Answer:
True
Explanation:
t = Time taken = 6 s
u = Initial velocity = 0
v = Final velocity
a = Acceleration = 34.3 m/s²
[tex]v=u+at\\\Rightarrow v=0+34.3\times 6\\\Rightarrow v=205.8\ m/s[/tex]
After the six seconds the acceleration from the rocket stops. After the fuel is exhausted the velocity of the rocket will be 205.8 m/s which will reduce by the acceleration due to gravity (i.e., 9.81 m/s²).
Free fall is the state of a body where the only the force of gravity is acting on it other forces are not acting on it.
Hence, the statement here is true.
The weights of bags of baby carrots are normally distributed, with a mean of 32 ounces and a standard deviation of 0.32 ounce. Bags in the upper 4.5% are too heavy and must be repackaged. What is the most a bag of baby carrots can weigh and not need to be repackaged?
Answer:
[tex]x= 32.544[/tex]
Explanation:
given,
mean weight of bag (μ) = 32
standard deviation (σ) = 0.32
percentage of bag heavier = 4.5%
weight of the bag less than 4.5 % = 100 - 4.5
= 95.5%
we have to determine the z- value according to 95.5% or 0.955
using z-table
z-value = 1.70
now, using formula
[tex]Z = \dfrac{x-\mu}{\sigma}[/tex]
[tex]1.70 = \dfrac{x-32}{0.32}[/tex]
[tex]x-32 = 1.70\times {0.32}[/tex]
[tex]x= 32.544[/tex]
The most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.
To determine the most a bag of baby carrots can weigh and not need to be repackaged, we need to find the weight corresponding to the upper 4.5% of the normal distribution.
The weights are normally distributed with a mean of 32 ounces and a standard deviation (σ) of 0.32 ounces.
We first find the z-score that corresponds to the top 4.5% of the distribution.
The cumulative probability up to this z-score is 1 - 0.045 = 0.955.Using a z-table or calculator,
the z-score for 0.955 is approximately 1.645.Next, we use the z-score formula:
Z = (X - μ) / σSolving for X (the weight we need):
1.645 = (X - 32) / 0.32Multiplying both sides by 0.32:
1.645 * 0.32 = X - 320.5264 = X - 32Adding 32 to both sides:
X = 32 + 0.5264 X = 32.5264Thus, the most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.
A planar electromagnetic wave of frequency 8.5×1014Hz is propagating in a vacuum in the +y direction. At a certain point P and at a given instant, the electric field of the wave is given by E⃗ =(0.041V/m)z ^. What is the magnetic vector of the wave at the point P at that instant?
Answer:
[tex]B = 1.37 \times 10^{-10} T[/tex]
Explanation:
As we know that in electromagnetic waves the two fields will induce each other
So here we have
[tex]B = \frac{E}{c}[/tex]
we have
[tex]E = 0.041 V/m[/tex]
[tex]c = 3\times 10^8 m/s[/tex]
so we have
[tex]B = \frac{0.041}{3 \times 10^8}[/tex]
[tex]B = 1.37 \times 10^{-10} T[/tex]
Which of the following is an example of potential rather than kinetic energy?
a. a crawling beetle foraging for food
b. light flashes emitted by firefly water
c. rushing over Niagara Falls
d. a molecule of glucose
A molecule of glucose is an example of potential energy rather than kinetic energy.
What is potential energy and kinetic energy?Potential energy is a form of stored energy that is dependent on the relationship between different system components. Example are when a spring is compressed or stretched, its potential energy increases. If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy.
The energy an object has as a result of motion is known as kinetic energy. A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed.
From the given option, crawling beetle foraging for food, light flashes emitted by firefly water, rushing over Niagara Falls and a molecule of glucose. Molecule of glucose have potential rather than kinetic energy
because there is no motion in the molecule of glucose.
To learn more about potential energy refer to the link
brainly.com/question/24284560
#SPJ2
A molecule of glucose is an example of potential energy because it has energy stored within its chemical bonds waiting to be used, whereas the other examples are instances of kinetic energy in action.
Potential vs. Kinetic Energy
When discussing energy, it is vital to understand the difference between potential energy and kinetic energy. Potential energy is stored energy, waiting to be released, whereas kinetic energy is the energy of motion. An example of potential energy is a molecule of glucose, which contains energy in its chemical bonds that is not currently in motion but has the possibility to be used in the future. This contrasts with kinetic energy, which would include objects or substances that are actively moving or releasing energy, such as a crawling beetle, light flashes from a firefly, or water rushing over Niagara Falls.
To answer the student's question, the example representing potential energy rather than kinetic energy is:
A molecule of glucose because it has chemical energy stored within its bonds.The examples of a crawling beetle, light from a firefly, and rushing water over Niagara Falls all represent kinetic energy as they are all in motion and actively using or releasing energy.
