A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And oil are 0.5kj/kg.k and 3.5kj/kg.k calculate the entropy change in steel,oil and in the universe.
Answer:
700J
Explanation:
Place the two thermometers outdoors in different locations. Place one in direct sunlight and the other in shade. Make sure that the thermometers are in a secure location so they will not fall or blow away. Do you think that there will be a difference in temperature between these two locations over a period of one week? Write down your prediction.
Answer:
YES THERE WILL BE A DIFFERENCE
Explanation:
THE THERMOMETER WHICH IS KEPT UNDER SUNLIGHT WILL BREAK DUE TO THE HEAT OF THE SUN
AND THE ONE KEPT UNDER SHADE WILL BE SAFE.
Final answer:
Yes, there will likely be a difference in temperature between a thermometer placed in direct sunlight and one placed in the shade due to the energy carried by sunlight that converts to heat when absorbed.
Explanation:
If a student places one thermometer in direct sunlight and another in the shade and records the temperatures over a period of one week, yes, there will likely be a difference in temperature readings between the two locations. Direct sunlight warms the surface it strikes more than an area in the shade does. This is due to the fact that sunlight carries energy which converts to heat when it is absorbed by objects, including thermometers. If the thermometer in the shade is shielded from this direct infusion of solar energy, it will generally record a lower temperature.
Several experiments suggested, such as shielding a thermometer with aluminum foil, wrapping it with a handkerchief soaked in nail polish remover, or simply measuring the air temperature in a sunny spot, all showcase the effects of sunlight on temperature readings. It is important to use an appropriate laboratory thermometer that can handle specific temperature ranges to avoid breakage and ensure accurate results. Additionally, when conducting the experiment, it is vital to ensure that the thermometer is upright to avoid incorrect readings.
1. Suppose you give a shopping cart a push and it begins t
give a shopping cart a push and it begins to roll across the parking
lot. Once you have stopped pushing it, it slows down and comes to a sto
observation consistent with Newton's First Law? Why or why not?
Newton's first law says the cart should keep going if there's no force acting on it to slow it down. But it slows down and stops. Is there any force acting on it ? You bet ! There's friction in the wheels, and a little bit of air resistance too.
What are five of the most important things you learned in physics? ASAP
Answer:
The definitions of mass, energy, time and space as used in Physics are circular
The Conservation of Mass and Energy
The definition of zero is not sufficiently defined in Physics
Newton’s Laws of Motion are related by Calculus
Knowing the terms of Physics is most important.
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How much pressure is applied to the ground
by a 93 kg man who is standing on square
stilts that measure 0.04 m on each edge?
Answer in units of Pa.
003 (part 2 of 2) 10.0 points
What is this pressure in pounds per square
inch?
Answer in units of lb/in.
Answer:
Pressure applied by the man= 285103.125 [tex]Pa[/tex] or 41.35 [tex]lb/in^{2}[/tex]
Explanation:
Pressure is defined as the perpendicular force applied per unit area.
i.e. [tex]Pressure=\frac{Force}{Area}[/tex]
Now, [tex]Force= mg[/tex]
where, [tex]m[/tex] = mass of the body(man) = 93 kg
[tex]g[/tex] = acceleration due to gravity of Earth = 9.81 [tex]m/{s^{2}}[/tex]
[tex]Area[/tex] covered is equal to the area of both stilts(a man generally stands on two feet)
therefore [tex]Area=2(0.04)^{2}[/tex] [tex]m^{2}[/tex]
and putting in the values, we get,
[tex]Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}[/tex]
Now we need to convert to our required units:
[tex]1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}[/tex]
(We can get the above result by individually converting kg to lb and meters to inches respectively)
Using the above relations we get,
[tex]Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}[/tex]
Based on the formula for pressure, the pressure applied on the ground is 570206.25 N/m² or 82.7 pound per square inch
What is pressure?Pressure is defined as the perpendicular force applied per unit area.
