Answer:
Explanation:
The first part of question is about the height of the rock from which it falls and hit the ground with speed of 11 m/s. Lets find out that height.
We will use the formula,
[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]
As the initial velocity of the rock was zero. [tex]v_{f} = 0[/tex]
[tex]v^{2} _{f} = 2gh\\ h = v^{2} _{f} / 2g\\h = \frac{(11 m/s)^{2} }{2(9.8 m/s^{2} )} \\h = 6.17 m[/tex]
Now we have to find the height from which the rock should be dropped and it's speed on hitting the ground should be 22 m/s.
Again we will use the same formula, same calculation but the value of velocity now should be 22 m/s.
[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]
[tex]v^{2} _{f} = 2gh\\ h = \frac{(22m/s)^{2} }{2(9.8 m/s^{2}) } \\h = 24.69 m[/tex]
To double the speed of the rock when it hits the ground (from 11 m/s to 22 m/s), the height from which it is dropped should be quadrupled. Hence, the rock should be dropped from a height of 24.5 meters.
Explanation:The question is about finding the height from which a rock should be dropped so that its speed on hitting the ground is 22 m/s, given that when it is dropped from height h, its speed is 11 m/s. To solve this, we can use the physics equation for motion under constant acceleration, which is v² = 2gh, where v is the final velocity, g is the acceleration due to gravity, and h is the height of fall.
First, let us find the height h in the initial scenario: (11)² = 2*9.8*h => h = 6.125 m. Generally, the height h is proportional to the square of the speed, so if we double the final speed, the height should be quadrupled: h'(new height) = 4 * h = 4 * 6.125 m = 24.5 m.
Therefore, the rock should be dropped from a height of 24.5 m so that its speed on hitting the ground is 22 m/s.
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Identify the mathematical relationship that exists between pressure and volume, when temperature and quantity are held constant, as being directly proportional or inversely proportional. Explain your answer and write an equation that relates pressure and volume to a constant, using variables, not the mathematical equation from the best fit line.
Answer:
BOYLE'S Law
Explanation:
Boyle's law states that at constant temperature the volume of a fixed quantity of an ideal gas is inversely proportional to the pressure of the gas.
Mathematically:
From the universal gas law we have:
[tex]P.V=n.R.T[/tex]
where:
P = pressure of the gas
V = volume of the gas
n = no. of moles of gas
T = temperature of the gas
R = universal gas constant
when the mass of gas is fixed i.e. n is constant and temperature is also constant.
[tex]PV=constant[/tex]
[tex]P\propto\frac{1}{V}[/tex]
[tex]P_1.V_1=P_2.V_2[/tex]
here the suffix 1 and 2 denote two different conditions of the same gas.
Calculate the initial (from rest) acceleration of a proton in a 5.00 x 10^6 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
Answer:
Acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]
Explanation:
We have given electric field [tex]E=5\times 10^6N/C[/tex]
Mass of proton is equal to [tex]m=1.67\times 10^{-27}kg[/tex]
And charge on proton is equal to [tex]e=1.6\times 10^{-19}C[/tex]
Electrostatic force will be responsible for the motion of proton
Electrostatic force will be equal to [tex]F=qE=1.6\times 10^{-19}\times 5\times 10^6=8\times 10^{-13}N[/tex]
According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration
This newton force will be equal to electrostatic force
So [tex]1.67\times 10^{-27}\times a=8\times 10^{-13}[/tex]
[tex]a=4.79\times 10^{14}m/sec^2[/tex]
So acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]
Final answer:
The initial acceleration of a proton in a [tex]5.00 * 10^6 N/C[/tex] electric field is approximately [tex]4.79 * 10^1^5 m/s^2[/tex], calculated by applying the electric force formula and Newton's second law.
Explanation:
Calculating Proton Acceleration in an Electric Field
To calculate the initial acceleration of a proton in a [tex]5.00 * 10^6 N/C[/tex] electric field, we follow the steps in the Problem-Solving Strategy for electrostatics:
Identify the known quantities. The electric field (E) is [tex]5.00 * 10^6 N/C[/tex], and the charge of a proton (q) is approximately [tex]1.60 * 10^-^1^9[/tex] C.
Write down the formula for electric force (F = qE).
Calculate the force on the proton: F = [tex](1.60 * 10^-^1^9 C)(5.00 * 10^6 N/C) = 8.00 * 10^-^3 N.[/tex]
Use Newton's second law (F = ma) to find the acceleration (a), knowing the mass of a proton (m) is approximately [tex]1.67 * 10^-^2^7 kg[/tex].
Solve for acceleration: a = F/m =[tex](8.00 * 10^-^3 N) / (1.67 * 10^-^2^7 kg) = 4.79 * 10^1^5 m/s^2[/tex].
Thus, the initial acceleration of the proton is approximately[tex]4.79 * 1015 m/s^2.[/tex]
A 5.00-kg block is in contact on its right side with a 2.00-kg block. Both blocks rest on a horizontal frictionless surface. The 5.00-kg block is being pushed on its left side by a horizontal 20.0-N force. What is the magnitude of the force that the 5.00-kg block exerts on the 2.00-kg block?
Explanation:
The relation between force, mass and acceleration is as follows.
F = ma
or, a = [tex]\frac{F}{m}[/tex]
As two blocks are in contact with each other. Hence, total mass will be as follows.
mass = 5 kg + 2 kg
= 7 kg
Now, we will calculate the acceleration as follows.
a = [tex]\frac{F}{m}[/tex]
= [tex]\frac{20}{7}[/tex]
Hence, force exerted by mass of 2 kg on a mass of 5 kg will be calculated as follows.
