The mass of the ion is 5.96 X 10⁻²⁵ kg
Explanation:
The electrical energy given to the ion Vq will be changed into kinetic energy [tex]\frac{1}{2}mv^2[/tex]
As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force [tex]\frac{mv^2}{r}[/tex].
So,
[tex]Vq = \frac{1}{2}mv^2[/tex]
and
[tex]Bqv = \frac{mv^2}{r}[/tex]
Right from these eliminating v, we can derive
[tex]m = \frac{B^2r^2q}{2V}[/tex]
On substituting the value, we get:
[tex]m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\[/tex]
m = 5.96 X 10⁻²⁵ kg.
The mass of the ion is found using the principles of kinetic and magnetic forces, is approximately 9.65 x 10⁻²⁷ kg.
To find the mass of the ion, we use the principles of energy and magnetic force. When a singly charged positive ion is accelerated through a voltage (V), it gains kinetic energy equal to the electrical potential energy provided by the voltage:
Kinetic Energy (KE) = qV
Since the ion starts from rest, this kinetic energy is also given by:
Kinetic Energy (KE) = (1/2)mv²
Equating these two expressions for kinetic energy gives:
qV = (1/2)mv²
Solving for the velocity (v), we get:
v =√(2qV/m)
When the ion enters the magnetic field (B) and bends into a circular path, the centripetal force required is provided by the magnetic force:
Magnetic Force = qvB
The centripetal force is also given by:
Centripetal Force = mv²/r
Equating these two expressions for the forces, we get:
qvB = mv²/r
Solving for the mass (m) of the ion:
m = qBr/v
We already have v = √(2qV/m), substituting this into the equation above to eliminate v, we get:
m = (qBr) / √(2qV/m)
Square both sides to simplify and solve for m:
m² = (q²B²r²) / (2qV)
m = (qB²r²) / (2V)
Using the given values (q = 1.602 x 10⁻¹⁹ C as charge of a singly charged positive ion, V = 2000 V, B = 0.400 T, and r = 0.305 m), we get:
m = (1.602 x 10⁻¹⁹ C x (0.400 T)² x (0.305 m)²) / (2 x 2000 V)
Calculating this gives the mass of the ion approximately equal to:
m ≈ 9.65 x 10⁻²⁷ kg
Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that apply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.
Complete Question:
An object oscillates back and forth on the end of a spring.
Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that apply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.
Answer:
a) the object can have zero acceleration and, simultaneously, nonzero velocity
b) the object can have zero velocity and, simultaneously, nonzero acceleration
d) the object can have nonzero velocity and nonzero acceleration simultaneously
Explanation:
For an object oscillating back and forth on the end of a spring, when the object swings to the extremes, it momentarily stops and the velocity becomes zero. Rather than being zero, acceleration is maximum at these points because the net force at these extremes is maximum. This justifies option B
At the middle, which is the equilibrium position, the net force is zero because the object is motionless, and hence the acceleration is zero. At this point, the velocity is maximum and not zero. This justifies option A
When the object is swinging and it is neither at the middle nor at the extremes, both acceleration and velocity are not zero. This justifies option D
It is not possible for both the acceleration and the velocity of the swinging object to be simultaneously zero. Option C is wrong
An object can have zero acceleration and non-zero velocity or have non-zero velocity and acceleration simultaneously at some point during its motion. However, an object can also have zero velocity and zero acceleration at all times during its motion.
Explanation:Both the statements "The object can have zero acceleration and, simultaneously, nonzero velocity" and "The object can have nonzero velocity and nonzero acceleration simultaneously" are true at some time during the course of motion.
For example, when an object is thrown upwards, at the highest point of its trajectory, its velocity becomes zero, but it still experiences the acceleration due to gravity, which is nonzero. This satisfies the first statement. Additionally, when an object moves in a circular path at a constant speed, it has a nonzero velocity and nonzero acceleration directed towards the center of the circle. This satisfies the second statement.
The statement "The object can have zero velocity and, simultaneously, zero acceleration." is true at all times during the course of motion. For instance, when an object is at rest, both its velocity and acceleration are zero.
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A block of mass 500 g is attached to a spring of spring constant 80 N/m. The other end of the spring is attached to a support while the mass rests on a rough surface with a coefficient of friction of 0.20 that is inclined at angle of 30°. The block is pushed along the surface till the spring compresses by 10 cm and is then released from rest. (a) How much potential energy was stored in the block-spring-support system when the block was just released? (b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched. (c) Determine the position of the block where it just comes to rest on its way up the incline.
Answer:
a) Vki = 0.4 J
b) v = 0.98 m/s
c) final position = 0.11745 m from initial position
Explanation:
Given:-
- The mass of block, m = 500 g
- The spring constant, k = 80 N/m
- The coefficient of the surface, u = 0.20
- The inclination of the block, θ = 30°
- The block is initially compressed, xi = 10 cm
- The block is initially at rest, vi = 0 m/s
Find:-
(a) How much potential energy was stored in the block-spring-support system when the block was just released?
(b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched.
