Answer:
Light takes 14.12 minutes to travel from the Sun to tatoone.
Explanation:
The equation for the average velocity can be used to estimate the time that light will take to travel from the Sun to tatoone. The average velocity is defined as:
[tex]v = \frac{d}{t}[/tex] (1)
Where v is the velocity, d is the covered distance and t is the time.
Therefore, t can be isolated from equation 1:
[tex]t = \frac{d}{v}[/tex]
It is necessary to express the speed of light in terms of minutes:
[tex]3x10^{8} \frac{m}{s} . \frac{60 s}{1 min}[/tex] ⇒ [tex]1.8x10^{10} m/min[/tex]
An astronomical unit is defined as the distance between the Earth and the Sun ([tex]1.496 x 10^{11} m[/tex]).
[tex]d = 1.7 AU \cdot \frac{1.496 x 10^{11} m}{1AU}[/tex] ⇒ [tex]2.5432x10^{11}[/tex]m
Finally, equation 2 can be used:
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{2.5432x10^{11}m}{1.8x10^{10} m/min}[/tex]
t = 14.12 min
Hence, light takes 14.12 minutes to travel from the Sun to tatoone.
A juggler throws a bowling pin straight up in the air. After the pin leaves his hand and while it is in the air, which statement is true?(a) The velocity of the pin is always in the same direction as its acceleration.(b) The velocity of the pin is never in the same direction as its acceleration.(c) The acceleration of the pin is zero.(d) The velocity of the pin is opposite its acceleration on the way up. (e) The velocity of the pin is in the same direction as its acceleration on the way up.
Answer:
The velocity of the pin is opposite its acceleration on the way up.
(d) option is correct.
Explanation:
when the juggler throws a bowling pin straight in the air, the acceleration working on the pin is in the downward direction due to the gravitational force of the earth.
According to Newton's Universal Law of Gravitation
''The gravitational force is a force that attracts any objects with mass''
Final answer:
The acceleration due to gravity is always downward, so when the pin is thrown up, its velocity is opposite to its acceleration. At the peak, the velocity is zero and then aligns with the direction of gravity on the descent.
Explanation:
When a juggler throws a bowling pin straight up in the air, the acceleration due to gravity is always directed towards the ground, which means it is downwards. As the pin moves upwards, its velocity is in the opposite direction of the acceleration. At the peak of its motion, the velocity of the pin is zero, after which it starts to fall back down, and its velocity is then in the same direction as acceleration. Thus, the correct statement is (d) The velocity of the pin is opposite its acceleration on the way up.
A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south.
What is the charge of object A?
the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]
We can use the formula for electric field intensity ( E ) to calculate the charge of object A. The electric field intensity is given by:
[tex]\[ E = \frac{k \cdot |q|}{r^2} \][/tex]
Where:
- ( E ) is the electric field intensity (40.0 N/C),
- ( k) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),[/tex]
- [tex]\( |q| \)[/tex] is the magnitude of the charge on object A (what we're trying to find), and
- ( r ) is the distance from object A to point P (0.250 m).
First, rearrange the formula to solve for[tex]\( |q| \)[/tex]:
[tex]\[ |q| = \frac{E \cdot r^2}{k} \][/tex]
Now, substitute the given values into the formula:
[tex]\[ |q| = \frac{40.0 \, \text{N/C} \cdot (0.250 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \]\[ |q| = \frac{40.0 \times 0.0625}{8.99 \times 10^9} \, \text{C} \]\[ |q| = \frac{2.5}{8.99 \times 10^9} \, \text{C} \]\[ |q| = 2.78 \times 10^{-10} \, \text{C} \][/tex]
So, the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]
In a liquid with a density of 1400 kg/m3 , longitudinal waves with a frequency of 370 Hz are found to have a wavelength of 8.40 m . Calculate the bulk modulus of the liquid.
Answer:
Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa
Explanation:
given data
density = 1400 kg/m³
frequency = 370 Hz
wavelength = 8.40 m
solution
we get here bulk modulus of the liquid that is
we know Bulk Modulus = [tex]v^2*\rho[/tex] ...............
here [tex]\rho[/tex] is density i.e 1400 kg/m³
and v is = frequency × wavelength
v = 370 × 8.40 = 3108 m/s
so here bulk modulus will be as
Bulk modulus = 3108² × 1400
Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa
A negative charge of -0.510 μC exerts an upward 0.600-N force on an unknown charge that is located 0.300 m directly below the first charge. What are (a) the value of the unknown charge (magnitude and sign); (b) the magnitude and direction of the force that the unknown charge exerts on the −0.550μC charge?
By employing Coulomb's Law and additional physical principles like Newton's third law, it becomes possible to determine the characteristics of an unknown electrical charge from the force it experiences and exerts.
Explanation:The question concerns two charges, one of which is negative and known (-0.510 µC), and the other is unknown. You're asked to find the value of this unknown charge and the force it exerts on the known charge. To do this, we would use Coulomb's Law, which states that the electric force between two charges is directly proportional to the product of their charges, and inversely proportional to the square of the distance separating them.