[tex]20 - F_{1} = 5 \times \frac{20}{7}[/tex]
[tex]F_{1}[/tex] = 5.714 N
Thus, we can conclude that magnitude of the force that the 5.00-kg block exerts on the 2.00-kg block is 5.714 N.
If a 430 mL ordinary glass beaker is filled to the brim with ethyl alcohol at a temperature of 6.00°C, how much (in mL) will overflow when their temperature reaches 22.0°C?
Answer : The volume of ethyl alcohol overflow will be, 7.49 mL
Explanation :
To calculate the volume of ethyl alcohol overflow we are using formula:
[tex]\Delta V=V_o(\alpha \Delta T)\\\\\Delta V=V_o\times \alpha \times (T_2-T_1)[/tex]
where,
[tex]\Delta V[/tex] = volume expand = ?
[tex]\alpha[/tex] = volumetric expansion coefficient = [tex]0.00109/^oC[/tex]
[tex]V_o[/tex] = initial volume = 430 mL
[tex]T_2[/tex] = final temperature = [tex]22.0^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]6.00^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta V=(430mL)\times (0.00109/^oC)\times (22.0-6.00)^oC[/tex]
[tex]\Delta V=7.49mL[/tex]
Thus, the volume of ethyl alcohol overflow will be, 7.49 mL
Answer:
7.3094 ml
Explanation:
Initial volume of the glass, Vo = 430 ml
Initial temperature, T1 = 6°C
final temperature, T2 = 22°C
Temperature coefficient of glass, γg = 27.6 x 10^-6 /°C
Temperature ethyl alcohol, γa = 0.00109 /°C
Use the formula of expansion of substances
Expansion in volume of glass
ΔVg = Vo x γg x ΔT
ΔVg = 430 x 27.6 x 10^-6 x 16 = 0.1898 ml
Expansion in volume of ethyl alcohol
ΔVa = Vo x γa x ΔT
ΔVa = 430 x 0.00109 x 16 = 7.4992 ml
The amount of volume over flow is
ΔV = ΔVa - ΔVg
ΔV = 7.4992 - 0.1898
ΔV = 7.3094 ml
Thus, the amount of ethyl alcohol over flow is 7.3094 ml.
The position x, in meters, of an object is given by the equation:
x = A + Bt + Ct^2,
where t represents time in seconds.
1. What are the SI units of A, B, and C?
A. m, s, s
B. m, m/s, m/s^2
C. m, m, m
D. m/s, m/s^2, m/s^3
E. m, s, s^2
Answer:
The SI units of A, B and C are :
[tex]m,\ m/s\ and\ m/s^2[/tex]
Explanation:
The position x, in meters, of an object is given by the equation:
[tex]x=A+Bt+Ct^2[/tex]
Where
t is time in seconds
We know that the unit of x is meters, such that the units of A, Bt and [tex]Ct^2[/tex] must be meters. So,
[tex]A=m[/tex][tex]bt=m[/tex][tex]b=\dfrac{m}{s}=m/s[/tex]
[tex]Ct^2=m[/tex][tex]C=m/s^2[/tex]
So, the SI units of A, B and C are :
[tex]m,\ m/s\ and\ m/s^2[/tex]
So, the correct option is (B).
The SI units of A, B, and C is option B [tex]m, m/s, m/s^2[/tex]
The calculation is as follows;The position x, in meters, of an object is provided by the equation:
[tex]x = A + Bt + Ct^2[/tex]
Here
t should be t in seconds
As We know that the unit of x should be in meters, in such a way that the units of A, Bt and must be meters.
So,
A = m
bt = m
[tex]b = m\div s = m/s\\\\ Ct^2 = m\\\\ C = m/s^2[/tex]
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An object with a mass of 49.9 pounds is moving with a uniform velocity of 54.4 miles per hour. Calculate the kinetic energy of this object in joules.
Answer:
6698.03 J
Explanation:
Kinetic Energy: This is a form of mechanical energy that is due to a body in motion. The S.I unit of Kinetic Energy is Joules (J).
The formula for kinetic energy is given as
Ek = 1/2mv².......................... Equation 1
Where Ek = kinetic Energy, m = mass of the object, v = velocity of the object.
Given: m = 49.9 pounds, v = 54.4 miles per hours.
Firstly, we convert pounds to kilogram.
If 1 pounds = 0.454 kg,
Then, 49.9 pounds = (0.454×49.9) kg = 22.655 kg.
Secondly, we convert miles per hours to meters per seconds.
If 1 miles per hours = 0.447 meter per seconds,
Then, 54.4 miles per hours = (0.447×54.4) = 24.3168 meters per seconds.
Substitute the value of m and v into equation 1
Ek = 1/2(22.655)(24.3168)²
Ek = 6698.03 J.
Thus the Kinetic energy of the object = 6698.03 J
Can you find a vector quantity that has a magnitude of zero but components that are not zero? Explain. Can the magnitude of a vector be less than the magnitude of any of its components? Explain.
It is not possible to find a vector quantity of magnitude zero but components different from zero
The magnitude can never be less than the magnitude of any of its components
A ball is released from rest at the top of an incline. It is measured to have an acceleration of 2.2. Assume g=9.81 m/s2. What is the angle of the incline in degrees?