Please help me! Show work! Margaret, whose mass is 52 kg, experienced a net force of 1750 N at the bottom of a roller coaster loop during her school's physics field trip to Six Flags. What is her acceleration at the bottom of the loop?
Answer Choices
A. 4
B. 14
C. 24
D. 34
The acceleration at the bottom of the loop is 34m/[tex]\bold{s^2}[/tex]
Explanation:
Given:
Mass=52kg
Force=1750N
To calculate:
The acceleration at the bottom of the loop
Formula:
Force=Mass x Acceleration
1750=52 x Acceleration
1750/52=Acceleration
Therefore acceleration at the bottom of the loop is 34m/[tex]s^2[/tex]
Roller coasters are mainly based upon acceleration theory they have two types of acceleration one is at the top of the loop and the other is at the bottom of the loop.
Then the net forces and the values are given. In many problems the roller coaster concept is included and it gives another level of clarity to the problems including the net forces
Object A has mass mA = 8 kg and initial momentum pA,i = < 15, -8, 0 > kg · m/s, just before it strikes object B, which has mass mB = 11 kg. Just before the collision object B has initial momentum pB,i = < 2, 7, 0 > kg · m/s.
A) Consider a system consisting of both objects A and B. What is the total initial momentum of this system, just before the collision?
B) The forces that A and B exert on each other are very large but last for a very short time. If we choose a time interval from just before to just after the collision, what is the approximate value of the impulse applied to the two-object system due to forces exerted on the system by objects outside the system?
C) Therefore, what does the Momentum Principle predict that the total final momentum of the system will be, just after the collision?
D) Just after the collision, object A is observed to have momentum pA,f = < 13, 4, 0 > kg · m/s. What is the momentum of object B just after the collision?
Answer:
A)
<17, - 1, 0>
B)
0 Ns
C)
<17, - 1, 0>
D)
<4, - 5, 0>
Explanation:
A)
[tex]p_{A,i}[/tex] = initial momentum of object A = 15 i - 8 j + 0 k
[tex]p_{B,i}[/tex] = initial momentum of object B = 2 i + 7 j + 0 k
Total initial momentum of the system is given as the sum of initial momenta of A and B , hence
[tex]p_{i}[/tex] = [tex]p_{A,i}[/tex] + [tex]p_{B,i}[/tex]
[tex]p_{i}[/tex] = (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)
[tex]p_{i}[/tex] = 17 i - j + 0 k
B)
[tex]F_{ext}[/tex] = Net external force on the two objects = 0 N
[tex]t[/tex] = duration of the collision
[tex]I[/tex] = Impulse
Impulse is given as
[tex]I = F_{ext} t \\I = (0) t \\I = 0 Ns[/tex]
C)
[tex]p_{f}[/tex] = final momentum of the system
we know that the Impulse is nothing but change in momentum of the system of objects, hence
[tex]I = p_{f} - p_{i}\\0 = p_{f} - p_{i}\\p_{f} = p_{i}[/tex]
[tex]p_{f}[/tex] = 17 i - j + 0 k
D)
[tex]p_{A,f}[/tex] = final momentum of object A = 13 i + 4 j + 0 k
[tex]p_{B,f}[/tex] = final momentum of object B = ?
Total final momentum of the system is given as
[tex]p_{f} = p_{A,f} + p_{B,f}[/tex]
17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + [tex]p_{B,f}[/tex]
[tex]p_{B,f}[/tex] = 4 i - 5 j + 0 k
So final momentum of the object is <4, - 5, 0>
The total initial momentum of this system just before the collision is 17 i - j + 0 k.
What is Impulse?
This is defined as a term which quantifies the overall effect of a force acting over time.
Total initial momentum of this system just before the collision:Initial momentum of A = 15 i - 8 j + 0 k
initial momentum of B = 2 i + 7 j + 0 k
Total = sum of initial momentum of A and B
= (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)
= 17 i - j + 0 k
The approximate value of the impulseImpulse = Ft where F is force and t is time
Net external force on the two objects = 0 N
Impulse = (0)t
= 0Ns
The final momentum of the systemImpulse = change in momentum
= 17 i - j + 0 k
The momentum of object B just after the collisionFinal momentum of object A = 13 i + 4 j + 0 k
Final momentum of object B = ?
Total final momentum = Final momentum of object A + Final momentum of object B
17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + Final momentum of object B
= 4 i - 5 j + 0 k
Read more about Momentum here https://brainly.com/question/402617
A sample of nitrogen gas is inside a sealed container. The container is slowly compressed, while the temperature is kept constant. This is a(n) ________ process.
a. constant-volume
b. isothermal
c. adiabatic
d. isobaric
Answer:
b. isothermal
Explanation:
When a thermodynamic process occurs and the temperature remains constant, it is said to be an isothermal process.
The prefix "iso" means "equal", so the word isothermal indicates that the temperature remains equal in the process, and that is what happens in the situation described, the volume is varied (compressed) while maintaining the sample of nitrogen gas remains at a constant temperature.