So, using the formula F = k*q1*q2/r^2, where F is the force, k is Coulomb's constant (approximately 9 * 10^9 N.m^2/C^2), q1 and q2 are the charges, and r is the distance between them, we can solve for the unknown charge(q2) - which gives us q2 = Fr^2/(kq1). Substituting the provided values in the question, we should be able to solve it.
For part (b), the sign of the force that the unknown charge exerts on the known charge would be exactly opposite in direction to the force it experiences, as per Newton's third law (equal and opposite reactions). However, we'd need to know the magnitude of the unknown charge to calculate the actual force.
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How long does it take for a rotating object to speed up from 15.0 sto 33.3 rad/s if it has an angular acceleration of 3.45rad/s2?
a) 9.57 s
b) 5.30 s
c) 63.1 s
d) 4.35 s
Answer:
The time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds.
Explanation:
It is given that,
Initial angular speed of the object, [tex]\omega_o=15\ rad/s[/tex]
Final angular speed of the object, [tex]\omega_f=33.3\ rad/s[/tex]
Angular acceleration of the object, [tex]\alpha =3.45\ rad/s^2[/tex]
Angular acceleration of an object is object is defined as the change in angular velocity per unit time. It is given by :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]
[tex]t =\dfrac{\omega_f-\omega_i}{\alpha}[/tex]
[tex]t =\dfrac{33.3-15}{3.45}[/tex]
t = 5.30 seconds
So, the time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds. Hence, this is the required solution.
The time taken will be "5.30 seconds".
According to the question,
Initial angular speed,
[tex]\omega_i = 15 \ rad/s[/tex]Final angular speed,
[tex]\omega_f = 33.3 \ rad/s[/tex]Angular acceleration,
[tex]\alpha = 3.45 \ rad/s^2[/tex]As we know,
→ [tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
or,
→ [tex]t = \frac{\omega_f -\omega_i}{\alpha}[/tex]
By substituting the values, we get
[tex]= \frac{33.3-15}{3.45}[/tex]
[tex]= \frac{18.3}{3.45}[/tex]
[tex]= 5.30 \ seconds[/tex]
Thus the above answer i.e., "option b" is right.
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A 89.3 kg man sits on the stern of a 5.8 m
long boat. The prow of the boat touches the
pier, but the boat isn’t tied. The man notices his mistake, stands up and walks to the boat’s prow, but by the time he reaches the prow, it’s moved 4.17 m away from the pier.
Assuming no water resistance to the boat’s
motion, calculate the boat’s mass (not counting the man).
Answer in units of kg
Answer:
34.9 kg
Explanation:
Since there are no net external forces on the system, the center of gravity does not move.
Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.
When the man is at the stern, the distance between the center of gravity and the pier is:
CG = (m L + M (L/2)) / (m + M)
When the man reaches the prow, the distance between the center of gravity and the pier is:
CG = (m x + M (x + L/2)) / (m + M)
Since these are equal:
(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)
m L + M (L/2) = m x + M (x + L/2)
m L + M (L/2) = m x + M x + M (L/2)
m L = m x + M x
m L − m x = M x
m (L − x) = M x
M = m (L − x) / x
Plugging in values:
M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m
M = 34.9 kg
The required mass of boat is 34.9 kg.
The given problem is based on the concept of the center of mass. The point of analysis where the entire mass of the system is known to be concentrated is known as the center of mass.
Given data:
The mass of man is, m = 89.3 kg.
The length of boat is, L = 5.8 m.
The distance away from the pier is, d = 4.17 m.
Since there are no net external forces on the system, the center of gravity does not move. Let's say that m is the mass of the man, M is the mass of the boat
When the man is at the stern, the distance between the center of gravity and the pier is:
CG = (m L + M (L/2)) / (m + M)
When the man reaches the prow, the distance between the center of gravity and the pier is:
CG = (m d + M (d + L/2)) / (m + M)
Since these are equal:
(m L + M (L/2)) / (m + M) = (m d + M (d + L/2)) / (m + M)
m L + M (L/2) = m d + M (d + L/2)
m L + M (L/2) = m d + M d + M (L/2)
Further solving as,
m L = m d + M d
m L − m d = M d
m (L − x) = M x
M = m (L − x) / x
M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m
M = 34.9 kg
Thus, we can conclude that the required mass of boat is 34.9 kg.
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Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance between them is reduced to r0/4 ?
What is the electrostatic force between and electron and a proton separated by 0.1 mm?
Answer:
The electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]
Explanation:
It is given that, charge [tex]q_1[/tex] is placed at a distance [tex]r_o[/tex] from charge [tex]q_2[/tex]. The force acting between charges is given by :
[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]
We need to find the force if the distance between them is reduced to [tex]r_o/4[/tex]. It is given by :
[tex]F'=\dfrac{kq_1q_2}{(r_o/4)^2}[/tex]
[tex]F'=16\times \dfrac{kq_1q_2}{r_o^2}[/tex]
[tex]F'=16\times F[/tex]
So, if the the distance between them is reduced to [tex]r_o/4[/tex], the new force becomes 16 times of the previous force.