Answer:
θ=12.7°
Explanation:
Lets take the mass of the ball = m
The acceleration due to gravity = g = 9.81 m/s²
The acceleration of the block = a
a= 2.2 m/s²
Lets take angle of incline surface = θ
When block slide down :
The gravitational force on the block = m g sinθ
By using Newton's second law
F= m a
F=Net force ,a acceleration ,m=mass
m g sinθ = ma
a= g sinθ
Now by putting the values in the above equation
2.2 = 9.81 sinθ
[tex]sin\theta =\dfrac{2.2}{9.81}\\sin\theta=0.22\\\theta = 12.7\ degrees[/tex]
θ=12.7°
Given two vectors A⃗ = 4.20 i^+ 7.00 j^ and B⃗ = 5.70 i^− 2.60 j^ , find the scalar product of the two vectors A⃗ and B⃗ .
Applying the concept of scalar product. We know that vectors must be multiplied in their respective corresponding component and then add the magnitude of said multiplications. That is, those corresponding to the [tex]\hat {i}[/tex] component are multiplied with each other, then those corresponding to the [tex]\hat {j}[/tex] component and so on. Finally said product is added.
The scalar product between the two vectors would be:
[tex]\vec{A} \cdot \vec{B} = (4.2\hat{i}+7\hat{j})\cdot (5.7\hat{i}-2.6\hat{j})[/tex]
[tex]\vec{A} \cdot \vec{B} = (4.2*5.7) +(7*(-2.6))[/tex]
[tex]\vec{A} \cdot \vec{B} = 5.74[/tex]
Therefore the scalar product between this two vectors is 5.74
Final answer:
The scalar product of vectors A = 4.20 i + 7.00 j and B = 5.70 i - 2.60 j is calculated by multiplying corresponding components and adding them up, resulting in a scalar product of 5.74.
Explanation:
To find the scalar product (also known as the dot product) of two vectors, you multiply the corresponding components of the vectors and then add these products together. Given two vectors A = 4.20 i + 7.00 j and B = 5.70 i - 2.60 j, the scalar product A cdot B is calculated as follows:
Multiply the x-components together: (4.20)(5.70)
Multiply the y-components together: (7.00)(-2.60)
Add these two products together to get the scalar product.
Now let's do the calculations:
(4.20)(5.70) = 23.94
(7.00)(-2.60) = -18.20
23.94 + (-18.20) = 5.74
Therefore, the scalar product of vectors A and B is 5.74.
A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/m
Part A What work is done by the electric force when the charge moves a distance of 0.480 m to the right?
Part B What work is done by the electric force when the charge moves a distance of 0.660 m upward?
Part C What work is done by the electric force when the charge moves a distance of 2.50 m at an angle of 45.0° downward from the horizontal?
A) The work done by the electric field is zero
B) The work done by the electric field is [tex]9.1\cdot 10^{-4} J[/tex]
C) The work done by the electric field is [tex]-2.4\cdot 10^{-3} J[/tex]
Explanation:
A)
The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).
The work done by a force is given by the equation
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement of the particle
[tex]\theta[/tex] is the angle between the direction of the force and the direction of the displacement
In this problem, we have:
The force is directed vertically upward (because the field is directed vertically upward)The charge moves to the right, so its displacement is to the rightThis means that force and displacement are perpendicular to each other, so
[tex]\theta=90^{\circ}[/tex]
and [tex]cos 90^{\circ}=0[/tex]: therefore, the work done on the charge by the electric field is zero.
B)
In this case, the charge move upward (same direction as the electric field), so
[tex]\theta=0^{\circ}[/tex]
and
[tex]cos 0^{\circ}=1[/tex]
Therefore, the work done by the electric force is
[tex]W=Fd[/tex]
and we have:
[tex]F=qE[/tex] is the magnitude of the electric force. Since
[tex]E=4.30\cdot 10^4 V/m[/tex] is the magnitude of the electric field
[tex]q=32.0 nC = 32.0\cdot 10^{-9}C[/tex] is the charge
The electric force is
[tex]F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N[/tex]
The displacement of the particle is
d = 0.660 m
Therefore, the work done is
[tex]W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J[/tex]
C)
In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is
[tex]\theta=90^{\circ}+45^{\circ}=135^{\circ}[/tex]
Moreover, we have:
[tex]F=1.38\cdot 10^{-3} N[/tex] (electric force calculated in part b)
While the displacement of the charge is
d = 2.50 m
Therefore, we can now calculate the work done by the electric force:
[tex]W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J[/tex]
And the work is negative because the electric force is opposite direction to the displacement of the charge.
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A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 84.0 mL. The liquid solvent has a mass of 26.5 g and a density of 0.865 g/mL. Determine the mass of the solid given its density is 3.25 g/mL.
Answer:
The mass of solid is 173.45 g.
Explanation:
Given that,
Total volume of solid and liquid = 84.0 mL
Mass of liquid = 26.5 g
Density of liquid = 0.865 g/mL
Density of solid = 3.25 g/mL
We need to calculate the volume of liquid
Using formula of density
[tex]\rho_{l}=\dfrac{m_{l}}{V_{l}}[/tex]
[tex]V_{l}=\dfrac{m_{l}}{\rho_{l}}[/tex]
Put the value into the formula
[tex]V_{l}=\dfrac{26.5}{0.865}[/tex]
[tex]V_{l}=30.63\ mL[/tex]
We need to calculate the volume of solid
Volume of solid = Total volume of solid and liquid- volume of liquid
[tex]V_{s}=84.0-30.63[/tex]
[tex]V_{s}=53.37\ mL[/tex]
We need to calculate the mass of solid
Using formula of density
[tex]\rho_{s}=\dfrac{m_{s}}{V_{s}}[/tex]
[tex]m_{s}=\rho_{s}\timesV_{s}[/tex]
Put the value into the formula
[tex]m_{s}=3.25\times53.37[/tex]
[tex]m_{s}=173.45\ g[/tex]
Hence, The mass of solid is 173.45 g.