(c) Determine the position of the block where it just comes to rest on its way up the incline.
Solution:-
- The potential energy initially stored in a spring "Vki" with constant "k" which has undergone a displacement of "x" is given by:
Vki = 0.5*k*xi^2
Vki = 0.5*80*0.1^2
Vki = 0.4 J
- The block is released from rest when the energy stored in the spring is dispensed in three forms of energy.
- There would be an increase in the potential energy " Ep " of the block as it moves up.
- There would be an increase in kinetic energy for some time " Ek "
- There would be loss of total energy due to fictitious force ( Ff) the work done against friction will dissipate energy.
- The increase in potential energy "Ep" at displacement xo = 0 from mean position is given by:
ΔEp = m*g*Δh ... change in vertical height h
ΔEp = m*g*( xi - xo )*sin ( θ )
ΔEp = 0.5*9.81*(0.1)*sin ( 30 )
ΔEp = 0.24525 J
- The increase in kinetic energy "Ek" at displacement xi = 10 cm, vi = 0 m/s from initial position to mean position at xo = 0 ,its velocity is "vf" is given by:
ΔEk = 0.5*m*( vf^2 - vi^2 ) ... change in velocity
ΔEk = 0.5*m*( vf )^2
- There would be loss of total energy due to fictitious force ( Ff ) the work done against friction will dissipate energy. First apply the equilibrium conditions on the block normal to slope and determine the contact force ( Nc ):
Nc - m*g*cos ( θ ) = 0
Nc = m*g*cos ( θ )
- The friction force ( Ff ) is given by:
Ff = Nc*u
Ff = u*m*g*cos ( θ )
- The work done by block against friction is given by:
Wf = -Ff*( xi - xo )
Wf = -u*m*g*xi*cos ( θ )
Wf = -0.2*0.5*9.81*0.1*cos ( 30 )
Wf = -0.08495 J
- We can now express the work done principle for the block:
Vki = ΔEp + ΔEk + Wf
ΔEk = - ΔEp - Wf + Vki
0.5*m*( vf )^2 = -0.24525 + 0.08495 +0.4
vf^2 = 4*(-0.24525 + 0.08495 +0.4 )
vf = √0.9588
vf = 0.98 m/s
- We will denote the extension of spring at top most position from mean position as "x".
- From mean position xo = 0 m. The block will further move up the slope and expense all its kinetic energy "Ek" in the form of gain in potential energy and gain in elastic potential energy "Vk" and work is done against friction.
Vki = Ep2 + Wf + Vk2
0.4 = mg*x*sin ( 30 ) + 0.24525 + 0.08495 + u*m*g*x*cos ( θ ) + 0.5*k*x^2
0.5*80*x^2 + x*(0.5*9.81*sin(30) + 0.2*0.5*9.81*cos ( 30 ) ) + 0.3302-0.4 =0
40x^2 + 3.30207x - 0.0698 = 0
Solve the quadratic equation:
x = -0.1 m (10 cm) - initial compression at rest ;
x = 0.01745 m
- So the extension in spring at the rest position is x = 0.01745 m. The position of the next resting point is:
final position = xi + x
final position = 0.1 + 0.01745
final position = 0.11745 m from initial position.
x =
(a) The stored PE will be "0.40 J".
(b) The block's speed will be "0.48 m/s".
(c) The block's position will be at "12 cm".
Potential energy (P.E)According to the question,
Mass, m = 500 h
Spring constant = 80 N/m
Coefficient of friction = 0.20
Inclined angle = 30°
(a) We know the formula,
Spring P.E = [tex]\frac{1}{2}[/tex] kx²
By substituting the values,
= [tex]\frac{1}{2}[/tex] × 80 × (0.1)²
= 0.40 J
(b) We know that,
→ [tex]P.E_s[/tex] = ΔKE + Work done by friction + Δ[tex]PE_g[/tex]
or,
= [tex]\frac{1}{2}[/tex] mv² + [tex]\mu_R[/tex] mgx Cosθ + mgx Sinθ
By substituting the values,
0.40 = [tex]\frac{1}{2}[/tex] × 0.5 × v² + 0.2 × 0.5 × 9.8 × Cos30° × 0.1 + 0.5 × 9.8 × 0.1 × Sin30°
= 0.48 m/s
(c) By using the above formula,
→ 0.40 = [tex]\mu_R[/tex] mgl Cosθ + mgl Sinθ
= 0.2 × 0.5 × 9.8 × Cos30° × l + 0.5 × 9.8 × l × Sin30°
l = 0.12 m or,
= 12 cm
Thus the above responses are correct.
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1. Is the collision between the ball and the pendulum elastic or inelastic? Justify your answer by calculating the kinetic energy of the system before collision using the value of vxo in the experiment and the kinetic energy just after collision using the experimental value of h in Eq. 9.2.
Answer:
So energy is not conserved and inelastic shock
Explanation:
In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.
