How many neutrons does protium have
Answer:
I believe none
Explanation:
A balloon was filled to a volume of 2.50 LL when the temperature was 30.0∘C30.0∘C. What would the volume become if the temperature dropped to 11.0∘C11.0∘C.
Answer:
The volume would become 2.34 L if the temperature dropped to 11.0 °C
Explanation:
From the question, A balloon was filled to a volume of 2.50 L when the temperature was 30.0 °C.
To determine what the volume will become if the temperature dropped to 11.0 °C, we will use one of the Gas laws that relates Volume and Temperature. The Gas law that relates Volume and Temperature is the Charlse' law which states that, "the Volume of a fixed mass of gas is directly proportional to its Temperature ( in Kelvin) provided that the Pressure remains constant"
That is
V ∝ T ( at constant pressure)
Where V is the Volume
and T is the Temperature in Kelvin
Then, we can write that
V = kT
Where k is the constant of proportionality
Then,
[tex]\frac{V}{T} = k[/tex]
Hence, we can write that
[tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} } = \frac{V_{3} }{T_{3} } ...[/tex]
∴ [tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }[/tex]
Where [tex]{V_{1} }[/tex] is the initial volume
[tex]{T_{1} }[/tex] is the initial temperature
[tex]{V_{2} }[/tex] is the final volume
and [tex]{T_{2} }[/tex] is the final temperature
From the question,
[tex]{V_{1} }[/tex] = 2.50 L
[tex]{T_{1} }[/tex] = 30.0 °C = 30.0 + 273.15 K = 303.15 K
[tex]{V_{2} }[/tex] = ??
[tex]{T_{2} }[/tex] = 11.0 °C = 11.0 + 273.15 K = 284.15 K
Putting the values into the equation, we get
[tex]\frac{2.50}{303.15} = \frac{V_{2} }{284.15}[/tex]
∴ [tex]V_{2} = \frac{2.50 \times 284.15}{303.15}[/tex]
[tex]V_{2} = 2.34 L[/tex]
Hence, the volume would become 2.34 L if the temperature dropped to 11.0 °C.
The final volume of the balloon is equal to 0.92 Liter.
Given the following data:
Initial volume = 2.50 LInitial temperature = 30°CFinal temperature = 11°CTo determine the final volume of the balloon, we would apply Charles's law:
Mathematically, Charles law is given by the formula;
[tex]\frac{V}{T} =k\\\\\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Where;
T is the temperature of an ideal gas.V is the volume of an ideal gas.Substituting the given parameters into the formula, we have;
[tex]\frac{2.5}{30} = \frac{V_2}{11} \\\\2.5 \times 11 = 30V_2\\\\27.5 = 30V_2\\\\V_2 = \frac{27.5}{30}[/tex]
Final volume = 0.92 Liter.
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Give real-world examples of evidence that supports the evolution of Earth in each category:
Deposition -
Chemical weathering -
Volcanic eruption -
Answer:
real world examples
Explanation:
concave wave rock- The wave rock is formed by weathering of the surrounding area, which helps in proving the deposition part, as the wave rock was below the ground, occurred due to deposition of rock years over years. It is made from very tough material from its surrounding. The weathering reduced the surrounding terrain, while the bedrock remained to witness for the history. --Also, volcanic eruptions have changed earth greatly. EX- Ash and sulfur went into Earth's atmosphere because of the Tambora eruption which dimmed incoming sunlight. It lowered global temperatures by about 3°F. The Mount Pinatubo of 1991 that erupted in the Philippines cooled the planet by about 1°F.
A 200 – g object is tied to the end of a cord and it is turning in horizontal circle of radius 1.20 Cm at the constant 3 rev/sec. Calculate the centripetal acceleration of the object.
At angular speed of 3 rev/s, the object moves a distance equal to 3 times the circumference of the circle each second, or a distance of 3 • 2π (1.20 cm) ≈ 22.6 cm.
So, with a linear speed of 22.6 cm/s = 0.226 m/s, the object has a centripetal acceleration a of
a = (0.226 m/s)² / (0.012 m) ≈ 18.8 m/s²
directed toward the center of the circle.