A 7.0-N force parallel to an incline is applied to a 1.0-kg crate. The ramp is tilted at 20° and is frictionless.
(a) What is the acceleration of the crate?
(b) If all other conditions are the same but the ramp has a friction force of 1.9 N, what is the acceleration?
Final answer:
The acceleration of the crate on the frictionless ramp is 2.39 m/s², while the acceleration on the ramp with friction is 0.49 m/s².
Explanation:
To find the acceleration of the crate, we first need to determine the force component parallel to the incline. This can be done using the formula F_parallel = F * sin(theta), where F is the applied force and theta is the angle of the ramp. Substituting the values, we have F_parallel = 7.0 N * sin(20°) = 2.39 N. The acceleration can then be found using the equation a = F_parallel / m, where m is the mass of the crate. Substituting the values, we have a = 2.39 N / 1.0 kg = 2.39 m/s².
If the ramp has a friction force of 1.9 N, it will oppose the motion of the crate. In this case, we need to subtract the friction force from the parallel force to find the effective force. The effective force, Feff, is given by Feff = F_parallel - friction force. Substituting the values, we have Feff = 2.39 N - 1.9 N = 0.49 N. The acceleration can then be found using the equation a = Feff / m. Substituting the values, we have a = 0.49 N / 1.0 kg = 0.49 m/s².
Natural selection is defined as A. the changes that occur in an organism over its lifespan. B. the changes in the characteristics within a population that lead to survival. C. the changes that occur in a population that do not affect survival. D. the process by which humans select the most desirable organisms in a population.
Answer:
B:the changes in the characteristics within a population that lead to survival.
Explanation:
Answer:
B
Explanation:
study Island
In which one of the following situations is zero net work done?
a) a bunch of bananas are placed on a spring scale in the supermarket.
b) a horse pulls a wagon at a constant velocity
c) a sky diver falls from an airplane before opening her parachute
d) a snowball rolls down a hill e a skateboarder steps on a skateboard and begins to roll
Answer:
b) a horse pulls a wagon at a constant velocity
Explanation:
As we know that ,work done is the dot product of force vector and displacement vector.
W= F.d
W=work
F=Force
d=Displacement
We also know that
F = m a
m= mass ,a = acceleration
When velocity is constant then rate of change in the velocity will be zero,then we can say that acceleration will be zero.
When a= 0 Then F= 0
W= F.d ( F=0)
W = 0
Therefore option b i correct because horse is going with constant velocity.
Zero net work is done when a bunch of bananas is placed on a spring scale in the supermarket, as there is no displacement of the bananas despite the force exerted by the scale.
Explanation:The situation in which zero net work is done is when a bunch of bananas is placed on a spring scale in the supermarket. In this scenario, the force exerted by the scale is upward and counteracts the weight of the bananas, but since there is no displacement or movement of the bananas, the work done is zero. The concept of work in physics is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Since the bananas don't move, the displacement is zero, resulting in zero work.
In the other scenarios provided:
A horse pulls a wagon at a constant velocity has a force in the direction of motion, so work is done.A sky diver falls from an airplane before opening her parachute is experiencing a gravitational force and moving in the direction of that force, hence work is done.A snowball rolls down a hill is also subject to a force (gravity) and motion (down the hill), which means work is done.A skateboarder stepping on a skateboard and beginning to roll is exerting a force that causes the skateboard to move, doing work in the process.Meaning of geosynchronous satellite
Meaning of geosynchronous satellite:
A satellite above Earth placed in a geosynchronous circular orbit, inclined at an angle with the equatorial plane of Earth and orbiting at an orbital period equal to that of earth’s rotational period is termed as a geosynchronous satellite. Geosynchronous orbit is a High Earth orbit at a distance of 42,164 kilometres from the centre of earth such that the gravitational pull from earth is fair enough for the object in orbital motion to match with the speed of earth’s rotational period (24 hrs)
As an observer from ground, a geosynchronous satellite appears to be at a stationary point, at any part of the day; since it's velocity synchronizes with the angular velocity that of earth. A geosynchronous satellite orbiting in an equatorial plane without any inclinations is termed as a geostationary satellite.
What is the potential difference between the terminals of the battery?
Answer:
Explained
Explanation:
Potential difference in a battery across its terminals is the force that pushes the charge to move around the circuit. This means that work id neede to be done to move the charge. And potential difference is the work done per unit charge. The unit of potential difference is V and defined as one joule per coulomb.
V= W/q J/C
The equation for the speed of a satellite in a circular orbit around the earth depends on mass. Which mass?
The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?
a. The mass of the sun
b. The mass of the satellite
c. The mass of the Earth
Answer:The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.
Option c
Explanation:Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence [tex]F_{G} = F_{C}[/tex].