Let's find the kinetic energy just before the crash
K₀ = ½ m vₓ₀²
After the crash we can use the law of conservation of energy
Starting point. Right after the crash, before starting to climb
Em₀ = K = ½ (m + M) v₂²
Final point. At the maximum height of the pendulum
[tex]Em_{f}[/tex] = U = (m + M) g h
Where m is the mass of the bullet and M is the mass of the pendulum
Em₀ = Em_{f}
½ (m + M) v₂² = (m + M) g h
v₂ = √ 2g h
Now we can calculate the final kinetic energy
K_{f} = ½ (m + M) v₂²²
K_{f} = ½ (m + M) (2gh)
The relationship between these two kinetic energies is
K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)
K₀ / K_{f} = m / (m + M) vₓ₀² / 2 g h
We can see that in this relationship the Ko> Kf
So energy is not conserved and inelastic shock
To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.
Explanation:An elastic collision is one that conserves kinetic energy, while an inelastic collision does not conserve kinetic energy. To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.
In this case, we can calculate the kinetic energy of the system before the collision using the value of vxo and the kinetic energy just after the collision using the experimental value of h in Eq. 9.2.
If the kinetic energy of the system is the same before and after the collision, then it is an elastic collision. If the kinetic energy decreases after the collision, then it is an inelastic collision.
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The limit to the eye's acuity is actually related to diffraction by the pupil. What is the angle between two just‑resolvable points of light for a 3.75 mm 3.75 mm diameter pupil, assuming the average wavelength of 554 nm 554 nm ?
Answer: [tex]1.8(10)^{-4} rad[/tex]
Explanation:
This problem is related to the Rayleigh Criterion, which provides the following formula to find the acuity or limit of resolution of an optic system with circular aperture (the eye in this case):
[tex]\theta=1.22\frac{\lambda}{D}[/tex]
Where:
[tex]\theta[/tex] is the angle of resolution (related to the acuity)
[tex]\lambda=554 nm=554(10)^{-9} m[/tex] is the wavelength of the light
[tex]D=3.75 mm=3.75(10)^{-3} m[/tex] is the diameter of the pupil
Solving:
[tex]\theta=1.22 \frac{554(10)^{-9} m}{3.75(10)^{-3} m}[/tex]
[tex]\theta=1.8(10)^{-4} rad[/tex]
The human eye's acuity is limited by light's diffraction through the pupil, which determines the minimum angle, Δθ, between two resolvable points. This is calculated using the formula Δθ = 1.22 (λ / D), where λ is the light's wavelength and D the diameter of the pupil.
Explanation:The resolution of human visual acuity is related to the diffraction of light by the eye's pupil. This diffraction effect can be worked out using a formula that quantifies resolving power (Δθ), when the diameter of the aperture (in this case, the pupil size) and the wavelength of incident light are known.
In your example, a 3.75 mm diameter pupil encountering light of 554 nm wavelength can identify two points of light provided they subtend an angle (Δθ) at the pupil calculated by using the formula: Δθ = 1.22 (λ / D). For all light's range of visible wavelengths, the angle is extremely tiny - on the order of tens of thousands of a degree. This shows that the eye’s ability to distinguish two separate points of light is limited by the diffraction of light by the eye’s pupil.
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An electron is placed in a magnetic field that is directed along a z axis. The energy difference between parallel and antiparallel alignments of the z component of the electron's spin magnetic moment with is 7.50 × 10-25 J. What is the magnitude of ?
Answer:
B = 0.0404 т
Explanation:
Given
ΔE = 7.50 x 10⁻²⁵ J
μb = 927.4 x 10⁻²⁶ J/т
ΔE = 2 * μb * B
B = ΔE / 2 μb
B = 7.50 x 10⁻²⁵ / 2 * 927.4 x 10⁻²⁶
B = 7.50 / 1854.8 x 10⁻²⁵⁻⁽⁻²⁶⁾
B = 0.00404 x 10¹
B = 0.0404 т
A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the bicycle rests on. Assume that the rear tire is rotating at ω = 32k rad/s. What are the angular velocities of the two cylinders? Consider r1 = 460 mm and r2 = 46 mm.
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 = [tex]\frac{460\times 32}{34\times 2}[/tex]
ω2 = 216.47
A comet has a period of 324 years; in other words, it orbits the Sun in 324 years. Most likely, this comet came from...(Hint: 324 years define a long period comet)
A. Somewhere outside the solar system
B. Oort's cloud
C. The ecliptic plane
D. The asteroid belt
E. The Kuiper belt
Comets are divided into two types. Long-period comets are the comets that take more than two hundred years to finish an orbit throughout the Sun originate from the Oort Cloud.
Explanation:
The Oort Cloud has sufficient distance apart of the Sun than the Kuiper Belt, it seems that the Oort Cloud objects were made closer to the Sun than the Kuiper Belt things.Long-period comets have highly eccentric orbits.The Oort cloud is considered to own a large range commencing from among 2,000 and 5,000 AU to 50,000 AU.The chemical makeup of long-period and short-period comets is quite alike.
Final answer:
A comet with a 324-year orbit is likely from the Oort Cloud, the distant spherical region of icy bodies surrounding our solar system, known as the source of long-period comets. The Kuiper Belt, on the other hand, sources short-period comets.
