9.8m/s^2 and 1 food Calorie is 10^3 calories. Answer in units of km.
Given:
the mass of the weight-watcher is
[tex]m=63.1\text{ kg}[/tex]The work off equivalent to
[tex]W=797\text{ food}[/tex]Required: height climbed by the person
Explanation:
first we need to change the work into calories.
it is given that
[tex]1\text{ food=10}^3\text{ calories}[/tex]then the work done is
[tex]W=797\times10^3\text{ calories}[/tex]now change this work done from calories to joules.
we know that
[tex]1\text{ calorie = 4.2 J}[/tex]Then the work done is ,
[tex]\begin{gathered} W=797\times10^3\times4.2 \\ W=3347.4\times10^3\text{ J} \end{gathered}[/tex]as the person climbs to the mountain the work done is stored as potential energy.
we assume that a person attained some height h,
then the work-energy relation,
[tex]W=mgh[/tex]Plugging all the values in the above relation and solve for h, we get
[tex]\begin{gathered} 3347.4\times10^3\text{ J=63.1 kg}\times9.8\text{ m/s}^2\times h \\ h=\frac{3347.4}{618.38} \\ h=5.41\text{ m} \end{gathered}[/tex]Thus, the height climbed by the person is
[tex]5.4\text{1 m}[/tex]Ms. Cast's husband calculates the height of their house as 10.0 meters. The actual height is 11.0
meters. What is the experimental (%) error?
Answer:
mrs.casts
Explanation:
she's married
A vehicle of mass 1,084 kg is traveling in the +y-direction along a straight, horizontal road at a speed of 18.2 m/s when the driver applies the brakes, stopping the car in a distance of 244 m. Find the y-component of the constant net force applied to the car to bring it to a stop.
[Note: Remember that vector components can be positive or negative.]
The y-component of the constant net force applied to the car to bring it to a stop is mathematically given as
F_y =666.38N
This is further explained below.
What is force?Generally, In the field of physics, an influence that may alter the motion of an object is referred to as a force.
An object having mass may experience a change in its velocity, often known as acceleration, when subjected to a force.
Intuitively, force may also be conceptualized as either a push or a pull. Because it may be measured in both magnitude and direction, a force is considered a vector quantity.
When considering coplanar force of a point or a body we tend to consider the x components and the y components of force as they come together to give the Net Force or the resultant for and directions of force.
But for this case we shall strictly consider the y company of force as it pertains to this body and its force independent of the x components force as regards to Resultant force
In conclusion,
v^2 =2 a s
[tex]a &=\frac{v^2}{25}=\frac{(18.4)^2}{2 \times 268} \\\\[/tex]
a =0.631ms^2
F_y =m a
F_y=1055* 0.631
F_y =666.38N
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procidure for tapping a blind hole
Answer:
HOW TO DO IT…
Use the correct cutting oil on the tap when cutting threads.
Turn the tap clockwise one-quarter to one-half turn, then turn back three-quarters of a turn to break the chip.
When tapping a blind hole, use the taps. in the order starting, plug, and then bottoming.
Why are the accelerations due to gravitational force on the moon and the Earth different? Do you think you could shield a gravitational field in a vacuum?
The difference in the acceleration due to gravity on the moon and the Earth are different due to the difference in their masses and radius.
Yes, can shield a gravitational field in a vacuum by coming in between the mass and gravitational source.
What is Newton's law of universal gravitation?
Newton's law of universal gravitation states that every two objects in the universe attracts each other with a force directly proportional to the product of the masses and inversely proportional to the distance between the two objects.
F = GmM/R²
where;
G is universal gravitation constantm and M are the massesR is the distance between the two objectsFrom Newton's second law of motion,
F = mg
where;
g is acceleration due to gravitymg = GmM/R²
g = GM/R²
Thus, the acceleration due to gravity on the moon and the Earth are different because of difference in their masses and radius.
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Drag is usually ignored because its effect on the horizontal velocity is usually negligible due to the short time of flight.
An object's surface area and geometry, along with the object's surrounding wind speed will affect the drag force.
In most cases, drag force will cause the object to land horizontally closer to the predicted landing point as drag is a resistive force.
Drag is ignored in projectile predictions because projectiles usually have a relatively short time of flight.
The surface area of the object, the wind speed, as well as the relative velocity of the airplane will affect the drop of relief packages.
The drag force will cause the projectile to take a longer time to land and may cause to land far from its expected drop point.
What is drag?Drag is a force that acts in opposition to the motion of an object moving through a fluid.
Drag can be thought of as friction in fluids because similar to friction, it acts in an opposite direction of the relative motion of a moving object.
For example, airplanes moving through air experience a drag; ships and boats moving through water experience drag too.
Drag also occurs in projectiles moving through the air. However, because of the relatively short time of flight, it is usually ignored in projectile motion.
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Complete question:
While studying projectile motion, we consider ideal scenarios, where the projectile travels along its trajectory only under the influence of gravity. In real-world situations, however, other forces act on the projectile.
Consider a cargo plane that is dropping relief packages to flood victims. In predicting and studying this motion, we might consider gravity, but ignore the horizontal and vertical forces associated with drag (or air friction). Discuss this simplification. Specifically address these questions:
•Why do we often ignore drag in projectile predictions?
•What conditions (of the object, its surroundings, and its launch) do you think might make drag a significant factor in the relief package drop?
•How would drag affect the projectile's motion if it really were a significant factor in the relief package drop?
Richard has just completed his science experiment, and collected all of the data. What is the next step that Richard needs to complete?
a.background research
b.analyze results
c.recognize a problem
d.form a hypothesis
plane, starting from rest, accelerates at a rate of a = 5 m/s2 for a distance of 250 m. At this distance, calculate the plane's FINAL VELOCITY to the nearest 1 m/s.
The final velocity of the plane at 250 m distance is 50 m/s.
Equation :To calculate the final velocity of plane we need to use the formula from equation of motion,
In equation of motion we will use the formula,
v² = u² + 2as
where,
v is final velocity
u is initial velocity
a is the acceleration of plane
s is the distance
In this given data are,
s = 250 m
a = 5 m/s²
u = 0 m/s ( as the initial of plane is resting )
v = ?
Now, since we know the values, putting them into the formula,
v² = ( 0 )² m/s + 2 x (5 m/s²) x 250 m
v² = 2 x (5 m/s²) x 250 m
v² = 2,500 m/s
v = √2,500 m/s
v = 50 m/s
Hence, the final velocity of the plane at 250 m is 50 m/s.
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I need help finding the solution
The potential energy of an object of mass m at a height h from the ground on the Earth, where the gravitational acceleration is g, is:
[tex]U=\text{mgh}[/tex]As we can see, the velocity of the object is not a variable that needs to be taken into account to find the potential energy,
Then, the answer is:
[tex]\text{Velocity of the object}[/tex]The Italian greyhound jumps 0.85 m in the air to catch a frisbee. How much potential energy does the dog have at the height of its jump?
Answer:
U = 41.65 J
Explanation:
We have just to calculate it with the following expression of gravitational potential energy:
U = m*g*h
Where:
m: mass
g: gravity acceleration;
h: height.
So:
U = m * 9.8 * 0.85
U = 8.33 * m
Then, it is just putting the mass value of the dog in that equation to obtain the value of potential energy acquired by the dog when jumping.
The mass of this kind of dog is between 3.6kg - 5kg, so, the maximum value of the potential energy can be like this:
U = 5 * 9.8 * 0.85
U = 41.65 J
a walrus accelerates from7.0 km/h to 34.5 km/h over a distance of 95m. What is the magnitude of thee walrus's acceleration?
The magnitude of the walrus's acceleration that accelerates from 7.0 km/h to 34.5 km/h over a distance of 95m is 6 m / s²
( v + u ) / 2 = d / t
v = Final velocity
u = Initial velocity
t = Time
d = Distance
u = 7 km / h
v = 34.5 km / h
d = 95 m
( 34.5 + 7 ) / 2 = 95 / t
t = 95 / 20.75
t = 4.58 s
a = v - u / t
a = Acceleration
a = ( 34.5 - 7 ) / 4.58
a = 6 m / s²
Therefore, the magnitude of the walrus's acceleration is 6 m / s²
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