Gravitational force between Earth and Satellite: [tex]F_{G} = \frac{G \times M_e \times M_s}{R^2}[/tex]
Centripetal force of Satellite :[tex]F_C = \frac{M_s \times V^2}{R}[/tex]
Where G = Gravitational Constant
[tex]M_e[/tex]= Mass of Earth
[tex]M_s[/tex]= Mass of satellite
R= Radius of satellite’s circular orbit
V = Speed of satellite
Equating [tex]F_G = F_C[/tex], we get
Speed of Satellite [tex]V =\frac{\sqrt{G \times M_e}}{R}[/tex]
Thus the speed of satellite depends only on the mass of Earth.
Give a combination of four quantum numbers that could be assigned to an electron occupying a 5p orbital.
Answer:
n=5, l=1, m(l) = -1, m(s)= + 1/2
Explanation:
Quantum number are used to describe the position and spin of an electron inside an atom. There are four types of quantum number for describing an electron inside an atom. They are: the principal quantum number, spin quantum number, magnetic quantum number and angular momentum quantum number.
(1).PRINCIPAL QUANTUM NUMBER: denoted by n, and has possible values of n= 1,2,3,4,.... IN HERE, n= 5
(2).ANGULAR MOMENTUM QUANTUM NUMBER: it is denoted by l, and has possible values of l= 0,1,2,3,...,(n-1).
Our l here is one( that is, s-orbital=0, p-orbital=1, d-orbital= 3 and so on)
(3).MAGNETIC QUANTUM NUMBER: The magnetic quantum number, which is denoted by m subscribt l, specifies the exact orbital in which you can find the electron. It has values ranging from -l,...,-1,0,1,...,l.
Here, our value is -1 that is m(l)= -1
(4).SPIN QUANTUM NUMBER: describes the orientation of electrons. Electrons can only have two values here, either a positive one and the half(+1/2) that is the spin up electron or the negative one and half(-1/2) that is the spin down electron.
A set of quantum numbers for an electron in a 5p orbital could be: principal quantum number (n) = 5, angular momentum quantum number (l) = 1, magnetic quantum number (m) = 0, spin quantum number (m_s) = -1/2.
Explanation:
An electron in a 5p orbital can be described using four quantum numbers. Knowing that p orbitals correspond to the angular momentum quantum number l=1, and the principal quantum number n for the fifth energy level is 5, we can deduce the following. The magnetic quantum number, m, ranges from -l to +l, and thus it can have the values -1, 0, or +1 representing different p orbitals in the same shell. The spin quantum number, m_s, can either be -1/2 or +1/2, indicating the orientation of the electron's spin. Thus, an example of a set of four quantum numbers that could be assigned to an electron occupying a 5p orbital could be: (5, 1, 0, -1/2).
Learn more about Quantum Numbers here:https://brainly.com/question/27152536
#SPJ12
A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitude of the motion for which the block does not separate from the plate? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Answer:
maximum amplitude = 0.13 m
Explanation:
Given that
Time period T= 0.74 s
acceleration of gravity g= 10 m/s²
We know that time period of simple harmonic motion given as
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]0.74=\dfrac{2\pi}{\omega}[/tex]
ω = 8.48 rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
10= 8.48² A
A=0.13 m
maximum amplitude = 0.13 m
The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m
Given data :
period ( T ) = 0.74 secs
acceleration due to gravity ( g ) = 9.8 m/s62
Determine the maximum amplitude of the motionTime period of simple harmonic motion ( T ) = [tex]\frac{2\pi }{w}[/tex]
First step : solve for w
w = 2π / T
= 2π / 0.74
= 8.49 rad/sec
Next step : determine the maximum amplitude ( A )
a = w² A
where ; a = maximum acceleration = 9.8 , w = 8.49
therefore
A = a / w²
= 9.8 / ( 8.49 )²
= 0.136 ≈ 0.14 m
Hence we can conclude that The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m.
Learn more about simple harmonic motion : https://brainly.com/question/25849204
Technician A says that the cylinders should be bored and/or honed to match the size of the pistons to be used. Technician B says that bolt holes should be chamfered and cleaned with a thread chaser before assembly. Who is right?
A) Technician A only
B) Technician B only
C) Both technicians A and B
D) Neither technician A nor B
Final answer:
Both Technician A and Technician B are correct.
Explanation:
Technician A is correct. Boring and honing the cylinders is done to achieve the proper size, shape, and finish required for the installation of the pistons. This ensures a precise fit between the cylinders and pistons, allowing for optimal performance and reduced wear.
Technician B is also correct. Chamfering the bolt holes helps to ensure smooth insertion of the bolts and prevents damage to the threads. Cleaning with a thread chaser removes any debris or contaminants that could compromise the integrity of the assembly.
Therefore, the correct answer is C) Both technicians A and B.
Suppose we wanted to represent all of these objects on the 1-to-10-billion scale from Chapter 1, on which the Sun is about the size of a grapefruit. Approximately how large in diameter would the star Aldebaran be on this scale?
Answer:
6.62 m
About 44.14 grapefruits lined in a single line would be the length of the diameter of star Aldebaran
Explanation:
First we know,
Diameter of the Sun - 1.39 million km (Ds)
Diameter of Aldbaran - 61.40 million km (Da)
Thus,
[tex]\frac{Da}{Ds}[/tex] = [tex]\frac{61.40}{1.39}[/tex] ≈ 44.14
We know that in this scale the size of the sun is that of a grapefruit.
The average diameter of a grapefruit is about 15cm
Thus Ds in this scale is 15cm
Thus Da in this scale,
Da = Ds *44.14
Da = 0.15 * 44.14 ≈ 6.62 m
Thus the diameter of Aldebaran in this scale is 6.62 m
A student m1 = 71 kg runs towards his skateboard, which has a mass m2 = 2.8 kg and is d = 2.75 m ahead of him. He begins at rest and accelerates at a constant rate of a = 0.65 m/s2. When he reaches the skateboard he jumps on it. What is the velocity of the student and skateboard in meters per second?
Answer:
The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]
Explanation:
Given:
Mass of student [tex]m_1=71\ kg[/tex]
Mass of skateboard [tex]m_2=2.8\ kg[/tex]
Distance between student and skateboard [tex]d=2.75\ m[/tex]
Acceleration of student [tex]a=0.65\ ms^{-2}[/tex]
Finding velocity [tex]v_1[/tex] of the student before jumping on skateboard
Using equation of motion
[tex]v_1^2=v_0^2+2ad[/tex]
here [tex]v_0[/tex] represents the initial velocity of the student which is [tex]=0[/tex] as he starts from rest.
So,
[tex]v_1^2=0^2+2(0.65)(2.75)[/tex]
[tex]v_1^2=3.575[/tex]
Taking square root both sides:
[tex]\sqrt{v_1^2}=\sqrt[1.7875}[/tex]
∴ [tex]v_1=1.89[/tex]
Finding velocity [tex]v[/tex] of student and skateboard.
Using law of conservation of momentum.
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
Where [tex]v_2[/tex] is initial velocity of skateboard which is [tex]=0[/tex] as it is at rest.
Plugging in values.
[tex]71(1.89)+(2.8)(0)=(71+2.8)\ v[/tex]
[tex]134.19=73.8\ v[/tex]
Dividing both sides by [tex]73.8[/tex]
[tex]\frac{134.19}{73.8}=\frac{73.8\ v}{73.8}[/tex]
∴ [tex]v=1.82[/tex]
The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]
A roller-coaster car rolls down a frictionless track, reaching speed v0 at the bottom. If you want the car to go twice as fast at the bottom, by what factor must you increase the height of the track
Answer:
h should become four times.
Explanation:
The speed at the bottom depends only on the vertical height difference from the top ... if there is no friction ... then the path you take from top to bottom doesn't matter in any way, only the difference in height between the two
mgh at top = 1/2 mv^2 at bottom, so that v=sqrt[2gh]
You see, the final velocity, v, depends on the square root of h, so double v, quadruple h, or four times higher
A power station with an efficiency of 0.4 generates 108 W of electric power and dissipates 1.3×108 J of heat energy each second to the cooling water that flows through it, which increases its temperature by 3.0 Celsius degrees.
find what mass of warmed water flows through the plant each
Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]\frac{dm}{dt} = 10352 kg/s[/tex]
Explanation:
As we know that thermal energy given to the water to raise its temperature is given as
[tex]Q = ms\Delta T[/tex]
now rate of energy given to the system is
[tex]\frac{dQ}{dt} = \frac{dm}{dt} s \Delta T[/tex]
so we have
[tex]\frac{dQ}{dt} = 1.3 \times 10^8[/tex]
s = 4186 J/kg C
now we have
[tex]1.3 \times 10^8 = \frac{dm}{dt} (4186) 3[/tex]
[tex]\frac{dm}{dt} = 10352 kg/s[/tex]
With the given data, the mass of the warmed water is approximately 1.0×10⁴ kg each second.
To determine the mass of water that is warmed as it flows through the power plant, we can use the following steps:
Calculate the amount of heat transferred to the water:
The power station dissipates 1.3×10⁸ J of heat each second.
Use the specific heat capacity formula:
The formula to calculate heat transfer is
Q = m * c * ΔT,where:
Q = heat energy (1.3×10⁸ J)m = mass of water (kg)c = specific heat capacity of water (4,186 J/kg°C)ΔT = change in temperature (3.0°C)Rearrange the formula to solve for mass (m):
m = Q / (c * ΔT)Plugging in the values, we get:
m = (1.3×10⁸ J) / (4,186 J/kg°C * 3.0°C)m ≈ 1.04×10⁴ kgTherefore, the mass of warmed water flowing through the plant each second is approximately 1.0×104 kg.