To find the mass of the solid, subtract the volume of the liquid solvent from the total volume of the mixture. Then use the density of the solid to find its mass.The mass of the solid is 173.44 g.
Explanation:To find the mass of the solid, we can first find the total volume of the solid by subtracting the volume of the liquid solvent from the total volume of the mixture.
Since the density of the liquid solvent is given, we can use it to calculate its volume.
The volume of the liquid solvent is found by dividing its mass by its density: 26.5 g / 0.865 g/mL = 30.64 mL. Therefore, the volume of the solid is 84.0 mL - 30.64 mL = 53.36 mL.
Next, we can use the density of the solid to find its mass.
The density of the solid is given as 3.25 g/mL.
We can use the formula mass = density * volume to find the mass of the solid.
Plugging in the values, we get:
mass = 3.25 g/mL * 53.36 mL = 173.44 g.
Therefore, the mass of the solid is 173.44 g.
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When a honeybee flies through the air, it develops a charge of +18pC. How many electrons did it lose in the process of acquiring this charge?
The honeybee lost about 112 million electrons to acquire a charge of +18pC. This calculation involves converting the charge from picocoulombs to coulombs and then dividing by the charge of a single electron.
Explanation:To calculate the number of electrons a honeybee lost to acquire a charge of +18pC, we first need to understand that the fundamental unit of charge, often represented as e, is +1.602 x 10-19 C for a proton and -1.602 x 10-19 C for an electron.
The number of electrons lost (n_e) is the total charge divided by the charge per electron. Therefore, we convert the charge of the honeybee from picocoulombs (pC) to coulombs (C) by multiplying by 10-12, because 1pC = 10-12C. The +18pC charge is thus equivalent to 18 x 10-12 C.
In relation to the charge of an electron, the honeybee's charge is -18 x 10-12 C / -1.602 x 10-19 C/e-, which gives approximately 1.12 x 108 or 112,000,000 electrons.
So, the honeybee lost about 112 million electrons to get a positive charge of +18pC.
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Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ?
Explanation:
A nucleus that absorbs 11.45δ is less shielded than a nucleus that absorbs at 4.13δ.
the nucleus that absorbs at 11.45δ requires weaker applied field strength to come into resonance than the nucleus that absorbs at 4.13δ.
A nucleus absorbing at 4.13 δ is more shielded than one absorbing at 11.45 δ because it is in a weaker local magnetic field and resonates at a lower frequency.
Explanation:In the context of nuclear magnetic resonance (NMR) spectroscopy, the chemical shift is given in units of delta (δ) which represents the resonance frequency of a nucleus relative to a standard reference compound. When a nucleus is surrounded by a dense cloud of electrons, it is considered to be ‘shielded’. A shielded nucleus is influenced by a smaller local magnetic field because the electrons repel some of the external magnetic field. As a result, shielded nuclei resonate at a lower frequency (higher δ values) when compared to de-shielded nuclei. Therefore, a nucleus that absorbs at 4.13 δ is more shielded than a nucleus that absorbs at 11.45 δ because the latter is in a stronger local magnetic field and is, hence, de-shielded.
Aristarchus measured the angle between the Sun and the Moon when exactly half of the Moon was illuminated. He found this angle to be A greater than 90 degrees. B exactly 90 degrees. C less than 90 degrees by an amount too small for him to measure. D less than 90 degrees by an amount that was easy for him to measure.
Answer:
when the Sun illuminates half of the Moon it must be at 90°
Explanation:
The Moon has a circular motion around the Earth and the relative position of the sun, the earth and the moon create the lunar phases.
For this case when the Sun illuminates half of the Moon it must be at 90°, this angle changes with the movement of the moon, it is zero degree for the new moon and 180° for the full moon
Two movers push horizontally on a refrigerator. One pushes due north with a force of 150 N and the other pushes due east with a force of 200 N. Find the direction and magnitude of the resultant force on the refrigerator.
Answer:
R=250 N
α = 38.86⁰
Explanation:
Given that
Force ,F₁ = 150 N (Towards north )
Force ,F₂ = 200 N ( Towards east )
We know that the angle between north and east direction is 90⁰ .
The resultant force R is given as
[tex]R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta\\[/tex]
We know that
cos 90⁰ = 0
That is why
[tex]R=\sqrt{150^2+200^2}\\R=250\ N[/tex]
Therefore the magnitude of the forces will be 250 N.