A soccer ball is kicked with an initial velocity of 35 m/s at an angle of 60 degrees from the horizontal. How far down the field will the ball hit the ground
Answer:
The ball will hit the ground 108.25 m down the field
Explanation:
To determine how far down the field the ball will hit the ground, that is the Range of the ball. From formula to calculate Range in projectile motion,
[tex]R = \frac{u^{2} sin2\theta }{g}[/tex]
Where R is the Range
u is the initial velocity
θ is the angle of projection
and g is acceleration due to gravity (Take g = 9.8 m/s²)
From the question,
Initial velocity, u = 35 m/s
Angle, θ = 60°
Putting these values into the equation, we get
[tex]R = \frac{35^{2} ( sin2(60)) }{9.8}[/tex]
[tex]R = \frac{1225 \times sin(120)}{9.8}[/tex]
[tex]R = \frac{1225 \times 0.8660}{9.8}[/tex]
[tex]R = \frac{1060.85}{9.8}[/tex]
[tex]R = 108.25 m[/tex]
Hence, the ball will hit the ground 108.25 m down the field.
A 4.80 g bullet moves with a speed of 170 m/s perpendicular to the Earth's magnetic field of 5.00×10−5T.
Part A
If the bullet possesses a net charge of 1.06×10−8 C , by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.00 km ?
Answer:
[tex]3.24\times 10^{-7}\ \text{m}[/tex]
Explanation:
m = Mass of bullet = 4.8 g
v = Velocity of bullet = 170 m/s
B = Magnetic field of Earth = [tex]5\times 10^{-5}\ \text{T}[/tex]
q = Charge of bullet = [tex]1.06\times 10^{-8}\ \text{C}[/tex]
a = Acceleration
Time the bullet will be in the air for is [tex]t=\dfrac{1000}{170}=5.88\ \text{s}[/tex]
Force is given by
[tex]F=ma[/tex]
Magnetic force is given by
[tex]F=qvB[/tex]
So
[tex]ma=qvB\\\Rightarrow a=\dfrac{qvB}{m}\\\Rightarrow a=\dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\ \text{m/s}^2[/tex]
From the linear equations of motion we have
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}\times \dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\times 5.88^2\\\Rightarrow s=3.24\times 10^{-7}\ \text{m}[/tex]
The defelection of the bullet is [tex]3.24\times 10^{-7}\ \text{m}[/tex]
The temperature of a 2.0-kg block increases by 5°C when 2,000 J of thermal energy are added to the block. What is the specific heat of the block?
A fish is hung by two spring scales as shown in the diagram. Assume the spring scales are massless. if the bottom scale reads 5N, what will the top scale read?
Question 3 options:
10 N
15 N
5 N
0 N
Answer:
I'm going to guess 15 N
Explanation:
Describe which relationships in Ohm’s law are DIRECT and which are INVERSE. Use examples to support your answer. You can use calculations, drawings, or graphs to make your point more clear
Answer:
I like to memorize excerpt from articles to solve and answer questions like these. I hope this can help, it's from study.com: "The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r."
Your friend clams that objects do not have to be thouching for a magnetic force to cause motion. How would you support your friends claim? Due today
A mother and daughter press their hands together and then push apart while ice skating. Immediately after they push away from each other, how does the motion of the mother and daughter change?
Answer:
J
Explanation:
The daughter moves with greater acceleration backwards because of her weight.
Both mother and daughter move backward, but the daughter moves with greater acceleration. Therefore, option (J) is correct.
What is Newton's second law of motion?This law states that the acceleration of a body depends on two factors. The first one is the mass of the body and the net force acting on the body. The acceleration is inversely proportional to the mass and is directly proportional to the net force acting on the body and
Newton’s second law can be written in an expression as follows:
F = ma
or, a = F/m
According to Newton's third law of motion, there is an equal and opposite reaction for every action. So when the two skaters push each other. They both will move in a backward direction.
But the mass of the daughter is 50 Kg while the mass of the mother is 100 Kg. As the mass of the daughter is less. Therefore daughter moves in a backward direction with greater acceleration.