Pressure = Force/AreaAlso,
Force = mgwhere,
mass of the body = 93 kg
g = 9.81 m/s²
Force = 93 * 9,81
Force = 912.33 N
Area of square stilt = (0.04m)²
Area = 1.6 * 10⁻³ m²
Then;
Pressure = 912.33 N/1.6 * 10⁻³ m²
Pressure = 570206.25 N/m²
Converting to pounds per square inch:
1 pound per square inch = 6894.76 N/m²
570206.25 N/m² = 570206.25 N/m² * 1 pound per square inch/6894.76 N/m²
Pressure = 82.7 pound per square inch
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Imagine that Kevin can instantly transport himself between Planet X and Planet Y. Which statement could be said about Kevin in this situation?
Kevin’s weight may change between Planet X and Planet Y.
Kevin’s volume may change between Planet X and Planet Y.
Kevin’s mass may change between Planet X and Planet Y.
Kevin’s matter may change between Planet X and Planet Y.
The answer is Kevin's weight may change between Planet X and Planet Y.
Explanation:
The mass is the amount of matter contained in our body and it does not vary with other planets (remains constant). But the weight on the planet is the strength of gravity on the planet and result of mass which will vary with different planets. Your weight is a measure of the pull of gravity between the person and the planet.Every object in the universe will attract each other object with the same mass. When the gravity is high, your weight will be high because your weight will be depending on the gravitational pull of planets.For example, if an astronaut travels from Earth to Mars, his mass remains unchanged and he floats with the same size and mass but his weight will vary due to varying gravity. Thus your mass doesn't affect other planets it's only the weight that may change due to the varying gravitational pull or no gravity.Answer:
the answer is A
Explanation:
Which statement describes Newton’s conception of the Solar System?
1.) the Sun at the center, with the Earth and other planets orbiting the Sun
2.) the Earth orbiting the Sun, with other planets orbiting the Earth
3.) the Earth at the center, with the Sun and other planets orbiting Earth
4.) the Sun orbiting the Earth, with other planets orbiting the Sun
PLEASE HELP I NEED THIS ANSWER QUICK!
Answer:
1). the sun at the center, with the earth and other planets orbiting the sun.
Explanation:
Newton believes that it's gravity between the sun and the other planets even at the large distances that cause them to orbit around the sun. This force build up by the masses of the sun and the planets which keeps the planets in their orbit.
Answer:the sun at the center, with the earth and other planets orbiting the sun.
Explanation:
A force of 30N acts through a distance of 4m in the direction of the force, what is the work done
Answer:
120J
Explanation:
Work done = Force (N) X Distance (m)
Answer:
Work = 120Nm or 120J
Explanation:
Work is said to be done when a force moves a body over a distance, in the direction of force applied.
Work done = force x distance (in the direction of the force)
Therefore work = force x displacement
As displacement is distance in a specific or specified direction.
From the parameters given
Force = 30N
Displacement = 4m.
Therefore work = force x displacement
= 30N x 4m
Work = 120Nm or 120J
Note :J is joules which is the S. I unit of work
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Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts of potential and kinetic energy the bowling ball has:
• as it sits on top of a building that is 40 meters tall.
• as it is half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second.
• as it is just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second.
---------------------------------------------------------------------------------------------------------------------------
4. What is the potential energy of the bowling ball as it sits on top of the building?
5. What is the potential energy of the ball as it is half way through the fall, 20 meters high?
4) Potential energy: 784 J
5) Potential energy: 392 J
Explanation:
4)
The potential energy of an object is the energy possessed by the object due to its position in the gravitational field. It is given by
[tex]PE=mgh[/tex]
where
m is the mass of the object
g is the acceleration of gravity
h is the height of the object from the ground
Here, for the ball sitting on top of the building, we have:
m = 2 kg (mass)
[tex]g=9.8 m/s^2[/tex] (acceleration of gravity)
h = 40 m (height of the builiding)
So, the potential energy is
[tex]PE=(2)(9.8)(40)=784 J[/tex]
5)
As the ball is half way through the fall, the height of the ball is
h = 20 m
The potential energy of the ball now is
[tex]PE = mgh[/tex]
where
m = 2 kg is the mass of the ball
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Therefore, the potential energy now is
[tex]PE=(2)(9.8)(20)=392 J[/tex]
And we see that the potential energy has halved: this is because part of the potential energy, during the fall, has been converted into kinetic energy, increasing the speed of the ball (which is in fact accelerating).
An object is placed 12cm from a converging lens of focal length 18cm. Find the position of the image.
Answer:
The position of the image = 7.2 cm.