The angle from the horizontal of the resultant force = α
[tex]tan\alpha=\dfrac{150}{200}\\tan\alpha=0.75\\\alpha=38.86\ degrees[/tex]
R=250 N
α = 38.86⁰
the boiling point of sulfur is 444.6 celsius .sulfur melting point is 586.1 fahrenheit lower than its boiling?
point.
a. Determine the melting point of sulfur is degrees celsius.
b. FInd the melting and boiling Points in degrees fahrenheit.
c. FInd the melting and boiling points in kelvins
Answer:
(a) Melting point is 136.8°C
(b) Melting point is 278.24°F
Boiling point is 832.28°F
(c) Melting point is 409.8K
Boiling point is 717.6K
Explanation:
(a) 586.1°F = 5/9(586.1 - 32)°C = 307.8°C
Melting point = 444.6°C - 307.8°C = 136.8°C
(b) Melting point = 136.8°C = (9/5×136.8) + 32 = 278.24°F
Boiling point = 444.6°C = (9/5×444.6) + 32 = 832.28°F
(c) Melting point = 136.8°C = 136.8 + 273 = 409.8K
Boiling point = 444.6°C = 444.6 + 273 = 717.6K
To determine the melting point of sulfur in degrees Celsius and Fahrenheit, and in Kelvin.
Explanation:a. To determine the melting point of sulfur in degrees Celsius, we can subtract the Fahrenheit value from the boiling point of sulfur. Given that the melting point of sulfur is 586.1 Fahrenheit lower than its boiling point, we can calculate the melting point as follows:
Melting point in degrees Celsius = Boiling point in degrees Celsius - 586.1
b. To convert the melting and boiling points from degrees Celsius to Fahrenheit, we can use the formula:
Degrees Fahrenheit = (Degrees Celsius × 9/5) + 32
c. To convert the melting and boiling points from degrees Celsius to Kelvin, we can use the formula:
Kelvin = Degrees Celsius + 273.15
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A steel piano wire is 0.7 m long and has a mass of 5 g. It is stretched with a tension of 500 N. What is the speed of transverse waves on the wire? To reduce the wave speed by a factor of 2 without changing the tension, what mass of copper wire would have to be wrapped around the wire?
The speed of transverse waves on the steel piano wire can be calculated using the formula √(Tension / (Mass per unit length)). To reduce the wave speed by a factor of 2 without changing the tension, we can solve for the new wave speed, and then calculate the difference in mass per unit length with the copper wire.
Explanation:The speed of transverse waves on a steel piano wire can be calculated using the formula:
Speed = √(Tension / (Mass per unit length))
Where the tension in the wire is 500 N and the mass per unit length is calculated by dividing the mass of the wire by its length. Therefore, the mass per unit length is 5 g / 0.7 m = 7.14 g/m.
To reduce the wave speed by a factor of 2 without changing the tension, we can use the equation:
New wave speed = √(New tension / (Mass per unit length))
We can solve this equation for the new tension by rearranging it as:
New tension = (New wave speed)^2 * (Mass per unit length)
Since we want to reduce the wave speed by a factor of 2, the new wave speed is half the original speed. Substituting these values into the equation, we have:
(0.5 * Old wave speed)^2 * (Mass per unit length) = 500 N
Solving for the new mass per unit length gives:
New mass per unit length = 500 N / (0.5 * Old wave speed)^2
The difference between the new mass per unit length and the mass per unit length of copper wire is the mass of copper wire that needs to be wrapped around the steel wire. However, we would need additional information to calculate this difference.
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In what ways do observed extrasolar planetary systems differ from our own solar system?
Answer:
Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet, as well as temperature or amount of heat received in each planet.
Explanation:
An extrasolar planet is a planet outside the Solar System, while the Solar System orbit around the sun as a result of the gravitational pull of the sun.
Thus, we can say that the major difference between extrasolar planetary systems and solar system is that in solar system, planets orbit around the Sun, while in extrasolar planetary systems, planets orbit around other stars.
All of the planets in our solar system orbit around the Sun. Planets that orbit around other stars are called exoplanets or extrasolar.
Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet. They also differ in terms of temperature, because the temperature in each planet in solar system depends on its distance from the sun while that of the extrasolar depends on the activities of the star.
sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz. How far doesthe needlemove in one period?
If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m
If the displacement of a particle undergoing simple harmonic motion of amplitude A at a time is [tex]→x=Asin[/tex]ω[tex]t^i[/tex]
then the total displacement of the particle over one period of the oscillation from time t=0 is:= [tex](+A^i)+ (-2A^i)+(+A^i)[/tex]
= [tex]0^i.[/tex]
The total distance traveled by the particle during that one period is A+2A+A = 4AGiven:
Amplitude = 0.0127 m
frequency = 2.55 Hz
Solution:
To find the displacement of the needle in one period we need to put value in the formula:
The total distance traveled = 4A
= 4*0.0127 m
= 0.0508 m
Thus, If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m
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Final answer:
The distance moved by the sewing machine needle in one period of its simple harmonic motion is four times the amplitude, which equals 0.0508 meters.
Explanation:
The simple harmonic motion of a sewing machine needle is defined by its amplitude and frequency. The amplitude (0.0127 m) is the maximum displacement from its rest position, and the needle's frequency (2.55 Hz) indicates how often it repeats this motion in one second. To determine the distance the needle moves in one period, we need to consider that it moves from the rest position to the maximum amplitude, back through the rest position to the negative maximum amplitude, and back to rest again, completing one full cycle.
In simple harmonic motion, the distance moved in one period is four times the amplitude. So, the needle moves a distance of 4 x 0.0127 m in one period. Therefore, the needle moves 0.0508 meters in one cycle.
Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a 750-ft ramp at a high speed v0 and travels 540 ft in 6 s at constant deceleration before its speed is reduced to v0/2. Assuming the same constant deceleration, determine (a) the additional time required for the truck to stop, (b) the additional distance traveled by the truck.