Explanation:
A comet with a period of 324 years is considered a long-period comet. These comets originate from a region far beyond the inner solar system. The Oort Cloud is a vast, spherical shell of icy bodies that extends up to about 50,000 astronomical units (AU) from the Sun and contains trillions of potential comets. This makes it the most likely source of long-period comets like the one mentioned. In contrast, the Kuiper Belt is a disk-shaped region beyond Neptune, extending to about 50 AU, which is known to source short-period comets, also known as Jupiter-family comets.
Long-period comets like the one with a 324-year orbit are thought to be influenced by gravitational perturbations from nearby passing stars or galactic tides, which can send them towards the inner solar system. Once these comets enter the inner solar system, they can have dramatic interactions with planets or the Sun, sometimes leading to their disintegration or alteration in orbit. An example of such an interaction was when Comet Shoemaker-Levy 9 broke apart and collided with Jupiter in 1994.
1. Consider a head-on collision between two identical billiard balls. Ball 1 is initially in motion toward ball 2, which is initially at rest. After the collision, ball 2 departs with the same velocity that ball 1 originally had. Disregard any friction between the balls and the surface. What happens to ball 1? What happens to ball 2?
Answer:
first ball is at rest after collision and the kinetic energy of the first ball is transferred to the second ball after collision.
Explanation:
mass of each ball = m
initial velocity of first ball = u
initial velocity of second ball = 0 m/s
final velocity of second ball = initial velocity of first ball = u
Use conservation of momentum
let v is the final velocity of the first ball after collision
m x u + m x 0 = m x v + m x u
So, v = 0
It means after the collision the second ball moves with the velocity which is equal to the initial velocity of the first ball and the first ball comes at rest.
As there is no friction between the balls during the collision, the collision of two balls is perfectly elastic and thus the kinetic energy of the system is conserved.
The entire kinetic energy of the first ball is transferred to the second ball after the collision.
The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.
According to the law of conservation of momentum, the change in momentum of a body before the collision is equal to the change in momentum after the collision.
The formula for calculating momentum is expressed as:
[tex]\rho = mv \\[/tex]
According to the conservation of momentum
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
Since ball 2 is initially at rest [tex]u_2=0[/tex]
Also if after the collision, ball 2 departs with the same velocity that ball 1 originally had, hence [tex]v=u_1[/tex]
Substituting the given parameters into the formula:
[tex]m_1u_1+m_2(0)=(m_1+m_2)u_1\\m_1u_1=m_1u_1+m_2u_2\\m_2u_2=0[/tex]
This shows that the velocity of the second ball is also zero
Due to the absence of friction between the colliding object, the collision is elastic in nature showing that the kinetic energy of the system is conserved.
The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.
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A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s and at the same time causes the engine to ignite. When the engine fires, it exerts a constant 20 N horizontal thrust away from the wall. a. How far from the base of the wall does the rocket land?
Answer:
Explanation:
mass of rocket m = .5 kg
height of wall h = 40 m
initial horizontal velocity u = .5 m /s
horizontal acceleration = force / mass
= 20 / .5
a = 40 m /s²
Let rocket falls or covers 40 m vertically downwards in time t
h = 1/2 gt² , initial vertical velocity = 0
40 = 1/2 x 9.8 x t²
t = 2.8566 s
During this period it will cover horizontal distance with initial velocity of .5 m /s and acceleration a = 40m /s²
horizontal distance = ut + 1/2 at²
= .5 x 2.8566 + .5 x 40 x 2.8566²
= 1.4283 + 163.2033
= 164.63 m .
The distance from the base of the wall does the rocket land is 164.63 m .
Calculation of the distance:A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s
We know that
horizontal acceleration = force / mass
a = 20 / .5
a = 40 m /s²
Now
h = 1/2 gt² ,
Here initial vertical velocity = 0
So,
40 = 1/2 x 9.8 x t²
t = 2.8566 s
Now
horizontal distance = ut + 1/2 at²
= .5 x 2.8566 + .5 x 40 x 2.8566²
= 1.4283 + 163.2033
= 164.63 m .
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An electron is accelerated through 1.95 103 V from rest and then enters a uniform 1.50-T magnetic field. (a) What is the maximum magnitude of the magnetic force this particle can experience?
Answer: 6.29*10^-12 N
Explanation:
given,
Potential difference of the electron, v = 1950 V
Magnetic field of the electron, B = 1.50 T
If the electron is accelerated through 19500 V from rest its Potential Energy has to be
converted to Kinetic Energy. This allows us solve for the velocity.