Explanation:
Given:
A Converging lens:
Object distance [tex](d_o)[/tex] = 12 cm = -12 cm (sign convention)
Focal length [tex](f)[/tex] = 18 cm
We have to find the position of the image.
Let the position of the image be [tex](d_i)[/tex] .
Sign convention:
The focal length of converging (convex) lens is always positive,while object distance is negative.
Using lens formula:
⇒ [tex]\frac{1}{object\ distance} + \frac{1}{image\ distance} =\frac{1}{focal\ length}[/tex]
⇒ [tex]\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{-d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} + \frac{1}{d_o}[/tex]
⇒ [tex]d_i=\frac{d_o\times f}{d_o+f}[/tex]
⇒ [tex]d_i=\frac{12\times 18}{12+18}[/tex]
⇒ [tex]d_i=\frac{216}{30}[/tex]
⇒ [tex]d_i=7.2[/tex] cm
So the position of the image = 7.2 cm.
A jet plane manages to go forward by sucking in_____
through the_______
and shooting it out again at the back. (Note: one word for each field)
QUESTION 3
In terms of the action-reaction law, what is the reason why firemen have difficulty holding on to the nozzle of a fire
hose when water is gushing out at high speeds?
The hose's material contracts (action) resulting in a recoil (reaction)
It is an electrically regulated jolt (action) as a build-in feature for safety purposes, so that firemen recoil back
(reaction) from potential flares
The force by which the water is shot out (action) results in an equal counter-force by the water (reaction)
The water hitting on a heavy object or surface (action) results in a counterforce from the object or surface
conducted through the water to the firemen (reaction)
The push-back is only a perceptional effect from the stress which the firemen experience
Save All Answers
Save and Sub
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
Answer:
sucking in air through the turbine.
Explanation:
When an airplane turbine operates, it sucks air through it, compresses the air, causes its pressure to increase in the same way as its temperature and ejects it at high speed through the nozzle-shaped output, creating a high output speed and push force capable of moving the plane horizontally.
Vehicles of the fire department are equipped with water pumps of high pressure, in such a way that, when operating, send a flow of water with high pressure and speed through the hose, this force output of the water is sometimes much greater than the force with which the fireman holds the hose, and hence the difficulty to maneuver the hose.
Now when the fireman points the hose with the water at high pressure and speed coming out of it, into a wall close enough. The Fireman will experience newton's third law in all its splendor, which says that every force of action leads to a reaction, so that the reaction could hurt the fireman operating the hose.
A 10-newton box rests on a table. What is the net force on the box? A. 0 N B. 10 N C. more than 10 N D. between 0 N and 10 N
A. 0 N is the net force acting on the box
Explanation:
Newton's first law of motion gives an insight to this problem, if an object stays in its position of either motion or rest, it can only change its current of either motion or rest is by applying an external force on it. If an object is resting on a table, there is a normal force acting perpendicular to the object and at the same, there is the gravitational force acting on the object in the perpendicular direction opposite to the direction of the normal forceHence, these two forces cancel and the resultant force will be zero. so it is in rest and it moves only when the resultant force is non zero.She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observation deck is 1,353 feet high, how far apart are the boats?
Answer:
The boats are 934.65 feet apart
Explanation:
Given:
The angles of depression to the two boats are 42 degrees and 29 degrees
Height of the observation deck i = 1,353 feet
To Find:
How far apart are the boats (y )= ?
Solution:
Step 1 : Finding the value of x(Refer the figure attached)
We can use the tangent ratio to find the x value
[tex]tan(42^{\circ}) = \frac{1353}{x}[/tex]
[tex]x = \frac{1353}{tan(42^{\circ}) }[/tex]
x = 590.47 feet
Step 2 : Finding the value of z (Refer the figure attached)
[tex]tan(29^{\circ}) = \frac{1353}{z }[/tex]
[tex]z = \frac{1353}{tan(29^{\circ})}[/tex]
z = 1525.12 feet
Step 3 : Finding the value of y (Refer the figure attached)
y = z -x
y = 1525.12 - 590.47
y = 934.65 feet
Thus the two boats are 934.65 feet apart
two different examples of forces that could cause a change in motion
Answer:
a)Kicking a ball; b) A car that gets hit
Explanation:
a)
A classic example is when a soccer player kicks a moving ball in another direction by modifying the original direction and speed that the ball brought.
b)
Another example is when a car goes on a highway and suddenly is hit by another car on its side modifying the original direction and speed.