Final answer:
By using the equations of motion under constant acceleration, we can calculate the additional time and distance required for a truck, decelerating on a ramp, to come to a complete stop after having its speed reduced to half of its initial value.
Explanation:
A truck enters a 750-ft ramp at high speed v0 and travels 540 ft in 6 s at constant deceleration before its speed is reduced to v0/2. To solve for both the additional time required for the truck to stop and the additional distance traveled, we use the equations of motion under constant acceleration.
Given:
Initial distance traveled: 540 ft
Time taken: 6 seconds
Initial speed: v0
Final speed at this stage: v0/2
Solution:
Calculate the constant deceleration using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Using the determined deceleration, calculate the additional time required for the truck to stop using the formula: t = (v - u) / a.
To find the additional distance traveled, we use the formula: s = ut + 0.5at2.
Through these calculations, we can determine the additional time required for the truck to stop and the additional distance it will travel.
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 4.80×10−19 JJ . In what direction and through what potential difference Vb−VaVb−Va does the particle move?
1) Potential difference: 1 V
2) [tex]V_b-V_a = -1 V[/tex]
Explanation:
1)
When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by
[tex]\Delta U=q\Delta V[/tex]
where
q is the charge's magnitude
[tex]\Delta V[/tex] is the potential difference between the initial and final position
In this problem, we have:
[tex]q=4.80\cdot 10^{-19}C[/tex]is the magnitude of the charge
[tex]\Delta U = 4.80\cdot 10^{-19}J[/tex] is the change in kinetic energy of the particle
Therefore, the potential difference (in magnitude) is
[tex]\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V[/tex]
2)
Here we have to evaluate the direction of motion of the particle.
We have the following informations:
- The electric potential increases in the +x direction
- The particle is positively charged and moves from point a to b
Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)
This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:
[tex]V_b-V_a = - 1V[/tex]
Final answer:
The particle moves in the direction of decreasing potential and through a potential difference of 3.0 V.
Explanation:
The potential difference through which the particle moves can be calculated using the formula for work done by the electric force: W = q(Vb - Va), where W is the change in kinetic energy, q is the charge of the particle, and (Vb - Va) is the potential difference. Rearranging the formula, we have Vb - Va = W/q. Substituting the given values, Vb - Va = (4.80×10-19 J) / (-1.60×10-19 C) = -3.0 V.
This implies that the particle moves in the direction of decreasing potential. Therefore, the particle moves from point a to point b through a potential difference of 3.0 V in the direction of decreasing potential.
Assuming the same initial conditions as described in FNT 2.2.1-1, use the energy-interaction model in two different ways (parts (a) and (b) below) to determine the speed of the ball when it is 4 meters above the floor headed down:
a) Construct a particular model of the entire physical process, with the initial time when the ball leaves Christine’s hand, and the final time when the ball is 4 meters above the floor headed down.
b) Divide the overall process into two physical processes by constructing two energy-system diagrams and applying energy conservation for each, one diagram for the interval corresponding to the ball traveling from Christine’s hand to the maximum height, and then one diagram corresponding to the interval for the ball traveling from the maximum height to 4 meters above the floor headed down.
c) Did you get different answers (in parts (a) and (b)) for the speed of the ball when it is 4 meters above the floor headed down?
Answer:
(a). Vf = 7.14 m/s
(b). Vf = 7.14 m/s
(c). same answer
Explanation:
for question (a), we would be applying conservation of energy principle.
but the initial height is h = 1.5 m
and the initial upward velocity of the ball is Vi = 10 m/s
Therefore
(a). using conservation law
Ef = Ei
where Ef = 1/2mVf² + mghf ........................(1)
also Ei = 1/2mVi² + mghi ........................(2)
equating both we have
1/2mVf² + mghf = 1/2mVi² + mghi
eliminating same terms gives,
Vf = √(Vi² + 2g (hi -hf))
Vf = √(10² + -2*9.8*2.5) = 7.14 m/s
Vf = 7.14 m/s
(b). Same process as done in previous;
Ef = Ei
but here the Ef = mghf ...........(3)
and Ei = 1/2mVi² + mghi ...........(4)
solving for the final height (hf) we relate both equation 3 and 4 to give
mghf = 1/2mVi² + mghi ..............(5)
canceling out same terms
hf = hi + Vi²/2g
hf = 1.5 + 10²/2*9.8 = 6.60204m ............(6)
recalling conservation energy,
Ef = Ei
1/2mVf² + mghf = mghi
inputting values of hf and hi we have
Vf = √(2g(hi -hf)) = 7.14 m/s
Vf = 7.14 m/s
(c). From answer in option a and c, we can see there were no changes in the answers.
Two identical stones, A and B, are thrown from a cliff from the same height and with the same initial speed. Stone A is thrown vertically upward, and stone B is thrown vertically downward. Which of the following statements best explains which stone has a larger speed just before it hits the ground, assuming no effects of air friction?
a. Both stones have the same speed; they have the same change in Ugand the same Ki
b. A, because it travels a longer path.
c. A, because it takes a longer time interval.
d. A, because it travels a longer path and takes a longer time interval.
e. B, because no work is done against gravity.
Answer:
Option A
Explanation:
This can be explained based on the conservation of energy.
The total mechanical energy of the system remain constant in the absence of any external force. Also, the total mechanical energy of the system is the sum of the potential energy and the kinetic energy associated with the system.