The electrostatic force between and electron and a proton separated by 0.1 mm or [tex]10^{-4}\ m[/tex] is :
[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(10^{-4})^2}[/tex]
[tex]F=2.30\times 10^{-20}\ N[/tex]
So, the electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]. Hence, this is the required solution.
When the distance between two charges is reduced, the magnitude of the force on one charge due to the other increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.
Explanation:When the distance between two charges, q1 and q2, is reduced from r0 to r0/4, the magnitude of the force on q1 due to q2 increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.
For example, if we have charges q1 = 2 C and q2 = 4 C, and the original distance r0 = 2 m, the force between them would be F = k × (q1 × q2) / r0^2 = k × (2 × 4) / (2^2) = k × 4, where k is the Coulomb's constant.
If we reduce the distance to r0/4 = 0.5 m, the force would become F' = k × q1 × q2) / (r0/4)^2 = k × 4) / (0.5^2) = k ×64, which is 16 times greater than the original force.
describe one new piece of information that you learned about one of the featured scientist
Answer:
Development in science:
There a number of individuals in the world around us who has improved the field of science and made it easy for us to know this universe more easily then ever. This universe is still unexplored and there are going to be more then billions of stars which are still not studied and has an immense amount of information regarding the universe.
Stephen Hawking: The scientist who made it easy for each of us to know the universe and more over the cosmos in a more detailed form, as before this no one knew about the different patterns of entities and there properties.
A genius who was handicapped:
He was a genius in cosmology, mathematics, and other subjects of science as he was diagnosed by the amyotrophic lateral sclerosis(ALS), which made it unable for him to live a normal life.But, he communicated through a computer which detected his nerve signals and shared his thoughts about any thing or subject.
Why him?
Stephen hawking was just tremendous in making it more understandable about the black holes, time worms, and space etc.As it was known to very small number of people before he took the stage.
In his famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil drops in an electric field. With a field strength 0f 20 MN/C, what mass drop can be suspended when the drop carries a net charge of 10 elementary charges?
Answer:
[tex]3.26198\times 10^{-12}\ kg[/tex]
Explanation:
E = Electric field = 20 MN/C
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
g = Acceleration due to gravity = 9.81 m/s²
m = Mass of drop
The electrical force will balance the weight
[tex]Eq=mg\\\Rightarrow 20\times 10^{6}\times 10\times 1.6\times 10^{-19}=m\times 9.81\\\Rightarrow m=\dfrac{20\times 10^{6}\times 10\times 1.6\times 10^{-19}}{9.81}\\\Rightarrow m=3.26198\times 10^{-12}\ kg[/tex]
The mass that can be suspended is [tex]3.26198\times 10^{-12}\ kg[/tex]
From charge to mass ratio, the mass of the charges is 3.2 × 10^-12 Kg.
The charge to mass ratio experiment was used by R. A. Millikan to accurately determine the charge to mass ratio of the electron. We have the following information from the question;
Field strength = 20 MN/C
Number of charges = 10
Now;
The magnitude of electric field strength is obtained from;
E = F/q
F = Eq
Where;
F = electric force
E = electric field intensity
q = magnitude of charge
F = 10 × 20 × 10^6 × 1.6 × 10^-19 = 3.2 × 10^-11 N
Where the charges fall freely under gravity;
F = mg
m = F/g
m = 3.2 × 10^-11 N/10 ms-2
m = 3.2 × 10^-12 Kg
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Each of the following equations was given by a student during an examination. Do a dimensional analysis of each equation and explain why the equation can't be correct.
(a) 1 2 mv2 = 1 2 mv02 + mgh
(b) v = v0 + at2
(c) ma = v2
a) The equation 1/2 mv² = 1/2 mv0² + mgh is not dimensionally consistent; b) The equation v = v0 + at² is not dimensionally consistent; c) The equation ma = v² is not dimensionally consistent.
(a) The left-hand side of the equation has dimensions of energy (M L² T⁻²), while the right-hand side has dimensions of energy plus length (M L² T⁻² + ML).
This is because the term mgh represents the potential energy gained by an object of mass m when it is lifted to a height h above the ground. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.
(b)The left-hand side of the equation has dimensions of velocity (L T⁻¹), while the right-hand side has dimensions of velocity plus time squared (L T⁻¹ + T²).
This is because the term at² represents the change in velocity over time due to acceleration. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.
(c) The left-hand side of the equation has dimensions of force (M L T⁻²), while the right-hand side has dimensions of velocity squared (L² T⁻²).
This is because the term v² represents the square of the velocity of an object. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.
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The dimensional analysis shows that equation (a) is correct while equations (b) and (c) are incorrect. Equation (b) is missing a time component, while equation (c) incorrectly equates force to velocity squared.