In a PA projection of the chest being used for cardiac evaluation, the heart measures 15.2 cm between its widest points. If the magnification factor is known to be 1.3, what is the actual diameter of the heart?
A. 9.7 cm
B. 11.7 cm
C. 19.7 cm
D. 20.3 cm
The size of the real heart is B) 11.7 cm
Explanation:
In lenses, mirror and other optical systems, the magnification factor is given by:
[tex]M=\frac{y'}{y}[/tex]
where
y' is the size of the image
y is the size of the real object
In this problem, we have:
M = 1.3 is the magnification factor
y' = 15.2 cm is the size of the image (the image of the heart)
Solving for y, we find the actual size of the heart:
[tex]y=\frac{y'}{M}=\frac{15.2 cm}{1.3}=11.7 cm[/tex]
#LearnwithBrainly
) If it actually starts 6.5 m higher than your answer to the previous part (yet still reaches the top of the loop with the same velocity), how much energy, in joules, did it lose to friction? Its mass is 1700 kg.
The object initially had potential energy of 108,410 Joules when placed 6.5 meters higher. Since it maintains the same speed, this energy was used to overcome friction. Thus, the object lost 108,410 Joules to friction.
Explanation:To answer your question, we first need to calculate the initial potential energy of the object when it was 6.5 m higher. Potential energy (PE) is given by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. So, the initial potential energy is PE = (1700 kg)(9.8 m/s²)(6.5 m) = 108,410 Joules.
Since the object reaches the top of the loop with the same velocity in both scenarios, we can say that its kinetic energy remained the same. This implies all of the initial potential energy went into overcoming friction. Therefore, the object lost 108,410 Joules of energy to friction.
Learn more about Energy loss to friction here:
https://brainly.com/question/34199719
#SPJ6
The object lost 111,230 joules of energy to friction during its ascent.
Explanation:The question is asking how much energy, in joules, the object lost to friction if it starts 6.5 m higher than the previous part and still reaches the top of the loop with the same velocity. To calculate this, we need to find the change in gravitational potential energy, which is given by the formula: ΔPE = mgh.
The mass of the object is 1700 kg, and the change in height is 6.5 m. Substituting these values into the equation, we have:
ΔPE = (1700 kg)(9.8 m/s^2)(6.5 m) = 111,230 J.
Therefore, the object lost 111,230 joules of energy to friction during its ascent.
Learn more about Energy loss to friction here:https://brainly.com/question/29486155
#SPJ11
A spring (k = 802 N/m) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at a = 0.41 m/s2?
The total force on the spring is the sum of the weight of the 5-kg object and the additional force due to the elevator's upward acceleration. This force leads to a stretch in the spring, calculated using Hooke's law, of 0.064 m or 6.4 cm.
Explanation:To solve this problem we'll use Hooke's law (F = -kx), where F is the force, k is the spring constant (in our case, 802 N/m), and x is the displacement or stretch of the spring.
Here, the force isn't just from the weight of the 5.0-kg object (mg), but also from the additional force due to the upward acceleration of the elevator (ma). By adding these two forces together, we can find the total force on the spring (F = mg + ma).
First, let's calculate the weight of the 5-kg object: F_weight = mg = 5.0 kg * 9.81 m/s^2 = 49.05 N.
Next, let's calculate the additional force due to the elevator's acceleration: F_acceleration = ma = 5.0 kg * 0.41 m/s^2 = 2.05 N.
Adding these two forces together: F_total = F_weight + F_acceleration = 49.05 N + 2.05 N = 51.1 N.
Now, we can use Hooke's law to find the stretch of the spring x = F_total / k = 51.1 N / 802 N/m = 0.064 m, or 6.4 cm when converted to centimeters.
Learn more about Spring Stretch here:https://brainly.com/question/13633422
#SPJ12
The spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.
Explanation:The amount by which the spring stretches can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring. In this case, the force exerted by the spring is equal to the weight of the object attached to it. The equation for this can be written as F = kx, where F is the force, k is the spring constant, and x is the displacement. Rearranging this equation, we can solve for x:
x = F / k
Plugging in the values given in the question, we have:
x = (m * a) / k
Substituting the values m = 5.0 kg, a = 0.41 m/s^2, and k = 802 N/m:
x = (5.0 kg * 0.41 m/s^2) / 802 N/m
x ≈ 0.020 m
Therefore, the spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.
Learn more about Determining the displacement of a spring when an object is attached and the elevator is accelerating upward here:https://brainly.com/question/1612309
#SPJ11
One billiard ball is shot east at 1.8 m/s . A second, identical billiard ball is shot west at 0.80 m/s . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90∘ and sending it north at 1.37 m/s .
Answer:
Velocity of ball1 after the collision is 0.35 m/s at 53.87° due south of east.