In case of two stones thrown from a cliff one vertically downwards the other vertically upwards, the overall gravitational potential energy remain same for the two stones as the displacement of the stones is same.
Therefore the kinetic energy and hence the speed of the two stones should also be same in order for the mechanical energy to remain conserved.
Answer:
b. A, because it travels a longer path.
Explanation:
If the stone A is thrown is thrown vertically upwards and another stone is dropped down directly from the same height above the ground then the stone A will hit the ground with a higher speed because it falls down from a greater height above the earth surface.This can be justified by the equation of motion given below:
[tex]v^2=u^2+2\times a\times s[/tex]
where:
[tex]v=[/tex] final velocity
[tex]u=[/tex] initial velocity
[tex]a=[/tex] acceleration = g (here)
[tex]s=[/tex] displacement of the body
Now we know that at the maximum height the speed of the object will be zero for a moment. So for both the stones A and B the initial velocity is zero, stone B is also dropped from a height with initial velocity zero.Acceleration due to gravity is same for the stones so the only deciding factor that remains is s, displacement of the stones. Since stone A is thrown upwards it will attain a greater height before falling down.A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is 4.2 m/s due east. The river is 500 m wide. (a) What is your velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of your starting point will you reach the opposite bank?
The velocity of the boat relative to the earth is 7.2 m/s, 32.0° south of east. It will take the boat 0.118 seconds to cross the river. The boat will reach a point 0.236 meters south of the starting point.
Explanation:To find the velocity of the boat relative to the earth, we need to find the resultant of the boat's velocity relative to the water and the river's velocity relative to the earth. Using vector addition, we can find that the magnitude of the total velocity is 7.2 m/s and the direction is 32.0° south of east.
To find the time required to cross the river, we can use the formula t = d/v, where d is the width of the river and v is the horizontal component of the boat's velocity relative to the earth. Plugging in the values, we find that the time required is 0.118 seconds.
To find how far south of the starting point the boat will reach the opposite bank, we can use the formula d = v*t, where v is the vertical component of the boat's velocity relative to the earth and t is the time. Plugging in the values, we find that the boat will reach a point 0.236 meters south of the starting point.
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We use vector addition to find the magnitude of 4.65 m/s at an angle of 25.4° south of east. The time required to cross the river is approximately 119.05 seconds, and the boat will reach a point around 238.1 meters south of its starting point.
The question is about calculating the motion of a boat in a river flowing in a specific direction. Let's solve each part step-by-step.
(a) The boat's velocity relative to the water is 4.2 m/s due east, and the river flows due south at 2.0 m/s. We can use the Pythagorean theorem to find the magnitude:
[tex]Magnitude = \sqrt{(velocity_x^2 + velocity_y^2)}\\Magnitude = \sqrt{(4.22 + 2.02)}\\Magnitude = \sqrt{(17.64 + 4.00)}\\Magnitude = \sqrt{21.64}\\Magnitude = 4.65 m/s[/tex]
To find the direction, we need to calculate the angle θ relative to the east:
[tex]tan(\theta) = velocity_y / velocity_x = 2.0 / 4.2\\\theta = tan-1(2.0 / 4.2)\\\theta \approx 25.4\textdegree south\right \left of\right \left east[/tex]
(b) Given the river's width is 500 m and the boat's velocity relative to the water is 4.2 m/s, we use the formula:
Time = Distance / Velocity
Time = 500 m / 4.2 m/s
Time ≈ 119.05 s
(c) To find the southward distance, we use the river's flow velocity and the time calculated in part (b):
[tex]Distance_{south} = Velocity_{river} \times Time\\Distance_{south} = 2.0 m/s \times 119.05 s\\Distance_{south} \approx 238.1 m[/tex]
Why are jovian planets so much larger than terrestrial planets?
Explanation:
The temperature in the inner solar system was too high for light gases to condense, while in the outer solar system, the temperature was much lower, which allowed the Jovian planets to form, which grew enough to accumulate and retain the hydrogen gas that remained in the solar nebula, which led to its high levels of hydrogen and large size.
A jet fighter pilot wishes to accelerate from rest at a constant acceleration of 5 g to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 m/s for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5g before he blacks out?
Final answer:
The jet fighter pilot will black out during the acceleration period as it lasts longer than 5.0 s. The greatest speed the pilot can reach with an acceleration of 5g before blacking out is 245 m/s.
Explanation:
To determine whether the jet fighter pilot will black out, we need to calculate the time of acceleration. The speed of sound is v = 331 m/s. The speed the pilot wants to reach is 3 times the speed of sound, so the desired speed is 3 * 331 m/s = 993 m/s. We can use the equation:
v = u + at
Where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration (5 g = 5 * 9.8 m/s^2 = 49 m/s^2), and t is the time.
Substituting the values, we get:
993 m/s = 0 m/s + 49 m/s² x t
Rearranging this equation to solve for t, we get:
t = 993 m/s / 49 m/s^2 = 20.265 s
Since the acceleration lasts for longer than 5.0 seconds, the pilot will black out during the acceleration period.
To calculate the greatest speed the pilot can reach before blacking out, we can use the same equation, but rearrange it to solve for v:
v = u + at
Substituting the values, we get:
v = 0 m/s + 49 m/s^2 x 5.0 s = 245 m/s
Therefore, the greatest speed the pilot can reach with an acceleration of 5 g before blacking out is 245 m/s.