Explanation:Let's analyze each equation in terms of their dimensions:
(a) 1/2 mv² = 1/2 m(v0)² + mgh: This equation is actually dimensionally correct. It represents the conservation of mechanical energy in Physics, where 1/2 mv² is the kinetic energy and mgh is the potential energy.(b) v = v₀ + at²: This equation is not dimensionally correct. In Physics, the equation of motion is v = v₀ + at, where v is the final velocity, v0 is the initial velocity, a is acceleration and t is time.(c) ma = v²: This equation is also not dimensionally correct. Force (ma) has dimensions of [M][L][T]⁻² while velocity squared (v)² has dimensions of [L]²[T]⁻².By comparing dimensions on both sides of an equation, we can spot if the equations are incorrect or not.
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A raft of mass 199 kg carries two swimmers of mass 52 kg and 70 kg. The raft is initially floating at rest. The two swimmers simultaneously dive off opposite ends of the raft, each with a horizontal velocity of 4 m/s. With what velocity the raft start to move
To solve this problem we will apply the concept related to the conservation of the Momentum. We will then start considering that the amount of initial momentum must be equal to the amount of final momentum. Considering that all the objects at the initial moment have the same initial velocity (Zero, since they start from rest) the final moment will be equivalent to the multiplication of the mass of each object by the velocity of each object, so
Initial Momentum = Final Momentum
[tex](m_B+m_1+m_2)v_i = m_1v_1+m_2v_2+m_Bv_B[/tex]
Here,
[tex]m_B[/tex] = mass of Raft
[tex]m_1[/tex] = Mass of swimmers 1
[tex]m_2[/tex] = Mass of swimmers 2
[tex]v_i[/tex] = Initial velocity (of the three objects)
[tex]v_B[/tex] = Velocity of Raft
Replacing,
[tex](199+52+70)*0 = (52)(4)+(70)(-4)+199v_B[/tex]
Solving for [tex]v_B[/tex]
[tex]vB = \frac{72}{199}[/tex]
[tex]v_B = 0.3618m/s[/tex]
Therefore the velocity the rarft start to move is 0.3618m/s
You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.00×10−9C/m. What is the magnitude of the electric force that the wire exerts on the point charge?
The question is about calculating the electric force on a point charge near an infinitely long wire using the concepts of electric fields and Coulomb's law within the domain of Physics.
Explanation:The question involves calculating the magnitude of the electric force that an infinitely long, thin wire exerts on a point charge placed at a certain distance from it. This falls under the subject of Physics, specifically dealing with electrostatics and the concept of electric fields and forces.
First, we need to find the electric field due to the wire at the location of the point charge. The electric field E from an infinitely long wire is given by the formula E = (2kλ)/r, where k is Coulomb's constant (8.99 × 109 Nm2/C2), λ is the linear charge density of the wire, and r is the distance from the wire to the point where the electric field is being calculated.
After calculating E, we can find the force exerted on the charge q using F = qE. Since we're given q = -4.00 nC and λ = 3.00×10−9C/m, with r = 9.00 cm, we can substitute these values into the formulas to calculate the electric field and then the force acting on the point charge.
The magnitude of the electric force exerted by the wire on the point charge is 2.40 × 10⁻⁶ N.
To find the magnitude of the electric force that an infinitely long, thin wire with linear charge density λ = 3.00×10⁻⁹ C/m exerts on a point charge q = -4.00 nC placed a distance of 9.00 cm from the wire, we use the principle that an electric field E is created by a line of charge.
The electric field created by an infinitely long, thin wire is given by:
E = (2kλ)/r
where k = 8.99 × 10⁹ Nm²/C²
λ = 3.00×10⁻⁹ C/m , charge density
r = 9 × 10⁻² m
Substitute the given values:
E = (2 × 8.99 × 10⁹ Nm²/C² × 3.00×10⁻⁹ C/m ) / 9 × 10⁻² m
Calculating this, we get:
E ≈ 600 N/C
The force F on the point charge q due to the electric field E is given by:
F = q × E
Since q = -4.00 nC = -4.00 × 10⁻⁹ C:
F = (-4.00 × 10⁻⁹ C) × 600 N/C ≈ -2.40 × 10⁻⁶ N
Therefore, the magnitude of the electric force is 2.40 × 10⁻⁶N.
A block of mass 3.1 kg, sliding on a horizontal plane, is released with a velocity of 2.3 m/s. The blocks slides and stops at a distance of 1.9 m beyond the point where it was released.
How far would the block have slid if its initial velocity were quadrupled?