Explanation:
By conservation of the linear momentum:
[tex]m*V_{1o}+m*V_{2o}=m*V_{1f}+m*V_{2f}[/tex] Since both masses are the same, and expressing the equation for each axis x,y:
[tex]V_{1ox}+V_{2ox}=V_{1fx}[/tex] eqX
[tex]0=V_{1fx}+V_{2fx}[/tex] eqY
From eqX: [tex]V_{1fx}=1m/s[/tex]
From eqY: [tex]V_{1fy}=-1.37m/s[/tex]
The module is:
[tex]V_{1f}=\sqrt{1^2+(-1.37)^2}=0.35m/s[/tex]
The angle is:
[tex]\theta=atan(-1.37/1)=-53.87\°[/tex] This is 53.87° due south of east
The 1.30-kg head of a hammer has a speed of 7.3 m/s just before it strikes a nail and is brought to rest Estimate the temperature rise of a 14-g iron nail generated by 8.0 such hammer blows done in quick succession. Assume the nail absorbs all the energy. The specific heat of iron is 450 J/kg⋅C∘.
Answer:
The rise in temperature is [tex]43.98^{\circ}C[/tex]
Solution:
As per the question:
Mass of hammer, M = 1.30 kg
Speed of hammer, v = 7.3 m/s
Mass of iron, [tex]m_{i} = 14\ g[/tex]
No. of blows, n = 8
Specific heat of iron, [tex]C_{i} = 450\ J/kg.^{\circ}C = 0.45\ J/g^{\circ}C[/tex]
Now,
To calculate the temperature rise:
Transfer of energy in a blow = Change in the Kinetic energy
[tex]\Delta KE = \frac{1}{2}Mv^{2} = \frac{1}{2}\times 1.30\times 7.3^{2} = 34.64\ J[/tex]
For 8 such blows:
[tex]\Delta KE = n\Delta KE = 8\times 34.64\ = 277.12 J[/tex]
Now, we know that:
[tex]Q = m_{i}C_{i}\Delta T[/tex]
[tex]\Delta T= \frac{\Delta KE}{m_{i}C_{i}}[/tex]
[tex]\Delta T= \frac{277.12}{14\times 0.45} = 43.98^{\circ}C[/tex]
A spring with a force constant of 5.5 N/m hasa relaxed length of 2.54 m. When a mass isattached to the end of the spring and allowedto come to rest, the vertical length of thespring is 3.52 m.Calculate the elastic potential energystored in the spring.Answer in units of J.
Answer:
E = 2.6411 J
Explanation:
we know that elastic potential energy could be calculated by:
[tex]E = \frac{1}{2}kx^2[/tex]
where k is the constant of the spring and x is the deform of the spring.
Now x can be calculated as:
x = 3.52m-2.54m
x = 0.98 m
Finally, replacing the data, we get:
[tex]E = \frac{1}{2}(5.5 N/m)(0.98m)^2[/tex]
E = 2.6411 J
A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The wave pulse has an amplitude of A = 0.23 m and takes t = 0.478 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.48 Hz.
1) What is the speed of the wave pulse? m/s2) What is the tension in the slinky?N3) What is the wavelength of the wave pulse?mPLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS
Answer:
1. [tex]v=14.2259\ m.s^{-1}[/tex]
2. [tex]F_T=25.8924\ N[/tex]
3. [tex]\lambda=29.6373\ m[/tex]
Explanation:
Given:
mass of slinky, [tex]m=0.87\ kg[/tex]length of slinky, [tex]L=6.8\ m[/tex]amplitude of wave pulse, [tex]A=0.23\ m[/tex]time taken by the wave pulse to travel down the length, [tex]t=0.478\ s[/tex]frequency of wave pulse, [tex]f=0.48\ Hz=0.48\ s^{-1}[/tex]1.
[tex]\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel[/tex]
[tex]v=\frac{6.8}{0.478}[/tex]
[tex]v=14.2259\ m.s^{-1}[/tex]
2.
Now, we find the linear mass density of the slinky.
[tex]\mu=\frac{m}{L}[/tex]
[tex]\mu=\frac{0.87}{6.8}\ kg.m^{-1}[/tex]
We have the relation involving the tension force as:
[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex]
[tex]14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }[/tex]
[tex]202.3774=F_T\times \frac{6.8}{0.87}[/tex]
[tex]F_T=25.8924\ N[/tex]
3.
We have the relation for wavelength as:
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{14.2259}{0.48}[/tex]
[tex]\lambda=29.6373\ m[/tex]
Taipei 101 (a 101-story building in Taiwan) is sited in an area that is prone to earthquakes and typhoons, both of which can lead to dangerous oscillations of the building. To reduce the maximum amplitude, the building has a tuned mass damper, a 660,000 kg mass suspended from 42 m long cables that oscillate at the same natural frequency as the building. When the building sways, the pendulum swings, reaching an amplitude of 75 cm in strong winds or tremors. Damping the motion of the mass reduces the maximum amplitude of the oscillation of the building.a. What is the period of oscillation of the building?b. During strong winds, how fast is the pendulum moving when it passes through the equilibrium position?