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver notices a red light ahead and slows down with constant acceleration −a0. Just as the car comes to a full stop, the light immediately turns green, and the car then accelerates back to its original speed v0 with constant acceleration a0. During the same time interval, the train continues to travel at the constant speed v0.
Answer:
A)The time taken for the car to come to a full stop is:
[tex]t=\displaystyle \frac{v_0}{a_0}[/tex]
b) The time taken for the car to accelerate from full stop to its original cruising speed is:
[tex]t=\displaystyle \frac{v_0}{a_0}[/tex]
c) The separation distance between the car and the train is:
[tex]d=v_0^2/a_0[/tex]
Completed question:
a) How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0.
b) How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
c) The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed again? Express the separation distance in terms of v0 and a0. Your answer should be positive.
Explanation:
a) The car slows down with constant acceleration (-a₀). Therefore, this movement is a linearly accelerated motion:
[tex]v(t)=v_0+t\cdot (-a_0)\\0=v_0+t\cdot (-a_0)\\t=v_0/a_0[/tex]
b) for the acceleration process we use the same equation than before:
[tex]v(t)=v_{orig}+t\cdot (a_0)\\v_0=0m/s+t\cdot (a_0)\\t=v_0/a_0[/tex]
c) To determine the separation distance between the car and the train we can observe how much distance each of them travels in the time spend for the car to deaccelerate and accelerate again.
For the train:
[tex]x(t_{tot})=x_0+v_0\cdot(t_{tot})\\x(t_{tot})=v_0\cdot(t_{dea}+t_{acc})\\x(t_{tot})=2v_0^2/a_0[/tex]
For the car:
[tex]x(t_{tot})=x(t_{dea})+x(t_{acc})\\x(t_{tot})=v_0\cdot(t_{dea})+0.5(-a)(t_{dea})^2+0.5a(t_{acc})^2\\x(t_{tot})=v_0\cdot(t_{dea})\\x(t_{tot})=v_0^2/a_0[/tex]
Therefore the separation distance between the car and the train is:
[tex]d=|x_{car}-x_{train}|=v_0^2/a_0[/tex]
In a double-slit interference experiment, interference fringes are observed on a distant screen. The width of both slits is then doubled without changing the distance between their centers.
a. What happens to the spacing of the fringes?
b. What happens to the intensity of the bright fringes?
Final answer:
If the width of both slits in a double-slit interference experiment is doubled without changing the distance between their centers, the spacing of the fringes decreases and the intensity of the bright fringes decreases.
Explanation:
In a double-slit interference experiment, if the width of both slits is doubled without changing the distance between their centers:
a. The spacing of the fringes will decrease.
b. The intensity of the bright fringes will decrease.
When the width of the slits is increased, the interference fringes become wider and less compact. This means that the spacing between the fringes becomes smaller. Additionally, the intensity of the bright fringes decreases because spreading out the light over a wider area results in less light reaching a specific point on the screen.
What are continuous, emission, and absorption spectra? How are they produced?
Answer: An emission line occurs when an electron drops down to a lower orbit around the nucleus of an atom and looses energy.
An absorption line also occurs when any electron move to a higher orbit by absorbing energy.
Each atom has a unique way of using some space of the orbits and can absorb only certain energies or wavelengths.
Explanation:
Starting with the definition 1 in. = 2.54 cm, find the number of (a) kilometers in 1.00 mile and (b) feet in 1.00 km.
Answer :
(a) [tex]1.00\text{ mile}=1.61km[/tex]
(b) [tex]1.00km=3.28\times 10^3ft[/tex]
Explanation :
As are given:
1 inch = 2.54 cm
(a) Now we have to convert the number of kilometers in 1.00 mile.
Conversions used:
1 mile = 5280 ft
1 ft = 12 inch
1 inch = 2.54 cm
1 cm = 10⁻⁵ km
Thus,
[tex]1.00\text{ mile}=5280ft\times \frac{12in}{1ft}\times \frac{2.54cm}{1in}\times \frac{10^{-5}km}{1cm}[/tex]
[tex]1.00\text{ mile}=1.61km[/tex]
(b) Now we have to convert the number of feet in 1.00 km.
[tex]1.00km=10^5cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12inch}[/tex]
[tex]1.00km=3.28\times 10^3ft[/tex]
The unit 1 mile is equivalent to 1.61 km.
Part BThe unit 1 km is equivalent to 3280 ft.
How do you convert the miles into km and km into feet?Part A
The unit 1 miles will convert into km. We know that 1 mile = 5280 ft and 1 ft = 12 inch.
Given that 1 inch = 2.54 cm. Hence
1 mile = 5280 ft.
1 mile = [tex]5280 \times 12[/tex] inch
1 mile = [tex]63360 \times 2.54[/tex] cm
1 mile = 160934.4 cm
Now, 1 cm = 10^-5 km. Then
1 mile = [tex]160934.4 \times 10^{-5}[/tex] km
1 mile = 1.61 km
The unit 1 mile is equivalent to 1.61 km.
Part B
The unit 1 km will convert into feet. We know that 1 km = 10^5 cm
Given that 1 inch = 2.54 cm
1 km = [tex]10^5[/tex] cm
1 km = [tex]10^5 \times \dfrac {1}{2.54}[/tex] inch
1 km = [tex]10^5 \times \dfrac {1}{2.54}\times \dfrac{1}{12}[/tex] ft.
1 km = 3280 ft.
The unit 1 km is equivalent to 3280 ft.
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