To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity
Our values are given as,
[tex]\text{Initial speed} =V_i = 2.3 m/s[/tex]
[tex]\text{Final speed}= V_f = 0 m/s[/tex]
[tex]\text{Stopping distance = }d = 1.9 m[/tex]
[tex]a = acceleration[/tex]
[tex]\text{mass} = m = 3.1kg[/tex]
Using the kinematic equation of motion we have
[tex]V_f^2 = V_i^2 + 2 a d[/tex]
[tex]0^2 = 2.3^2 + 2 a (1.9)[/tex]
[tex]a = -1.39211 m/s^2[/tex]
Now if the initial velocity is quadrupled we have that,
[tex]\text{Initial speed} =V_i' = 2.3*4 m/s = 9.2m/s[/tex]
[tex]\text{Final speed}= V_f' = 0 m/s[/tex]
[tex]\text{Stopping distance = }d'[/tex]
[tex]V_f^2' = V_i^2' + 2 a d'[/tex]
Replacing the values
[tex]0^2 = 9.2^2+ 2 (-1.39211) d'[/tex]
[tex]d' = 30.44m[/tex]
Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.
suppose two masses are connected by a spring. compute the formula for the trajectory of the center of mass of the two mass oscillator
Answer:
The center mass (Xcm) of the two mass = (M₁X₁ + M₂X₂)/(M₁ +M₂)
Explanation:
let the first mass = M
let the position of second = M
The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring is x(t) = A * cos(ωt + φ), where A represents the amplitude, ω is the angular frequency, and φ is the phase constant.
Explanation:The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring can be derived using the principles of simple harmonic motion (SHM). Let's assume that the masses are m1 and m2, and the spring constant is k. The equation for the trajectory of the center of mass can be written as:
x(t) = A * cos(ωt + φ)
where:
x(t) is the displacement of the center of mass at time tA is the amplitude of the motion, which can be calculated using the initial conditionsω is the angular frequency, which is given by ω = √(k / (m1 + m2))φ is the phase constant, which can also be calculated using the initial conditions
According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the hair. The thickness of hair varies considerably, but let's use a diameter of 50.0 μm
Part A
What is the magnitude of the force giving this tensile stress?
F = ? N
Part B
If the length of a strand of the hair is 12.0 cm at its breaking point, what was its unstressed length? original length = ?cm
Answer:
(a). The magnitude of the force is 0.38416 N.
(b). The original length is 0.0869 m.
Explanation:
Given that,
Tensile strength = 196 MPa
Maximum strain = 0.380
Diameter = 50.0 μm
Length = 12.0 cm
We need to calculate the area
Using formula of area
[tex]A=\dfrac{\pi}{4}\times d^2[/tex]
Put the value into the formula
[tex]A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2[/tex]
[tex]A=1.96\times10^{-9}\ m^2[/tex]
We need to calculate the magnitude of the force
Using formula of force
[tex]F=\sigma A[/tex]
Put the value into the formula
[tex]F=196\times10^{6}\times1.96\times10^{-9}[/tex]
[tex]F=0.38416\ N[/tex]
(b). If the length of a strand of the hair is 12.0 cm at its breaking point
We need to calculate the unstressed length
Using formula of strain
[tex]strain=\dfrac{\Delta l}{l_{0}}[/tex]
[tex]\Delta l=strain\times l_{0}[/tex]
Put the value into the formula
[tex]\Delta l=0.380\times l_{0}[/tex]
Length after expansion is 12 cm
We need to calculate the original length
Using formula of length
[tex]l=l_{0}+\Delta l[/tex]
Put the value into the formula
[tex]I=l_{0}+0.380\times l_{0}[/tex]
[tex]l=1.38l_{0}[/tex]
[tex]l_{0}=\dfrac{l}{1.38}[/tex]
[tex]l_{0}=\dfrac{12\times10^{-2}}{1.38}[/tex]
[tex]l_{0}=0.0869\ m[/tex]
Hence, (a). The magnitude of the force is 0.38416 N.
(b). The original length is 0.0869 m.
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656(red) and 486 (blue). Light from a hydrogen lamp illuminates a diffraction grating with 500 lines per mm, and the light is observed on a screen 1.50 behind the grating. What is the distance between the first-order red and blue fringes?
To find the distance between the first-order red and blue fringes formed by a diffraction grating, we calculate the diffraction angles using the grating equation and then use trigonometry to determine the positions of the fringes on the screen. The distance apart is the absolute difference between these positions.
Explanation:The student is asking about the distance between the first-order red and blue fringes that appear on a screen as a result of a diffraction pattern formed by light from a hydrogen discharge lamp passing through a diffraction grating. To answer this question, we apply the formula for diffraction maxima, d sin( heta) = m ext{ extlambda}, where d is the distance between the grating lines, ext{ extlambda} is the wavelength of light, ext{ exttheta} is the diffraction angle, and m is the order of the maximum.
Given that the grating has 500 lines per mm or 500,000 lines per meter, we can calculate d as 1/500,000 meters. Since we are looking for the first-order maxima (m=1), we can find the angles for each color using the formula sin( heta) = ext{ extlambda}/d. We can then use basic trigonometry to find the positions of the red and blue fringes on the screen, and their distance apart.
For the 656 nm red light:
ext{ exttheta}_{red} = sin^{-1}( ext{ extlambda}_{red}/d)
ext{ exttheta}_{red} = sin^{-1}(656 ext{ exttimes}10^{-9} m / (1/500,000 m))
For the 486 nm blue light:
ext{ exttheta}_{blue} = sin^{-1}( ext{ extlambda}_{blue}/d)
ext{ exttheta}_{blue} = sin^{-1}(486 ext{ exttimes}10^{-9} m / (1/500,000 m))
After calculating the angles, the positions on the screen for first-order maxima are found by L tan( heta), where L is the distance from the grating to the screen (1.50 m in this case).