Therefore, both move backward but the daughter moves with greater acceleration in comparison to the mother.
Learn more about Newton's second law of motion, here:
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#SPJ2
how do I find the x and y components and the resultant force?
Explanation:
1) the diagonal force times the sine of the angle it makes will give you the vertical component or y component . use this to get the vertical components of all the diagonal forces . add all the vertical components as well as the vertical forces together
2) the diagonal force times the cosine of the angle it makes gives the horizontal component or x component. do this to get the x component of all the diagonal forces.
add all the x components as well as the horizontal forces together to get the final x component
3)using the triangle of vectors, the resultant force is calculated
sum of y component = 3.2N
sum of x component= 5.7N
resultant force = 6.5N
A motor lifts a 500kg elevator a height of 100 m at a constant speed in 50 secs. How much power did the motor supply
Answer:
9,800 watts
Explanation:
The first step is to calculate the force
F= mg
= 500 × 9.8
= 4,900 N
The next step is to calculate the work done
= 4,900 × 100
= 490,000 joules
Therefore the power can be calculated as follows
Power= work done /time
= 490,000/50
= 9,800 watts
a 15-n force and a 45-n force act on an object in opposite directions. what is the magnitude of the net force on the object?
Answer: 30 N (in the direction of the 45 N force)
Explanation:
You have to choose a positive and negative direction since the question says the forces are in opposite directions.
Lets assume:
- <----------> +
<------ 15 N
45 ------->
45-15=30
Just before it landed on the moon, the Apollo 12 lunar lander had a mass of 1.4×104 kg. What rocket thrust was necessary to have the lander touch down with zero acceleration?
Answer:
[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]
Explanation:
For a lunar lander acceleration in an upward direction, the net force that acts on it can be expressed as:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]
The equation for the net force that acts on the lunar lander is:
[tex]F_{net} = m_La[/tex]
For the lander touch down to be zero acceleration, it implies that the acceleration of the moving rocket is zero(a free body fall)
i.e. a = 0
We can now regard the Apollo 12 lunar as a freely falling body
However; the force of gravity as a result of the moon acting on the lunar rocket is:
[tex]F_{g(moon)} = m_Lg_m[/tex]
Then; the equation for the thrust force of the lunar rocket is:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]
[tex]m_La = F_{thrust}-m_Lg_m[/tex]
[tex]m_L(0)= F_{thrust}-m_Lg_m[/tex]
[tex]0= F_{thrust}-m_Lg_m[/tex]
[tex]-F_{thrust}= -m_Lg_m[/tex]
[tex]F_{thrust}= m_Lg_m[/tex]
Finally; the thrust force of the lunar rocket can be determined as:
[tex]F_{thrust}= m_Lg_m[/tex]
Acceleration due to gravity ot the surface of the moon = 1.625 m/s²
[tex]F_{thrust}=(1.4 \times 10^{4} \ kg) (1.625 \ m/s^2)[/tex]
[tex]F_{thrust}=2.275 \times 10^4 \ N[/tex]
[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]
A thrust force of 22680N will be necessary to to have the lander touch down with zero acceleration.
Given the data in the question;
Mass of Apollo 12 lunar lander; [tex]m_L = 1.4 * 10^4 kg[/tex]
Using the expression for the net-force that acts on the lander when it is accelerated in the upward direction:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex] ------- Let this be equation 1
Also, the net-force that acts on the lunar lander can be expressed as:
[tex]F_{net} = m_L*a[/tex] ------- Let this be equation 2
Where [tex]m_L[/tex] is the mass of the lunar lander and a is the acceleration ( 0 ),
( the acceleration of the moving rocket is zero and its freely falling )
Also, Force of gravity due to moon that acts on the lunar lander is expressed as:
[tex]F_{g(moon)} = m_L * g_m[/tex] ------- Let this be equation 3
Where [tex]m_L[/tex] is the mass of the lunar lander and [tex]g_m[/tex] is the moon's gravity ( [tex]1.62m/s^2[/tex] )
Lets substitute equation 2 and 3 into 1
[tex]m_L * a = F_{thrust} - (m_L * g_m)\\\\m_L * 0= F_{thrust} - (m_L * g_m)\\\\0=F_{thrust} - (m_L * g_m)\\\\ F_{thrust} = m_L * g_m[/tex]
We substitute our values into the equation
[tex]F_{thrust} = (1.4*10^4kg) * 1.62m/s^2\\\\F_{thrust} = 22680 kg.m/s^2\\\\F_{thrust} = 22680N[/tex]
Therefore, a thrust force of 22680N will be necessary to to have the lander touch down with zero acceleration.