How does slope on this graph indicate the amount of density?
Density comparison
Question 3 options:
A higher density is indicated by a slope that is not steep (yellow line)
The line in the middle (red line) has the highest density
The steeper slope (green line) indicates a higher density
Final answer:
The slope of a line on the graph represents the amount of density. A steeper slope indicates a higher density, while a flatter slope indicates a lower density.
Explanation:
The slope of a line on the graph represents the amount of density in this scenario. A steeper slope indicates a higher density, while a flatter slope indicates a lower density.
For example, in the given graph, the steeper slope (green line) indicates a higher density, while the slope of the yellow line is not steep, indicating a higher density as well.
Therefore, the statement 'The steeper slope (green line) indicates a higher density' is correct.
14. Which of the following is an example of changing acceleration?
a. A car stopped at a red light.
b. A boat sailing west at 5 knots.
c. A train traveling 65 miles per hour south.
d. A man jogging 3 meters/second along a winding path.
What type of heat transfer occurs in solids?
Answer: Conduction
Explanation:
The atoms in the solid move together and create heat
Answer:
conduction
Explanation:
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Generators may be set up to produce either direct or _____ current
Answer:
Generators may be set up to produce either direct or alternating current
Explanation:
Two forces whose resultant is 100N,are perpendicular to each other.if one of them makes an angle of 60° with the resultant, calculate it's magnitude
Calculate the absolute pressure of an ocean depth of 1200atm. Assuming the density of liquid is 1200kg/m^2amd that (pa=1.01×10^6pa N/m^2
Magnitude of the force = 50N
Explanation:
Let one force be X
And the second force be Y
Both the forces are perpendicular, so α = 90°
The resultant R of the two forces = 100N
Angle between resultant and 1 force, α = 60°
We know,
R² = X² + Y² + 2XYcos 90
100² = X² + Y²
If X makes an angle of Φ = 60°
Then, tan Ф = Y sin α / X + Y cos α
Putting the values we find,
tan 60 = Y/X
Y = √3X
Putting Y = √3X in equation 1 we get,
(X)² + (√3X)² = 100 X 100
4X² = 100 X 100
X² = 2500
X = 50N
Therefore, magnitude of the force = 50N
What are the results of inserting a crimp connector into the crimp tool facing the wrong way ?
If you insert a crimp pin incorrectly, the ratcheted crimp tool will not sufficiently crimp the tabs. As a result, the wire may not fully conduct with the pin and the pin will be damaged.
Explanation:
The general theory for crimping all types of connectors is to strip a little bit of insulation off the wire. Then, put the connector into a suitably sized space in the jaws, insert the wire, and crimp it down. For non-ratcheting pliers, it's suggested the connector be re-crimped with the next smallest hole in the jaws.
A good crimp connection is gas tight and won't wick: it is sometimes referred to as a “cold weld”. Like the solder method, it can be used on solid or stranded conductors, and provides a good mechanical and electrical connection.
Inserting a crimp connector into a crimp tool incorrectly can lead to poorly fastened connections, resulting in signal loss, intermittent connectivity, or failure of the electrical connection and may create a safety hazard.
Inserting a crimp connector into a crimp tool facing the wrong way can result in a poor connection as the connector will not be properly fastened to the wire. The specific results depend on the type of crimp connector and tool, but generally, you may end up with a connector that is not adequately compressed, leading to signal loss, intermittent connectivity, or complete failure of the electrical connection. Additionally, it increases the risk of short-circuiting or creating a potential safety hazard since the contact between the conductor and the terminal may not be secure.
How do you make an electromagnet
Answer:
The correct answer is D. wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.