The final distance between the red and blue fringes is the absolute difference between their positions on the screen.
A circular ring of charge with radius b has total charge q uniformly distributed around it.
What is the magnitude of the electric field at the center of the ring?
a) 0
b) kₑq/b²
c) kₑq²/b²
d) kₑq²/b
e) None of these
Answer:
Option A is correct (0) ( The electric field at the center of circular charged ring is zero)
Explanation:
Option A is correct (0) ( The electric field at the center of circular charged ring is zero)
The reason why electric field at the center of circular charged ring is zero because the fields at the center of the circular ring due to any point are cancelled by electric fields of another point which is at 180 degree to that point i.e opposite to that charge.
Answer: a) 0
The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero
Explanation:
Given that the ring is perfectly circular with radius b which has a uniform distributed charge q around it.
The electric field experienced at the centre of the ring from opposite directions are given as
E1 = kₑq/b²
E2 = -kₑq/b²
It experience the two electric field E1 and E2 from opposite directions at the centre. So the resultant electric field is given by:
E = E1 + E2
E = kₑq/b² - kₑq/b²
E = 0
The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero
Sound of thunder is heard 5 seconds after the flash of lightning is seen.How far away was the lightning flash? Note that sound travels in air at a speed of about 340 meter second?
Answer:
The distance covered by the lighting flash is 1700 meters.
Explanation:
Given that,
The speed of sound travels in air at a speed of about 340 meter second, v = 340 m/s
The sound of thunder is heard 5 seconds after the flash of lightning is seen, t = 5 s
To find,
The distance far away was the lightning flash.
Solution,
The distance covered by the lighting flash is given by the product of speed and time. It is given by :
[tex]d=v\times t[/tex]
[tex]d=340\ m/s\times 5\ s[/tex]
d = 1700 meters
So, the distance covered by the lighting flash is 1700 meters. Therefore, this is the required solution.
Mark Watney begins his day 15 km West and 25 km North of his Mars Habitat. a. Set up a co-ordinate system (draw labeled axis and tickmarks showing the scale) and draw a vector representing the initial position of Mark Watney. b. Mark spends his day driving his Mars Rover towards Schiaparelli Crater and manages to make it an additional 20 km West but has to go around a hill so ended up 5 km South of his starting point for the day. Draw a vector representing Mark's total position change for the day. c. Using vector addition find Mark's position relative to the Mars Habitat. Give both the numerical description of the vector and show it on your plot.
Answer:
a, b) part a and b on diagram attached
c) sf = -35 i + 20 j
35 km West and 20 km North
Explanation:
For part a and b refer to the attached co-ordinate system:
Note: unit vector i is in West/East direction and unit vector j is in North/South direction.
si = -15 i + 25 j
sf-si = -20 i - 5 j
Hence,
Mark relative position from habitat sf = si + sf/i
sf = ( -15 i + 25 j ) + ( -20 i - 5 j )
sf = -35 i + 20 j
35 km West and 20 km North
Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +18 pC, fairly typical values.
What is the potential at the surface of the raindrop?
To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is
[tex]V = \frac{kq}{r}[/tex]
k = Coulomb's constant
q = Charge
r = Distance between them
[tex]q = 18 pC \rightarrow q = 1.8*10^-11 C[/tex]
[tex]d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m[/tex]
Replacing,
[tex]V = \frac{kq}{r}[/tex]
[tex]V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}[/tex]
[tex]V = 135 V[/tex]
Therefore the potential at the surface of the raindrop is 135 V
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.62 m shorter than before.
(a) What is the average speed of the truck during the collision (that is, during the interval between the first contact with the wall and coming to a stop)?
(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)
(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?
(d) It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force mg on the truck. This large ratio shows why a collision is so damaging.
(e) What approximations were necessary in making this analysis? (Select all that apply.)
a. Neglect the horizontal component of the force of the road on the truck tires.
b. Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.
c. The deceleration of the truck is approximately equal to g.
Answer:
a) v_average = 11 m / s, b) t = 0.0627 s
, c) F = 7.37 10⁵ N
, d) F / W = 35.8
Explanation:
a) truck speed can be found with kinematics
v² = v₀² - 2 a x
The fine speed zeroes them
a = v₀² / 2x
a = 22²/2 0.69
a = 350.72 m / s²
The average speed is
v_average = (v + v₀) / 2
v_average = (22 + 0) / 2
v_average = 11 m / s
b) The average time
v = v₀ - a t
t = v₀ / a
t = 22 / 350.72
t = 0.0627 s
c) The force can be found with Newton's second law
F = m a
F = 2100 350.72
F = 7.37 10⁵ N
.d) the ratio of this force to weight
F / W = 7.37 10⁵ / (2100 9.8)
F / W = 35.8
.e) Several approaches will be made:
- the resistance of air and tires is neglected
- It is despised that the force is not constant in time
- Depreciation of materials deformation during the crash
If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because _____________.