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Answer my last post with all needed work for 75 POINTSSSS vist my profile to see it its on projectile motion !!!!!!!!!!
Answer:
ok
Explanation:
ok ok I will thanks for free
A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?
We are given:
The tuning fork vibrates at 15660 oscillations per minute
Period of one back-and forth movement:
the given data can be rewritten as:
1 minute / 15660 oscillations
60 seconds / 15660 oscillations (1 minute = 60 seconds)
dividing the values
0.0038 seconds / Oscillation
Therefore, one back and forth vibration takes 0.0038 seconds
Jeffrey is using a pulley to get water from a well. He loops a rope over the top of a pulley wheel, then ties the rope to a bucket. He lowers the bucket into the water. To bring the bucket and water up from the well, Jeffrey pulls down on his end of the rope. Why does Jeffrey apply a pulling force to the rope? A. because adding his force to the force generated by the pulley will get the job done faster B. to test the tension in the rope before he uses it to climb down into the well C. to signal to the pulley to start drawing water from the well D. because work must be done on the pulley before it can do work on the bucket
Answer:
D
Explanation:
Taranga lifts 400N load using first class lever. if Taranga applies effort of 100N at 30 cm from the fulcrum what should be the load distance to balance it?
Answer:
d = 7.5 [cm]
Explanation:
To be able to solve this problem we must understand well the concept of a simple lever of First Class, which consists of a support point (fulcrum) and a beam where a force is applied at one end, while at the other end the force exerted is multiplied.
In the attached image we can see an image that will help us better understand the concept of this first-class lever.
The moment created by the effort and the distance is equal to:
M = 100*30 = 3000 [N*cm]
Now the balance is created because we have the same moment exerted by the effort and the one created by the load.
3000 = 400*d
d = 7.5 [cm]
You apply 741 J of energy to lift a box a distance of 2.83 meters, what
is the weight of the box?
Answer:
261.84 N.
Explanation:
From the question given above, the following data were obtained:
Energy (E) = work (W) = 741 J
Distance (d) = 2.83 m
Force (F) =?
Work is simply defined as the product of force and the distance moved in the direction of the the force. Mathematically, it is expressed as:
Work (W) = Force (F) × distance (d)
W = F × d
Thus, we obtained the weight of the object using the above formula.
Work (W) = 741 J
Distance (d) = 2.83 m
Force (F) =?
W = F × d
741 = F × 2.83
Divide both side by 2.83
F = 741 / 2.83
F = 261.84 N
Since force and weight has the same unit of measurement i.e Newton (N), the weight of the object is 261.84 N
Help me guys please with this question
Answer:
[tex]\mid \vec C\mid=31.9[/tex]
Explanation:
Consider the vectors:
[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]
[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]
Calculate the magnitude of
[tex]\vec C=-2\vec B-\vec A[/tex]
Substitute the values of the vectors:
[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]
Operate and remove parentheses:
[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]
Operating both components separately:
[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]
Now find the magnitude of C:
[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]
[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]
[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]
Catching a wave, a 77 kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?
Answer:
Explanation:
The total work done by the wave is expressed as;
Workdone = Potential energy + Kinetic energy
Workdone = mgh + 1/2mv²
m is the mass = 77kg
g is the acceleration due to gravity = 9.8m/s²
v is the velocity = 8.2m/s
h is the height = 1.65m
Substitute into the formula;
Workdone = 77(9.8)(1.65) + 1/2(77)8.2²
Workdone = 1245.09 + 2588.74
Workdone = 3833.83Joules
Hence the amount of non conservative work done on the sofa is 3833.83Joules
Given:
Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kgWe know,
→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]
or,
[tex]= mgh +\frac{1}{2} mv^2[/tex]
By putting the values,
[tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]
[tex]= 1245.09+2588.74[/tex]
[tex]= 3833.83 \ Joules[/tex]
Thus the above approach is right.