Explanation:
a 3kg has 9 joules of kinetic energy how fast is it flying
Answer: 2.4 m/s
Explanation:use KE= 1/2 mv²
Derive to find v:
v = √ 2KE / m
= √ 2 ( 9 J ) / 3 kg
= 2.4 m/s
4) Consider a separate rocket, also in deep space with a mass of 30.0 kg. If the rocket is
observed to be travelling at v= 5.00 m/s at t= 3.00 s and then travelling at û = -2.00 m/s at t= 14.0 s
with constant acceleration, calculate:
a) the acceleration, a.
b) the force F, acting on the rocket from the thrusters,
Answer:
(a) -0.636 m/s²
(b) -19.08 N
Explanation:
Given:
Mass of the rocket (m) = 30.0 kg
Initial velocity of rocket (v) = 5.00 m/s
Initial time of rocket (t₁) = 3.00 s
Final velocity of the rocket (u) = -2.00 m/s
Final time of rocket (t₂) = 14.0 s
(a)
Acceleration is given as the rate of change of velocity. Therefore,
[tex]a=\frac{u-v}{t_2-t_1}\\\\a=\frac{-2.00-5.00}{14.00-3.00}\ m/s^2\\\\a=\frac{-7.00}{11.00}\ m/s^2\\\\a=-0.636\ m/s^2[/tex]
Therefore, the acceleration of the rocket is 0.636 m/s².
(b)
From Newton's second law, we know that, force acting on a body is equal to the product of its mass and acceleration.
So, the force acting on the rocket is given as:
[tex]F=ma\\\\F=(30.0\ kg)(-0.636\ m/s^2)\\\\F=-19.08\ N[/tex]
Therefore, the force acting on the rocket is -19.08 N.
The negative sign implies the force acts in the direction opposite to motion.
If you were able to jog one mile in 10 minutes, how would you progressively increase your
performance by using each of the following? Be specific by using numbers in your answers.
1. (F) frequency variable?
2. (1) intensity variable?
3. (T) time variable?
Answer:
1. (F) Increasing the frequency variable reduces the time required.
2. (I) Intensity variable can be related the the energy expended.
3.(T) Time variable would affect the speed.
Explanation:
1. (F) increasing the frequency variable implies that the distance would be covered in lesser time. As a wave of 250Hz cover more distance than that of 200Hz.
2. (I) Intensity variable. An increase in this variable implies that more energy is expended.
3.(T) Time variable can either give the required out come or not. As increasing the time would mean that the speed has reduced, while decreasing the time means the distance would be covered quickly. If the required speed is 20m/s, then increasing it to 30m/s imples that lesser time would be recorded. If the reverse is the case, then more time would be recorded. time = distance/speed
1. reduces the time required.
2.energy expended.
3.affect the speed.
What is Frequency, time period and intensity ?Frequency, time period and intensity are quantitative dimensions which describe the amount of physical activity taking place during a given time.
1. (F) frequency variable - Frequency refers to the number of waves that pass a fixed point in unit time.
i.e. [tex]F =\frac{1}{T}[/tex]
When frequency is increased then distance would be covered in lesser time.
For example, a wave of 550Hz cover more distance than that of 500Hz.
2. (I) Intensity variable- The quantity of energy the wave conveys per unit time across a surface of unit area.
i.e. [tex]I = \frac{P}{4\pi r^{2} }[/tex]
When Intensity increases it implies that more energy is expended.
3.(T) Time variable -It is define as the ratio of distance and time.
i.e. [tex]time = \frac{distance}{speed}[/tex]
When time variable is increased than the quantity of energy the wave conveys per unit time across a surface of unit area, speed is reduced.
When time variable is decreased the distance would be covered quickly.
For example : If the speed of car is 80m/s, then by increasing the speed to 100m/s shows that lesser time would be required.
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A 0.250 kg mass is attached to a spring with k=18.9 N/m. At the equilibrium position, it moves 2.89 m/s. What is the amplitude of the oscillation? (Unit=m)
Answer: y = R*sin(2.75*t - δ)
Here δ is just the time offset and for our purposes is pretty irrelevant. You can in fact set it to zero since we can say we begin timing when the mass crosses equilibrium. So
y = R*sin(2.75*t)
We want to find a way to use the information "At the equilibrium position, it moves 2.89 m/s." I am going to use some calculus here since it makes things so much easier. If you haven't taken calculus yet, most likely your course has given you a formula to use instead.
We know y=0 when t=0, so y is at equilibrium when t=0. To say it moves 2.89 m/s is then to say that
y'(0) = 2.89.
From here we can differentiate the displacement function, set t=0 and solve for R. Using the chain rule:
y'(t) = 2.75*R*cos(2.75*t)
y'(0) = 2.75*R
2.75*R = 2.89
R = 1.051
Explanation: Since this is harmonic motion we can assume there is no damping force. The frequency of the oscillation is given by ω=√(k/m)=√(18.9/2.5)=2.75. Keep in mind this is angular frequency, i.e. radians per second, not wavelengths per second.
What effect does Earth’s spherical shape have on the amount of sunlight that reaches the equator?
A.
The latitudes near the equator receive direct sunlight, which causes a high-pressure system.
B.
The latitudes near the equator receive indirect sunlight, which causes a low-pressure system.
C.
The latitudes near the equator receive direct sunlight, which causes a low-pressure system.
D.
The latitudes near the equator receive indirect sunlight, which causes a high-pressure system.
A
Explanation:
The earth is spherical. So it's middle part is bulgjng outside. So more sunlight will be incident on the latitude near the equator. This will heat the air and it will rise up. This will cause high pressure difference and polar disturbances.
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The Earth's spherical shape ensures that the equator receives the most direct sunlight, resulting in higher temperatures and a low-pressure system due to the rising warm air. This leads to the high precipitation in the equatorial region and plays a significant role in global atmospheric and oceanic circulation patterns.
The spherical shape of the Earth influences the way sunlight is distributed across different latitudes. At the equator, the sun's rays are most direct, resulting in a higher concentration of solar energy and higher temperatures. This intense sunlight is focused on a smaller area, leading to the warm conditions that are characteristic of equatorial regions. Due to the heat at the equator, the air warms up, becomes less dense and creates a low-pressure system.
As a consequence of the heat and low-pressure conditions, there is significant convection, which causes air to rise. This is part of the reason why the equatorial region, also known as the Intertropical Convergence Zone, is often characterized by high precipitation levels. The rising warm air can lead to cloud formation and frequent rainfall. In contrast, at higher latitudes, the sun's rays strike the surface at a more oblique angle, spreading the energy over a wider area and resulting in cooler temperatures and different atmospheric conditions.
Therefore, the correct answer to the question is: C. The latitudes near the equator receive direct sunlight, which causes a low-pressure system.
10,500 J of GPE, weight 539 N, how tall is the hill you are sitting on?
To calculate the height of the hill, use the formula h = GPE/Weight with the provided gravitational potential energy (10,500 J) and weight (539 N) of the object. The height equals the GPE divided by the weight, which, upon calculation, gives the height of the hill in meters.
The student is asking how to calculate the height of the hill, given the gravitational potential energy (GPE) and the weight of an object. In physics, the formula to calculate GPE is GPE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s2 on Earth or 10 m/s2 for simplified calculations), and h is the height.
To find the height, one can rearrange the GPE formula to solve for h, which gives us h = GPE/(mg). The weight of an object is the force due to gravity acting on it, which can be calculated by multiplying the mass by the acceleration due to gravity (Weight = mg). Given that the weight (Weight) of the object is known, we can simplify the formula for height to h = GPE/Weight.
Using the information provided by the student, the GPE is 10,500 J, and the weight is 539 N. Substituting these values into our height formula gives us h = 10,500 J / 539 N, which results in an answer after calculation.
Let's calculate the height:
GPE = 10,500 J
Weight = 539 N
Height h = GPE / Weight = 10,500 J / 539 N
=19.48
5) Consider pushing a 50.0 kg box through a 5.00m displacement on both a flat surface and up a
ramp inclined to the horizontal by 15.0°. In both cases, you apply a force of 100. N parallel to the
surface (parallel to the floor or parallel to the ramp). Calculate the work done by:
a) the gravitational force as the box is pushed across the flat ground
b) the gravitational force as the box is pushed up the ramp
c) the force you apply as the box is pushed across the flat ground
d) the force you apply as the box is pushed up the ramp
Explanation:
Work equals the force times the parallel distance.
a) Force gravity is in the downward direction. The box is moving on flat ground, so there's no displacement in that direction.
W = Fd
W = (50 kg) (-9.8 m/s²) (0 m)
W = 0 J
b) This time, there is a displacement in the vertical direction.
W = Fd
W = (50 kg) (-9.8 m/s²) (5.00 m sin 15.0°)
W = -634 J
c) The force is parallel to the displacement.