A. the indices of refraction matched
B. the light was reflecte
C. the light was absorbed
Answer:
A. the indices of refraction matched
Explanation:
The index of refraction, or refractive
index, is a measure of how fast light
rays travel through a given medium.
Alternatively, it could be said that
the refractive index is the measure of
the bending of a light ray when
passing from one medium to
another. Mathematically, it can be
represented as a ratio between two
different velocities – the velocity of
light in vacuum and the velocity of
light in a given medium.
For example, try putting a pencil in a jar full of water. If you look at the pencil from above, it would look as though the pencil has bent in the water. That happens due to the refraction of light. It occurs because as light rays enter water, they slow down, as the speed of light in water is lower than the speed of light in air. The magnitude of how much a medium refracts a light ray is determined by the index
If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched here. Thus, the correct option is A.
What is the refraction of light?Refraction is the phenomena of bending of light, which also happens with the rays of sound, water and other waves as it passes from one transparent substance into another substance or medium. This phenomena of bending of light waves by the refraction makes it possible for the people to have lenses, magnifying glasses, prisms, and rainbows. Even, our eyes depend upon this bending of light for the visibility and other effects of light.
If we place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched.
Therefore, the correct option is A.
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6 A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump.
To solve this problem we will start by differentiating the values in each of the states of matter. Subsequently through the thermodynamic tables we will look for the values related to the entropy, enthalpy and respective specific volumes. Through the relationship of Power defined as the product between mass and enthalpy and mass, specific volume and pressure, we will find the energetic values in the two states investigated. We will start defining the states
State 1
[tex]T_1 = 700\°C[/tex]
[tex]P_1 = 4 Mpa[/tex]
From steam table
[tex]h_1 =3906.41 KJ/Kg[/tex]
[tex]s_1 = 7.62 KJ/Kg.K[/tex]
Now
[tex]s_1 = s_2 = 7.62 KJ/Kg.K[/tex] As 1-2 is isentropic
State 2
[tex]P_2 = 20 Kpa[/tex]
[tex]s_2 = 7.62 KJ/Kg \cdot K[/tex]
From steam table
[tex]h_2 = 2513.33 KJ/Kg[/tex]
PART A) The power produced by turbine is the product between the mass and the enthalpy difference, then
[tex]Power = m \times (h_1-h_2)[/tex]
[tex]P = (50)(3906.41 - 2513.33)[/tex]
[tex]P = 69654kW[/tex]
b) Pump Work
State 3
[tex]P_3 = 20 Kpa[/tex]
[tex]\upsilon= 0.001 m^3/kg[/tex]
The Work done by the pump is
[tex]W= m\upsilon \Delta P[/tex]
[tex]W = (50)(0.001)(4000-20)[/tex]
[tex]W = 199kJ[/tex]
Suppose quantity s is a length and quantity t is a time. Suppose the quantities vand aare defined by v = ds/dt and a = dv/dt. (a) What is the dimension of v? (b) What is the dimension of the quantity a?
What are the dimensions of (c)vdt, (d) a dt, and (e) da/dt?
Explanation:
(a) Velocity is given by :
[tex]v=\dfrac{ds}{dt}[/tex]
s is the length of the distance
t is the time
The dimension of v will be, [tex][v]=[LT^-1][/tex]
(b) The acceleration is given by :
[tex]a=\dfrac{dv}{dt}[/tex]
v is the velocity
t is the time
The dimension of a will be, [tex][a]=[LT^{-2}][/tex]
(c) Since, [tex]d=\int\limits{v{\cdot}dt} =[LT^{-1}][T]=[L][/tex]
(d) Since, [tex]v=\int\limits{a{\cdot}dt} =[LT^{-2}][T]=[LT^{-1}][/tex]
(e)
[tex]\dfrac{da}{dt}=\dfrac{[LT^{-2}]}{[T]}[/tex]
[tex]\dfrac{da}{dt}=[LT^{-3}]}[/tex]
Hence, this is the required solution.
At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v = 239 m/s. Upon landing, the plane can produce an average deceleration of a = 16.5 m/s².
How long will it take the plane to circle the Earth at the equator?
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.
[tex]R= 6370*10^3 m[/tex]
[tex]v = 239m/s[/tex]
[tex]a = 16.5m/s^2[/tex]
The circumference of the earth would be
[tex]\phi = 2\pi R[/tex]
Velocity is defined as,
[tex]v = \frac{x}{t}[/tex]
[tex]t = \frac{x}{v}[/tex]
Here [tex]x = \phi[/tex], then
[tex]t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}[/tex]
[tex]t = 167463.97s[/tex]
Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds
An astronaut is on a 100-m lifeline outside a spaceship, circling the ship with an angular speed of
0.100 rad/s. How far inward can she be pulled before the centripetal acceleration reaches 5g = 49 m/s2?