PE = Vq
PE = 1950 * 1.6*10^-19
PE = 3.12*10^-16 J
Also, PE = 1/2mv²
3.12*10^-16 = 1/2mv²
v = 2.62*10^7 m/s
to get F(max), we use,
F(max) = qvB
F(max) = 1.6*10^-19 * 2.62*10^7 * 1.5
F(max) = 6.29*10^-12 N
6.3 x 10⁻¹²N
Explanation:As stated by Lorentz Force law, the magnitude of a magnetic force, F, can be expressed in terms of a fixed amount of charge, q, which is moving at a constant velocity, v, in a uniform magnetic field, B, as follows;
F = qvB sin θ ------------(i)
Where;
θ = angle between the velocity and the magnetic field vectors
When the electron passes through a potential difference, V, it is made to accelerate as it gains some potential energy ([tex]P_{E}[/tex]) which is then converted to kinetic energy ([tex]K_{E}[/tex]) as it moves. i.e
[tex]P_{E}[/tex] = [tex]K_{E}[/tex] ----------------(ii)
But;
[tex]P_{E}[/tex] = qV
And;
[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex] x m x v²
Therefore substitute these into equation (ii) as follows;
qV = [tex]\frac{1}{2}[/tex] x m x v²
Make v subject of the formula;
2qV = mv²
v² = [tex]\frac{2qV}{m}[/tex]
v = [tex]\sqrt{\frac{2qV}{m} }[/tex] ---------------(iii)
From the question;
q = 1.6 x 10⁻¹⁹C (charge on an electron)
V = 1.95 x 10³V
m = 9.1 x 10⁻³¹kg
Substitute these values into equation (iii) as follows;
v = [tex]\sqrt{\frac{2*1.6*10^{-19} * 1.95*10^{3}}{9.1*10^{-31}} }[/tex]
v = [tex]\sqrt{\frac{6.24*10^{-16}}{9.1*10^{-31}} }[/tex]
v = 2.63 x 10⁷m/s
Now, from equation (i), the magnitude of the magnetic force will be maximum when the angle between the velocity and the magnetic field is 90°. i.e when θ = 90°
Substitute the values of θ, q v and B = 1.50T into equation (i) as follows;
F = qvB sin θ
F = 1.6 x 10⁻¹⁹ x 2.63 x 10⁷ x 1.50 x sin 90°
F = 6.3 x 10⁻¹²N
Therefore the maximum magnitude of the magnetic force this particle can experience is 6.3 x 10⁻¹²N
A spring with a constant of 16 N/m has 98 J of energy stored in it when it is extended. How far is the spring extended?
The spring has been extended for 3.5 m
Explanation:
We have the formula,
PE =1/2 K X²
Rewrite the equation as
PE=1/2 K d²
multiply both the sides by 2/K to simplify the equation
2/k . PE= 1/2 K d² . 2/K
√d²=√2PE/K
Cancelling the root value and now we have,
d=√2PE/k
d=√2×98 J / 16N/m
d=√12.25
d=3.5 m
The spring has been extended for 3.5 m
Final answer:
To find the extension of a spring with a constant of 16 N/m and 98 J of energy stored, we use the formula for potential energy. The spring is extended by 3.5 meters.
Explanation:
The question involves calculating the extension of a spring based on the energy stored in it and the spring's constant. The formula for the potential energy stored in a spring is given by
U = 1/2 kx², where U is the potential energy, k is the spring constant, and x is the extension of the spring from its equilibrium position. In this case, the energy (U) is 98 J and the spring constant (k) is 16 N/m.
The spring in question has an energy of 98 J stored in it.
Using the formula for potential energy stored in a spring, we have:
[tex]PE = 1/2 k x^2[/tex]
By substituting the known values, we find that the spring is extended by 0.7 m (70 cm
We rearrange the formula to solve for x:
x = √(2U/k).
Substituting the given values,
x = √(2*98/16)
= √(196/16)
= √(12.25)
= 3.5 meters. Therefore, the spring is extended by 3.5 meters.
Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant hor-izontal force on a charged object. A 15 g charged projectile is launched with a speed of 6.0 m/s at an angle 35° above the hori-zontal. It lands 2.9 m in front of the launcher. What is the magni-tude of the electric force on the projectile?
Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
Magnitude of electric force = 0.03345 N
Explanation:
We are given;
Mass; m = 15g = 0.015kg
Angle above horizontal; θ = 35°
Speed; v = 6 m/s
Horizontal displacement; d = 2.9m
Now formula for time of flight is given as;
time of flight; t = (2Vsinθ)/g
Thus, plugging in values, we have
t = (2 x 6.0 x sin35)/9.8
t = (12 x 0.5736)/9.8
t = 0.7024 s
Now, let's find the acceleration
The formula for horizontal displacement is given by;
d = (Vcosθ)t + (1/2)at²
Plugging in the relevant values ;
2.9 = [6(cos35) x 0.7024] + (1/2)a(0.7024)²
2.9 = (4.2144 x 0.8192) + (0.2467)a
2.9 = 3.45 + (0.2467)a
(0.2467)a = 2.9 - 3.45
a = -0.55/0.2467
a = -2.23 m/s²
Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²
We know that F = ma
Thus,Force = 0.015kg x 2.23m/s² =
= 0.03345 N
What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? View Available Hint(s) What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude.
Answer:
It has no effect on the amplitude.
Explanation:
When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.
Answer:
It decreases the amplitude.
Explanation:
If it is located at the end, its kinetic energy will be equal to zero, so the system as a whole would only have potential energy, which is defined as:
Ep = (1/2) * k * A²
Where A is the amplitude of the oscillation.
According to the energy principle, said energy must remain conserved, therefore, the total energy is equal to:
Etot = (1/2) * k * A²
Since the system is in equilibrium, according to the expression of conservation of energy, the energy in the position located at the end would be equal to the energy in equilibrium. As the sandbag falls, its kinetic energy decreases, which is why the total energy of the system also decreases.
According to the total energy expression, it is directly proportional to the square of the oscillation amplitude. If the total energy is decreased, the amplitude of the oscillation would also decrease.
"An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has an internal resistance of 10.0 Ω and the open-circuit voltage across its terminals is 50.0 V. The leads have no appreciable resistance. At time t = 0, the switch is suddenly closed." "When does the maximum current occur?"
Answer:
1.04x[tex]10^{-3}[/tex] s
Explanation:
->The maximum current through the resistor is
[tex]I_{max}[/tex] = V/R = V/[tex]Re^{-t/RC}[/tex]= V/R×[tex]e^{0}[/tex] = V/R
Voltage 'V'=50V
Effective resistance 'R'= 25.0-Ω+ 10.0 Ω= 35.0 Ω
Therefore, [tex]I_{max}[/tex]=50/35=> 1.43 A
->The maximum charge can be determined by
Q = CV
where,
Capacitance of the capacitor 'C' = 30.0µF = 30×10-⁶F
Therefore,
Q=30×10-⁶ x 50=>1.5 x [tex]10^{-3}[/tex]
In order to find that when does the maximum current occur, the time taken given the quantity of charge and the electric current is:
t= Q / I=> 1.5 x [tex]10^{-3}[/tex]/ 1.43
t= 1.04x[tex]10^{-3}[/tex] s
A loop of wire lies flat on a horizontal surface. A bar magnet is held above the center of a loop with south pole pointing down. Magnet is released.
As magnet approaches the loop of wire, the induced current in this wire is:
a) Clockwise
b) Counterclockwise
c) Zero
Answer:
A
Explanation:
Current, magnetic field and motion are mutually dependent and perpendicular to one another
Answer:
b) clockwise
Explanation:
The right hand rule states that to determine the direction of the magnetic force on a positive moving charge, point the thumb of the right hand in the direction of the potential v, the fingers in the direction of the magnetic field, B, and a perpendicular to the palm points in the direction of the force, F.
Since the south pole of the bar magnet is pointing downwards, the induced current developed in the wire repels it and moves in the clockwise direction.
A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.25 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet
Answer:
2.59 T
Explanation:
Parameters given:
Current flowing through the wire, I = 29 A
Angle between the magnetic field and wire, θ = 90°
Magnetic force, F = 2.25 N
Length of wire, L = 3 cm = 0.03 m
The magnetic force, F, is related to the magnetic field, B, by the equation below:
F = I * L * B * sinθ
Inputting the given parameters:
2.25 = 29 * 0.03 * B * sin90
2.25 = 0.87 * B
=> B = 2.25/0.87
B = 2.59 T
The magnetic field strength between the poles is 2.59 T
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reach a target 22 m above the marble's position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble-Earth system during the 22 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?
Answer:
a) [tex]\Delta U_{g} = 12.945\,J[/tex], b) [tex]\Delta U_{k} = 12.945\,J[/tex], c) [tex]k = 2930.059\,\frac{N}{m}[/tex]
Explanation:
a) The change in the gravitational potential energy of the marble-Earth system is:
[tex]\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)[/tex]
[tex]\Delta U_{g} = 12.945\,J[/tex]
b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:
[tex]\Delta U_{k} = 12.945\,J[/tex]
c) The spring constant of the gun is:
[tex]\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}[/tex]
[tex]k = \frac{2\cdot \Delta U_{k}}{x^{2}}[/tex]
[tex]k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}[/tex]
[tex]k = 2930.059\,\frac{N}{m}[/tex]
Electromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when they broadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to check your teeth. Even though they have different names and different applications, these types of radiation are really all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillating electric and magnetic fields.
Answer:
Explanation:
1. They have different wavelengths - Because These radiations form a spectra that differs by the size of the wavelength.
2. They have different Frequencies (f) that is frequency = 1/ wavelength
(f = 1/wavelength)
3. They propagate at different speed though a non vacuum media (non vacuum media affect the speed based on the wavelength)
A circus act involves a trapeze artist and a baby elephant. They are going to balance on a teeter-totter that is 10 meters long and pivoted at the middle. The lady has a weight of 500 newtons and is standing on one end. The elephant has a weight of 2500 N and is walking toward her on the beam. How far does the elephant have to walk from the end toward the middle to balance the beam
Answer:
Explanation:
Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman
weight x distance from pivot
= 500x 5
= 2500 Nm
torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance
= 2500x d
for equilibrium
2500 d = 2500
d = 1 m
So elephant will have to walk up to 1 m close to pivot or middle point.
A spherical gas-storage tank with an inside diameter of 8.0 m is being constructed to store gas under an internal pressure of 1.62 MPa. The tank will be constructed from steel that has a yield strength of 295 MPa. If a factor of safety of 3.0 with respect to the yield strength is required, determine the minimum wall thickness tmin required for the spherical tank.
Answer:
The minimum wall thickness Tmin required for the spherical tank is 65.90mm
Explanation:
Solution
Recall that,
Tmin = The minimum wall thickness =PD/2бp
where D = diameter of 8.0 m
Internal pressure = 1.62 MPa
Then
The yield strength = 295MPa/3.0 = 98.33
thus,
PD/2бp = 1.62 * 8000/ 2 *98.33
= 12960/196.66 = 65.90
Therefore the wall thickness Tmin required for the spherical tank is 65.90mm
A 6200 line/cm diffraction grating is 3.14 cm wide. If light with wavelengths near 624 nm falls on the grating, how close can two wavelengths be if they are to be resolved in any order
Answer:
[tex]1.6026299569\times 10^{-11}\ m[/tex]
Explanation:
Grating constant
[tex]d=\dfrac{1}{6200}=0.000161\ cm=0.000161\times 10^{-2}\ m[/tex]
Number of slits
[tex]N=3.14\times 6200=19468[/tex]
Order
[tex]m=\dfrac{d}{\lambda}\\\Rightarrow m=\dfrac{0.000161\times 10^{-2}}{624\times 10^{-9}}\\\Rightarrow m\approx 2[/tex]
At m = 1
[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{1\times 19468}\\\Rightarrow \Delta\lambda=3.2052599137\times 10^{-11}\ m[/tex]
At m = 2
[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{2\times 19468}\\\Rightarrow \Delta\lambda=1.6026299569\times 10^{-11}\ m[/tex]
The wavelengths can be close by [tex]1.6026299569\times 10^{-11}\ m[/tex]
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.785 of the escape speed from Earth and (b) its initial kinetic energy is 0.785 of the kinetic energy required to escape Earth
Answer:
Explanation:
Given that,
We have a projectile that is shot from the earth
Then, it's escape velocity is
Ve = √(2GM / R)
Ve² = 2GM / R
M is the mass of earth
R is the radius of Earth
Using conservation of energy
Ki + Ui = Kf + Uf
The final kinetic energy is zero since the body is finally at rest
Then,
Ki + Ui = Uf
Kinetic energy can be determine using
K = ½mv²
Potential energy can be determine using
U = -GmM / R
Where m is mass of the body
Then,
Ki + Ui = Uf
½mv² - GmM / R = -GmM / r
r is the maximum height reached
m cancel out
½ v² - GM / R = -GmM / r
A. We want to find the maximum height reached when initial speed (v) is equal to 0.785 escape velocity
So, v = 0.785•Ve
So,
½v² - GM / R = -GM / r
½(0.785•Ve)² - GM / R = -GM / r
0.308•Ve² - GM / R = -GM / r
0.308 × 2GM / R - GM / R = -GM / r
Divide through by GM
0.616 / R - 1 / R = -1 / r
-0.384 / R = -1 / r
Cross multiply
-0.384r = -R
r = -R / -0.384
r = 2.6 R
B. When it initial kinetic energy is is 0.785 of the kinetic energy required to escape Earth
Ki = 0.785 Ke
Ki = 0.785 ½mVe²
Ki = 0.785 × ½m× 2GM / R
Ki = 0.785•G•m•M / R
So,
Ki + Ui = Uf
0.785•G•m•M / R - GmM / R = -GmM / r
Let G•m•M cancels out
0.785 / R - 1 / R = -1 / r
-0.215 / R = -1 / r
Cross multiply.
-0.215r = -R
r = -R / -0.215
r = 4.65 R
The gold-like mineral in the metamorphic rock pictured below is known as pyrite. When pyrite is exposed to air and water it can react
to form sulfate and an acid. The acid can discolor and eat away the surrounding rock.
Image courtesy of David Usher, USGS
Pyrite breaks down rocks in which of the following processes?
A
deposition
B. physical weathering
C.
erosion
D.
chemical weathering
Answer: C. Chemical Weathering
Explanation:
Answer:
Chemical Weathering
Explanation:
A laser beam is incident on a plate of glass that is 2.8 cm thick. The glass has an index of refraction of 1.6 and the angle of incidence is 36°. The top and bottom surfaces of the glass are parallel. What is the distance b between the beam formed by reflection off the top surface of the glass and the beam reflected off the bottom surface of the glass?
Final answer:
The distance between the reflected light beams can be found using reflection and refraction laws, applying Snell's law to calculate the angle of refraction, and subsequently using trigonometry to determine the light path within the glass and the shift caused by both reflections.
Explanation:
To determine the distance b between the light beam reflected from the top and the bottom surfaces of a glass plate, we need to use the laws of reflection and Snell's law for the refraction of light. The given information includes the thickness of the glass (2.8 cm), the index of refraction of the glass (1.6), and the angle of incidence (36°). When the laser beam strikes the top surface at an angle of incidence of 36°, it reflects at the same angle (law of reflection).