Answers:
a) 13 s
b) 0.362 m/s
Explanation:
We have the following data:
[tex]m=660000 kg[/tex] is the mass of the mass damper
[tex]L=42 m[/tex] is the length of the pendulum
[tex]A=75 cm \frac{1m}{100 cm}=0.75 m[/tex] is the amplitude
a) Period of oscillation:This can be solved by the following equation:
[tex]T=2 \pi \sqrt{\frac{L}{g}}[/tex] (1)
Where:
[tex]T[/tex] is the period
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]T=2 \pi \sqrt{\frac{42 m}{9.8 m/s^{2}}}[/tex] (2)
[tex]T=13 s[/tex] (3)
b) Maximum Velocity:The velocity in a pendulum is maximum [tex]V_{max}[/tex] when the pendulum is in its mean position and the amplitude is maximum. So, the equation in this case is:
[tex]V_{max}=A \frac{2 \pi}{T}[/tex] (4)
[tex]V_{max}=0.75 m \frac{2 \pi}{13 s}[/tex] (5)
[tex]V_{max}=0.362 m/s[/tex]
What type of bonds form between the monomers that make up carbohydrates?
Answer:
Glycosidic Bonds
Explanation:
When two adjacent mono-saccharide units link to form di-saccharides and this repeated chain link forms poly-saccharide. These reactions are called dehydration or condensation reactions because upon the formation of the bond the water molecule is released in the form of a byproduct. Glycosidic bonds are covalent bonds that link ring-shaped sugar molecules to other molecules.
The flowers of the bunchberry plant open with astonishing force and speed, causing the pollen grains to be ejected out of the flower in a mere 0.30 ms at and acceleration of 2.5×104m/s2. If the acceleration is constant, what impulse in 10−10kgm/s is delivered to a pollen of mass 1.0×10−7g? (do not include unit in answer)
Impulse : 7.5.10⁻¹⁰N.s
Further explanationMomentum is the product of the mass of an object and its velocity
p = mvWhereas impulse is defined is the product of the force with the time interval when the force is acting on the object.
I = F. Δtthe pollen grains to be ejected out of the flower in 0.30 ms and acceleration of 2.5 × 10⁴ m/s², then
Δt = 0.3 ms = 3.10⁻⁴ s
a = 2.5 × 10⁴ m/s²
mass of pollen: 1.0 × 10⁻⁷g = 1.0 x 10⁻¹⁰ kg
then:
A force that works according to Newton's second law:
F = m. a[tex]\rm F=1.0\times 10^{-10}\times 2.5\times 10^4\\\\F=2.5\times 10^{-6}\:N[/tex]
Impulse (I) :
[tex]\rm I=2.5 \times 10^{-6}\times 3\times 10^{-4}\\\\I=7.5\times 10^{-10}N.s[/tex]
Learn moreCollisions between 2 objects
https://brainly.com/question/1578457
displacement of A skateboarder
brainly.com/question/1581159
The distance of the elevator
brainly.com/question/8729508
Final answer:
To calculate the impulse delivered to a pollen grain, the mass is converted to kilograms and then multiplied with the given acceleration to find the force. This force is then multiplied by the time over which it acts to obtain the impulse, which is 75 in units of 10⁻¹⁰ kgm/s.
Explanation:
The question asks us to calculate the impulse delivered to a pollen grain by the forceful opening of a bunchberry plant flower. To find the impulse (I), which is the change in momentum, we can use the formula I = Ft, where F is the force and t is the time that the force is applied. Alternatively, since the question provides the acceleration (a) and initial time, we can utilize the link between force, mass, and acceleration (F = ma) and then apply that result to find the impulse using the time.
Firstly, we need to convert the mass of the pollen grain to kilograms (kg) which is the standard unit of mass in physics equations:
1.0 × 10⁻⁷ g = 1.0 × 10⁻¹⁰ kg.
Now, we calculate the force on the pollen grain:
F = ma = (1.0 × 10⁻¹⁰ kg) × (2.5 × 10⁴ m/s²) = 2.5 × 10⁻⁵ N.
Next, with force and the time of application known, we can find the impulse:
I = Ft = (2.5 × 10⁻⁵ N) × (0.30 × 10⁻³ s) = 7.5 × 10⁻⁹ kg·m/s.
In terms of 10⁻¹⁰ kg·m/s, the impulse is 75.
An aluminum cup of 150 cm3 capacity is completely filled with glycerin at 23°C. How much glycerin will spill out of the cup if the temperature of both the cup and glycerin is increased to 41°C? (The linear expansion coefficient of aluminum is 23 × 10-6 1/C°. The coefficient of volume expansion of glycerin is 5.1 × 10-4 1/C°.
Answer:
1.19cm^3 of glycerine
Explanation:
Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:
Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum
Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature
coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1
Change in temperature = 41-23 = 18oC
Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3