Answer:
D = 72.68 m
Explanation:
given,
R = 100 m
angular speed = 0.1 rad/s
distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².
using conservation of Angular momentum
[tex]I_i\omega_i= I_f\omega_f[/tex]
[tex]mr_i^2\omega_i=m r_f^2\omega_f[/tex]
[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]
[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]
we know,
centripetal acceleration
[tex]a = \dfrac{v^2}{r}[/tex]
v = r ω
[tex]a =\omega_f^2 r_f [/tex]
[tex]a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f [/tex]
[tex]a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}[/tex]
[tex]r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}[/tex]
[tex]r_f^3=20408.1632[/tex]
[tex]r_f = 27.32\ m[/tex]
distance she has reached inward is equal to
D = 100 - 27.23
D = 72.68 m
A car, on a straight road, is stopped at a traffic light. When the light turns to green the car accelerates with a constant acceleration. It reaches a speed of 19.1 m/s (68.8 km/h) in a distance of 101 m. Calculate the acceleration of the car.
The car accelerates from 0 m/s to 19.1 m/s in 101 m, with an acceleration of approximately 1.81 m/s².
Given that,
The car is stopped at a traffic light on a straight road.
When the light turns green, the car accelerates.
The car reaches a speed of 19.1 m/s (68.8 km/h).
The distance covered during acceleration is 101 m
To calculate the acceleration of the car,
Use the kinematic equation:
v² = u² + 2as
Where,
v is the final velocity (19.1 m/s in this case),
u is the initial velocity (0 m/s since the car is stopped),
a is the acceleration that we're trying to find,
s is the distance covered (101 m).
Plugging in the known values, we have:
19.1² = 0² + 2a(101)
Simplifying further:
365.81 = 202a
Now, let's solve for a:
a = 365.81 / 202
Calculating that, we find:
a ≈ 1.81 m/s²
Hence,
The acceleration of the car is approximately 1.81 m/s².
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The acceleration of the car is 1.85 m/s^2.
Explanation:To calculate the acceleration of the car, we can use the equation:
v^2 = u^2 + 2as
Where:
v is the final velocityu is the initial velocity (0 m/s in this case)a is the accelerations is the displacementPlugging in the given values, we have:
19.1^2 = 0 + 2a(101)
Simplifying the equation gives us:
a = (19.1^2) / (2 * 101) = 1.85 m/s^2
Therefore, the acceleration of the car is 1.85 m/s^2.
Calculate the average linear momentum of a particle described by the following wavefunctions: (a) eikx, (b) cos kx, (c) e−ax2 , where in each one x ranges from −[infinity] to +[infinity].
Answer:
a) p=0, b) p=0, c) p= ∞
Explanation:
In quantum mechanics the moment operator is given by
p = - i h’ d φ / dx
h’= h / 2π
We apply this equation to the given wave functions
a) φ = [tex]e^{ikx}[/tex]
.d φ dx = i k [tex]e^{ikx}[/tex]
We replace
p = h’ k [tex]e^{ikx}[/tex]
i i = -1
The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero
p = 0
b) φ = cos kx
p = h’ k sen kx
The average sine function is zero,
p = 0
c) φ = [tex]e^{-ax^{2} }[/tex]
d φ / dx = -a 2x [tex]e^{-ax^{2} }[/tex]
.p = i a g ’2x [tex]e^{-ax^{2} }[/tex]
The average moment is
p = (p₂ + p₁) / 2
p = i a h ’(-∞ + ∞)
p = ∞
A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 14.5 m/s, and the distance from the limb to the level of the saddle is 3.29 m.
(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
(b) How long is he in the air?
Answer:
For Part A:
ΔX=11.8813 m
Part B:
[tex]t=0.8194 s[/tex]
Explanation:
Note: In order to find Part A we first have to find Part B i.e time
Data given:
[tex]V_i=0[/tex]
a=-9.8m/s^2 (-ve is because ranch is falling down)
Δy=-3.29m (-ve is because ranch is falling down)
Second equation of Motion:
Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]
V_i=0, Equation will become
Δy=[tex]\frac{1}{2}g*t^2[/tex]
[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]
For Part A:
Again:
Second equation of Motion:
ΔX=[tex]V_i*t+\frac{1}{2}a*t^2[/tex]
Since velocity is constant a=0
V_i=14.5m/s, t=0.8194 sec
ΔX=[tex]V_i*t[/tex]
ΔX=[tex]14.5*0.8194[/tex]
ΔX=11.8813 m
Part B: (Calculated above)
Data given:
[tex]V_i=0[/tex]
a=-9.8m/s^2 (-ve is because ranch is falling down)
Δy=-3.29m (-ve is because ranch is falling down)
Second equation of Motion:
Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]
V_i=0, Equation will become
Δy=[tex]\frac{1}{2}g*t^2[/tex]
[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]
For the merry-go-round problem, do the magnitudes of the position, velocity, and acceleration vectors change with time?
Answer:
No
Explanation:
Although the direction of position, velocity or acceleration of an object in marry-go-round problem changes continuously,however the magnitude of the position, velocity and acceleration do remains the same. Marry-go-round is nothing but a machine found in fairs that turn round in circular motion. So, the laws of circular motion are applicable in it.