The angle of refraction can be found using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the indices of refraction of the air and glass, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively. Once the angle of refraction is determined, the actual path of the beam within the glass can be calculated using trigonometry, and subsequently the shift b caused by the refraction at the two interfaces (again using trigonometry).
A mass M is attached to an ideal massless spring. When the system is set in motion, it oscillates with a frequency f. What is the new oscillation frequency if the mass is doubled to 2M? g
Answer:
[tex]\frac {f_i}{\sqrt 2}[/tex]
Explanation:
Frequency is given by [tex]\frac {1}{2\pi}\sqrt{{\frac {k}{m}}[/tex]
Let the initial frequency be denoted by [tex]f_i[/tex]
[tex]f_i=\frac {1}{2\pi}\sqrt{{\frac {k}{M}}[/tex]
When M is 2M then
[tex]f_f=\frac {1}{2\pi}\sqrt{{\frac {k}{2M}}=\frac {f_i}{\sqrt 2}[/tex]
Therefore, the final frequency is
[tex]\frac {f_i}{\sqrt 2}[/tex]
20 POINTS! TRUE OR FALSE:
To increase the acceleration of an object, you would reduce its mass or increase the applied force.
Answer: TRUE
Explanation:
(a) For what frequencies does a 23.0-μF capacitor have a reactance below 160 Ω? f Correct: Your answer is correct. 6.95 Incorrect: Your answer is incorrect. Hz (b)What is the reactance of a 41.0-μF capacitor over this same frequency range? Xc Correct: Your answer is correct. Ω
Answer:
(a) So range of frequency [tex]f > 43.27[/tex] Hz
(b) the reactance is 89.75 Ω
Explanation:
Given:
(a)
Capacitance of a capacitor [tex]C= 23 \times 10^{-6}[/tex] F
Reactance of capacitive circuit [tex]X_{C} =[/tex] 160 Ω
From the formula of reactance,
[tex]X_{C} = \frac{1}{\omega C}[/tex]
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]f = \frac{1}{2\pi X_{C} C }[/tex]
[tex]f = \frac{1}{6.28 \times 160 \times 23 \times 10^{-6} }[/tex]
[tex]f = 43.27[/tex] Hz
So range of frequency [tex]f > 43.27[/tex] Hz
(b)
Capacitance [tex]C = 41 \times 10^{-6}[/tex] F
Frequency [tex]f = 43.27[/tex] Hz
From the formula of reactance,
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]X_{C} = \frac{1}{6.28 \times 43.27 \times 41 \times 10^{-6} }[/tex]
[tex]X_{C} =[/tex] 89.75 Ω
Therefore, the reactance is 89.75 Ω
A plate of glass with parallel faces having a refractive index of 1.57 is resting on the surface of water in a tank. A ray of light coming from above in air makes an angle of incidence 37.5 ∘ with the normal to the top surface of the glass.
θ₃ is 27.12°
Explanation:
Using Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
where,
n₁ = refractive index of material with incident light
n₂ = refractive index of material with refracted light
(a)
1 → air
2 → glas
3 → water
n(water) = n₃ = 1.333
From air to glass,
n₁ sin θ₁ = n₂sin θ₂
The angle of refraction 2 is the angle of incidence when the light comes from glass to water.
From glass to water:
n₁ sin θ₁ = n₃ sin θ₃
1 X sin 37.5° = 1.333 sin θ₃
sin θ₃ = sin (37.5)/ 1.333
sin θ₃ = 0.456
θ₃ = 27.12°
Therefore, θ₃ is 27.12°
A kite weighs 2.50 lb and has an area of 8.00 ft2 . The tension in the kite string is 6.60 lb when the string makes an angle of 45⁰ with the horizontal. For a wind of 20 mph (29.3 ft/s), what are the coefficients of lift and drag (based on the kite’s area)? Use air density of 0.00234 slug/ft3 .
Answer:
the coefficient of lift is 3.5561
the coefficient of drag is 2.3153
Explanation:
the solution is in the attached Word file
An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.
The final velocity of the electrons is _____.
The mass of the electron is 9.1x10^(-31) kg and its charge is 1.6x10^(-19) C.
Answer:
[tex]9.38\times 10^7 m/s[/tex]
Explanation:
We are given that
Potential ,V=25 kV=[tex]25\times 10^3 V[/tex]
Distance,r =1 cm=[tex]\frac{1}{100}=0.01 m[/tex]
1 m=100 cm
Mass of electron, m=[tex]9.1\times 10^{-31} kg[/tex]
Charge, q=[tex]1.6\times 10^{-19} C[/tex]
We have to find the final velocity of the electron.
Speed of electron,[tex]v=\sqrt{\frac{2qV}{m}}[/tex]
Using the formula
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}[/tex]
v=[tex]9.38\times 10^7 m/s[/tex]
Hence, the final velocity of the electron=[tex]9.38\times 10^7 m/s[/tex]