W = Fd
W = (100. N) (5.00 m)
W = 500. J
d) Again, the force is parallel to the displacement.
W = Fd
W = (100. N) (5.00 m)
W = 500. J
A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N.
Complete Question:
A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N. (8 marks)
a) What is the downward force of gravity on the rocket at blast-off?
b) What is the unbalanced force on the rocket at blast-off?
c) What is the acceleration of the rocket as it leaves the launching pad?
d) As the rocket travels upwards, the engine thrust remains constant, but the mass of the rocket decreases. Why?
e) Does the acceleration of the rocket increase, decrease, or remain the same as the engines continue to fire?
Answer:
a) [tex]-2.8616 \times 10^{7} N[/tex] is the downward force of gravity on the rocket at blast-off.
b) [tex]4.784 \times 10^{6} N[/tex] is the unbalanced force on the rocket at blast-off
c) [tex]1.638 \mathrm{m} / \mathrm{s}^{2}[/tex] is the acceleration of the rocket as it leaves the launching pad
d) Because the propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases when the rocket moves upward.
e) The acceleration of the rocket increases when engines continue to fire
Explanation:
Given:
Mass (m) = [tex]2.92 \times 10^{6} \mathrm{kg}[/tex]
a) In physics, weight can be defined as the applied force on a body by gravity. It is the product of mass (m) and gravity [tex]\left(g=9.8 \mathrm{m} / \mathrm{s}^{2}\right)[/tex]
[tex]\text { weight }(W)=m \times g=2.92 \times 10^{6} \times(-9.8)=28.616 \times 10^{6}=-2.8616 \times 10^{7} N[/tex]
The negative sign indicates the downward force of gravity.
b) To find the unbalanced force on the rocket at blast-off,
Accelerating force,
[tex]F_{a}=F+W=3.34 \times 10^{7}+\left(-2.8616 \times 10^{7}\right)=(3.34-2.8616) \times 10^{7}[/tex]
[tex]F_{a}=0.4784 \times 10^{7}=4.784 \times 10^{6} N[/tex]
c) Newton’s second law of motion states that the object’s acceleration depends on two variable:
Directly proportionate to the object’s force existed Inversely proportionate to the mass of the objectsThe equation can be given as below,
[tex]Force =m \times acceleration[/tex]
[tex]\text { Acceleration }=\frac{F_{a}}{m}=\frac{4.784 \times 10^{6}}{2.92 \times 10^{6}}=1.638 \mathrm{m} / \mathrm{s}^{2}[/tex]
d) The pushing of rocket upward will happen as long as the engine gets fired. The propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases (taotal mass) when the rocket moves upward.
e) The acceleration of the rocket increases when engines continue to fire
Let consider [tex]F_{a}[/tex] is constant, mass gets decreasing, then the acceleration would be increasing (as mass and acceleration are inversely proportionate to each other) .
How is instantaneous speed calculated?
Answer:
Explanation:
Instantaneous speed is the speed of an object at a given point of time. While the speed we usually calculate is the average speed of the object throughout its journey, instantaneous speed can be determined by calculating the derivative of the average velocity of the object in question.
And the formula for average velocity is
V(avg)= Metre/ second, which is distance/time
Hence its derivative would be
dV(avg)=d(Distance)/d(time) which is the instantaneous speed of the object
Final answer:
Instantaneous speed is calculated by taking the absolute value of instantaneous velocity, which is the magnitude of velocity at a specific moment in time, without considering direction.
Explanation:
The calculation of instantaneous speed requires understanding that it is the magnitude of instantaneous velocity. Instantaneous velocity is a vector quantity that has both magnitude and direction, but when we are interested in the speed at an exact moment in time, we disregard the direction. To calculate the instantaneous speed, one typically takes the absolute value of the instantaneous velocity. For instance, if an object has an instantaneous velocity of -3.0 m/s, indicating motion towards the rear of a plane, its instantaneous speed is 3.0 m/s. Unlike average speed, which is calculated by dividing the total distance traveled by the total time taken, instantaneous speed is concerned solely with the rate of motion at a particular instant.
Instantaneous speed is a scalar quantity and, unlike velocity, does not include any directional information. The instantaneous speed can be determined from the instantaneous velocity by simply looking at its magnitude, without regard for its direction.