Learn more about work done here:
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Bob roller skates with a constant speed of 6 mi/hr. How long will it take him to travel 6 mi.? t=d/s
Answer:
1 hour
Explanation:
t = 6/6
6/6 = 1
You just said that his speed enables him to cover 6 miles in each hour. From this information, I calculated that it will take him one hour to travel 6 miles.
HELP ASAP !!!!!!!!!!!!!!
Answer:
big bang theory
Explanation:
Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.
please help
A toy rocket having a mass of 10 kg and weighing 100 N provides an applied upward thrust force of 400 N. What is the upward acceleration of the rocket? Fair = 400N Fw = 100N 5 m/s2 50 m/s2 40 m/s2 4 m/s² 30 m/s2 3 m/s2
We are given:
Mass of the rocket = 10 kg
Weight of the Rocket = 100 N
Upward thrust applied by the rocket = 400 N
Net upward force on the rocket:
We are given that gravity pulls the rocket with a force of 100 N
Also, the rocket applied a force of 400N against gravity
Net upward force = Upward thrust - Force applied by gravity
Net upward force = 400 - 100
Net upward force = 300 N
Upward Acceleration of the Rocket:
From newton's second law:
F = ma
replacing the variables
300 = 10 * a
a = 30 m/s²
2. Heather and Matthew walk with an average velocity of +0.87 m/s eastward.
If it takes them 27 minutes to walk to the store, what is their
displacement? (include direction)
(5 points)
A weight lifter lifts a 280-N set of weights from ground level to a position over his head, a vertical distance of 1.60 m. How much work does the weight lifter do, assuming he moves the weights at constant speed?
Given :
A weight lifter lifts a 280-N set of weights from ground level to a position over his head, a vertical distance of 1.60 m.
To Find :
How much work does the weight lifter do, assuming he moves the weights at constant speed.
Solution :
We know, work done is given by :
[tex]W= Fdcos\ \theta[/tex]
Here, F = 280 N and d = 1.60 m
Since, weight lifer is applying and upward force and displacement is also in upward direction.
[tex]\theta = 0^{\circ}\\cos\ 0^{\circ} = 1[/tex]
Putting all these in above equation, we get :
[tex]W = 280\times 1.6\times 1\\\\W = 448\ J[/tex]
Therefore, work done by weight lifter 448 J.
Hence, this is the required solution.
The girl climbs the ladder with the painting in 5 seconds. How much work does
she perform? How much power does she use? *
A 30 kg box is being pulled with a force of 125 N. The coefficient of static friction between the box and the floor is 0.35. What is the minimum downward force on the box that will keep it from slipping?
Answer:
The minimum downward force on the box that will keep it from slipping is 63.14 N
Explanation:
Given;
mass of the object, m = 30 kg
applied force, f = 125 N
coefficient of static friction, μ = 0.35
Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.
The net vertical forces on the box is given by;
R - mg - F = 0
F = R - mg
Now, determine normal reaction, R
f = μR
R = f / μ
R = 125 / 0.35
R = 357.14 N
Finally, determine the minimum downward force on the box that will keep it from slipping;
F = R - mg
F = 357.14 - (30 x 9.8)
F = 357.14 - 294
F = 63.14 N
Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N
Everything in the Universe originated from a single point. How was it possible for everything in the Universe to have occupied a
single point?
OA. Only protons and electrons existed in the Universe, and they were pulled together by gravity.
OB. Only single atoms existed in the Universe, and they were squeezed together by magnetic fields.
OC.
All mass in the Universe was in the form of gases, and gases take up minimal space.
OD
All mass in the Universe was once energy, and energy does not take up space.
Answer:
d
Explanation:
