Answer:
a) x = 0.33 A , b) K = ¾ Em , c) the kinetic energy increases 9 times
Explanation:
a) In a simple harmonic motion the mechanical energy is conserved and is expressed by the relation
Em = ½ k A²
At all points of movement the mechanical energy and
Em = K + U
remember that potential energy is
U = ½ k x²
they ask us for the point where
U = 1/9 Em
we substitute
½ k x² = 1/9 (½ k A²)
x = √1/9 A
x = 0.33 A
b) At all points the equation for mechanical energy is
Em = K + U
K = Em - U
K = ½ k A² - ½ k x²
K = ½ k (A² -x²)
at point x = ½ A
K = ½ k (A² - ¼ A²)
K = (½ k A²) ¾
K = ¾ Em
U = ½ k (½A) 2
U = ¼ Em
the fraction of energy e
U / K = 1/3
c) Kinetic energy is
K = ½ k v²
the system is described by the expression
x = A cos (wt + Ф)
speed is defined
v = dx / dt
v = - A w sin (wt + Ф)
we substitute
K = ½ k (- A w sin ( wt + Ф))²
We write this equation for the initial amplitude A
K₀ = ½ k (w sin (wt + Ф))² A²
we write it for the new amplitude A´ = 3 A
K = ½ k (w sin (wt +Ф))² (3A)²
the relationship between this energies is
K / K₀ = 9
whereby the kinetic energy increases 9 times
The displacement for 1/9 of mechanical energy is A/3. When the displacement is half amplitude, 3/4 of total mechanical energy becomes kinetic energy. If the amplitude increases by a factor of 3, the maximum kinetic energy increases by a factor of 9.
Explanation:In this context, we know that the total mechanical energy (E) of the system is constant and is the sum of kinetic and potential energy. The potential energy (U) of a mass-spring system is given by the formula U = 1/2 kx².
Part (a)
We want to find the displacement x such that the potential energy is 1/9 total mechanical energy, thus we set U = E/9. From substituting U = 1/2kx² and E = 1/2kA² in the equation,
we find that x²/A² = 1/9, which gives x = A/3.
Part (b)
We know that when the particle is at x = A/2 the potential energy is U = 1/2 k(A/2)² = 1/8 kA².
The total energy E = 1/2 kA², so the kinetic energy (K) is given by E - U = 1/2kA² - 1/8kA² = 3/8 kA².
The fraction of the total energy which is kinetic is (K/E) = (3/8)/(1/2) = 3/4.
Part (c)
The maximum kinetic energy is the total mechanical energy (since potential energy is zero at maximum kinetic energy), which is E = 1/2*kA².
If we triple the amplitude, the new energy E' = 1/2*k(3A)² = 9/2*kA².
The factor by which the maximum kinetic energy changes is (E'/E) = (9/2)/(1/2) = 9.
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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 31.5 kg. The child grabs and clings to a bar that is 1.25 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 51.0 rpm to 17.0 rpm . What is the moment of inertia of the merry‑go‑round with respect to its central axis?
Answer:
24.609375 kgm²
Explanation:
I = Moment of inertia of the merry go round
[tex]N_1[/tex] = Initial speed = 51 rpm
[tex]N_2[/tex] = Final speed = 17 rpm
m = Mass of child = 31.5 kg
r = Radius = 1.25 m
In this system the angular momentum is conserved
[tex]IN_1=(I+mr^2)N_2\\\Rightarrow I\times 51=(I+31.5\times 1.25^2)17\\\Rightarrow I\dfrac{51}{17}-I=31.5\times 1.25^2\\\Rightarrow 2I=49.21875\\\Rightarrow I=\dfrac{49.21875}{2}\\\Rightarrow I=24.609375\ kgm^2[/tex]
The moment of inertia of the merry go round is 24.609375 kgm²
A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much work is done in stretching the spring 5.5 m from its equilibrium position? b. How much work is done in compressing the spring 1.5 m from its equilibrium position?
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer:
a) W = 1815 J
b) W = 135 J
Explanation:
We need to model the force using Hooke's law
We first get the elastic constant, k
A force of 60 N causes an extension of 0.5 m
F = 60 N, e = 0.5 m
F = k * e
60 = 0.5 k
k = 60/0.5
k = 120 N/m
Therefore the force in terms of the extension, x is:
F(x) = 120x
a) Work done in stretching the spring 5.5 m from its equilibrium position
b = 0 m, a = 5.5 m
[tex]W = \int\limits^a_b {F(x)} \, dx[/tex]
[tex]W = \int\limits^a_b {120x} \, dx \\a =0, b = 5.5[/tex]
[tex]W = 60x^{2} \\W = 60 [5.5^{2} -0^{2} ][/tex]
[tex]W = 1815 J[/tex]
b) Work required to compress the spring 1.5 m from its equilibrium position
[tex]W = \int\limits^a_b {F(x)} \, dx[/tex] b = 0, a = -1.5
[tex]W = 60x^{2} \\W = 60 [(-1.5)^{2} -0^{2} ]\\[/tex]
W = 135 J
The work done to compress the spring 1.5 m from its equilibrium position is 135 J
The Earth produces an inwardly directed electric field of magnitude 150V/m near its surface. (a) What is the potential of the Earth's surface relative to V=0 at r=[infinity]? (b) If the potential of the Earth is chosen to be zero, what is the potential at infinity? (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth's net charge; how would this affect your answer?)
Answer:
a
The potential of the earth surface is [tex]V_E= - 9.6*10^9 V[/tex]
b
when the potential of the earth is zero the charge is choose to be negative this implies that the potential at infinity would be [tex]V_\infty=+9.6*10^9V[/tex] and (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth's net charge the potential at the surface increase and the electric field remains the same
Explanation:
From the question we are told that
The magnitude electric field is [tex]E = 150V/m[/tex]
The potential of the earth surface when V = 0 and [tex]r = \infty[/tex] is mathematically represented as
[tex]V_E = - \frac{q}{4 \pi \epsilon_o r} = - E *R[/tex]
Where q is the charge on the surface of the earth which is negative
R is the radius of the earth
[tex]V_E = 150 * 64 *10^6[/tex]
[tex]V_E= - 9.6*10^9 V[/tex]
when the potential of the earth is zero the charge is choose to be negative
Then the potential at infinity would be [tex]+9.6*10^9V[/tex]
The goal in a command economy is economic ____,
Answer:
A- Equality
Explanation:
A track consists of a frictionless arc XY, which is a quarter-circle of radius R, and a rough horizontal section YZ. Block A of mass M is released from rest at point X, slides down the curved section of the track, and collides instantaneously and inelastically with identical block B at point Y. The two blocks move together to the right, sliding past point P, which is a distance l from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is ì. Express your answers in terms of M, R, ì, R, and g.
The question is not complete, so i have attached an image with the complete question.
Answer:
A) speed of block a before it hits block B;v = √2gR
B) speed of combined blocks after collision;V_ab = ½√2gR
C) kinetic energy lost = ½MgR
D) temperature change; ∆t = ½gR/c
E) Additional thermal energy;W_f = 2μMgL
Explanation:
A) From conservation of energy, potential energy = kinetic energy.
Thus; P.E = K.E
So, mgr = ½mv²
m will cancel out to give;
gr = v²
So, v = √2gR
B) from conservation od momentum, momentum before collision = momentum after collision.
Thus;
M_a•V_a = M_ab•V_ab
Where;
M_a is the mass of block A
V_a is speed of block A
M_ab is the mass after collision
V_ab is speed after collision
So, making V_ab the subject, we have;
V_ab = M_a•V_a/M_ab
Now from answer in a above,
V_a = V = √2gR
Also, M_a = M and M_ab = 2M because it's the sum of 2 masses after collision.
Thus;
V_ab = (M√2gR)/(2M)
M will cancel out to give;
V_ab = ½√2gR
C) From the work-kinetic energy theorem, the net work done on the object is equal to the change in the kinetic energy of the object.
Thus;
W_net = K_f - K_i = ½m_ab•v_ab²- ½m_a•v_a²) = ∆K.
We have seen that:
M_a = M and M_ab = 2M
V_a = √2gR and V_ab = ½√2gR
Thus;
∆K = ½(2M•(½√2gR)²) - ½(M•(√2gR)²)
∆K = ½MgR - MgE
∆K = -½MgR
Negative sign means loss of energy.
Thus kinetic energy lost = ½MgR
D) The formula for heat energy is given by;
Q = m•c•∆t
Thus, change in temperature is;
∆t = Q/Mc
Q is the heat energy, thus Q = ½MgR
Thus;∆t = ½MgR/(Mc)
M will cancel out to give;
∆t = ½gR/c
E) Additional thermal energy is gotten from;
Work done by friction;W_f = F_f x d
Where;
F_f is frictional force given by μF_n. F_n is normal force
d is distance moved
Thus;
W_f = μF_n*d
Mass after collision was 2M,thus, F_n = 2Mg
We are told to express distance in terms of L
Thus;
W_f = μ2Mg*L
W_f = 2μMgL
The problem involves principles of energy conservation and kinetic friction. As the block slides down, it converts potential energy to kinetic, collides with another block, and moves together along a rough surface. The work done against friction equals energy at the collision point yielding the expression for velocity after collision V' = sqrt(2µgl).
Explanation:The important concept here is conservation of energy. As block A slides down, it is converting potential energy into kinetic energy. At point X, the potential energy is maximum because it's at the highest point. The potential energy is given by M*g*R. By the time it reaches point Y, the potential energy has turned into kinetic energy which results in the block's speed (½MV²) at point Y.
After the collision, the two blocks move together with a new mass of 2M and new velocity V': MV = 2MV'. In the horizontal section YZ, due to kinetic friction, the blocks lose energy as work done against friction which causes them to stop.
The work done against friction is: F *d = µ*2M*g*l. Equating the energy at point Y after collision to work done against friction, we get: 2*½MV'² = µ*2M*g*l. Resolving gives the expression V' = sqrt(2µgl).
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Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 780 m diameter rotating about its central axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity?
Answer:1.513 rps
Explanation:
Given
Diameter of cylindrical space [tex]d=780\ m[/tex]
When the station rotates it creates centripetal acceleration which is given by
[tex]a_c=\omega ^2r[/tex]
Now it must create the effect of gravity so
[tex]g=\omega ^2r[/tex]
[tex]\omega =\sqrt{\frac{g}{r}}[/tex]
[tex]\omega =0.158\ rad/s[/tex]
and [tex]\omega =\frac{2\pi N}{60}[/tex]
Thus [tex]N=\frac{4.755}{3.142}[/tex]
[tex]N=1.513\ rps[/tex]
A string under tension τi oscillates in the third harmonic at frequency f3, and the waves on the string have wavelength λ3. If the tension is increased to τf = 5.0τi and the string is again made to oscillate in the third harmonic, what then are (a) the ratio of frequency of oscillation to f3 and (b) the ratio of the wavelength of the waves to λ3?
Answer:
(a). [tex]\dfrac{f_3'}{f_3} =\sqrt{5}.[/tex]
(b). The wavelength remains unchanged.
Explanation:
The speed [tex]v[/tex] of the waves on the string with tension [tex]T_i[/tex] is given by
[tex]v = \sqrt{\dfrac{T_iL}{m} }[/tex]
And if the string is vibrating, its fundamental wavelength is [tex]2L[/tex], and since the frequency [tex]f[/tex] is related to the wave speed and wavelength by
[tex]f = v/\lambda[/tex]
the fundamental frequency [tex]f_1[/tex] is
[tex]f_1 =\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}[/tex]
and since the frequency of the third harmonic is
[tex]f_3 = 3f_1[/tex]
[tex]f_3 = 3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L},[/tex]
and the wavelength is
[tex]\lambda_3 = \dfrac{2L}{3}.[/tex]
(a).
Now, if we increase to the string tension to
[tex]T_f = 5.0T_i[/tex]
the third harmonic frequency becomes
[tex]f_3' = 3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L},[/tex]
The ratio of this new frequency to the old frequency is
[tex]\dfrac{f_3'}{f_3} = \dfrac{3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L}}{3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}}[/tex]
[tex]\boxed{\dfrac{f_3'}{f_3} =\sqrt{5}.}[/tex]
(b).
The wavelength of the third harmonic remains unchanged because [tex]\lambda_3 = \dfrac{2L}{3}.[/tex] depends only on the length of the string
A long, straight, cylindrical wire of radius R carries a current uniformly distributed over its cross section.
a) At what location is the magnetic field produced by this current equal to third of its largest value? Consider points inside the wire.
r/R=???
b) At what location is the magnetic field produced by this current equal to third of its largest value? Consider points outside the wire.
r/R=???
Answer:
Explanation:
We shall solve this question with the help of Ampere's circuital law.
Ampere's ,law
∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire
we shall find magnetic field at distance x . current enclosed in the area of circle of radius x
= I x π x² / π R²
= I x² / R²
B x 2π x = μ₀ x current enclosed
B x 2π x = μ₀ x I x² / R²
B = μ₀ I x / 2π R²
Maximum magnetic B₀ field will be when x = R
B₀ = μ₀I / 2π R
Given
B = B₀ / 3
μ₀ I x / 2π R² = μ₀I / 2π R x 3
x = R / 3
b ) The largest value of magnetic field is on the surface of wire
B₀ = μ₀I / 2π R
At distance x outside , let magnetic field be B
Applying Ampere's circuital law
∫ B dl = μ₀ I
B x 2π x = μ₀ I
B = μ₀ I / 2π x
Given B = B₀ / 3
μ₀ I / 2π x = μ₀I / 2π R x 3
x = 3R .
The location inside the wire where the magnetic field equals a third of its maximum value is at r/R = 1/3. There is no point outside the wire where the magnetic field reaches a third of its maximum value because it monotonically decreases.
Explanation:To determine at what location inside the wire the magnetic field produced by the current is equal to a third of its largest value, we need to apply Ampère's Law. Inside a conductor carrying uniform current, the magnetic field B increases linearly with the distance r from the center of the wire due to the proportion of current enclosed. Thus, the magnetic field is given by B = μ_0 J r / 2, where J is the current density and μ_0 is the permeability of free space. Since the value at the surface (r=R) will be maximum (B_max), for B to be a third of B_max, we must have (1/3)B_max = (μ_0 J r / 2). Solving for r, we find r/R = 1/3.
For locations outside the wire, Biot-Savart Law or Ampère's Law show that the magnetic field decreases with 1/r. However, since the field is maximum at the surface, and we do not have an equation that varies outside the wire, we cannot directly calculate when the field will be a third without additional information on how the field varies with r beyond the wire's surface. Normally, the largest value of B is at the surface, and it decreases monotonically outside, never increasing again to allow for a location at which it is a third of its largest value.
The magnetic field for points inside the wire is proportional to the distance from the center, and for points outside the wire, it is inversely proportional to the distance from the center. However, the magnetic field does not reach a third of its largest value at any given point outside the cylindrical wire since it monotonically decreases.
The two spheres are rigidly connected to the rod of negligible mass and are initially at rest on the smooth horizontal surface. A force F is suddenly applied to one sphere in the y-direction and imparts an impulse of 9.8 N·s during a negligibly short period of time. As the spheres pass the dashed position, calculate the magnitude of the velocity of each one.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The velocity of the both spheres is [tex]v=4.62 m/s[/tex]
Explanation:
The free body diagram of this question is shown on the second uploaded image
Looking at this diagram we can deduce that there is no impulse force along the horizontal direction so
The mathematical equation for the impulse force along the horizontal axis is
[tex]\int\limits {\sum F_s} \, dt = 0[/tex]
The mathematical equation for the impulse force along the horizontal axis is
[tex]\int\limits {\sum F_y} \, dt = \Delta I_y[/tex]
Where [tex]I_y[/tex] is the impulse momentum along the y-axis and this is mathematically given as
[tex]\Delta I_y = 2mv_y[/tex]
substituting 9.8 N.s for [tex]\Delta I_y[/tex] and 1.5kg for m (mass of sphere) and the making [tex]v_y[/tex] the subject
[tex]v_y = \frac{9.8}{2 *1.5}[/tex]
[tex]=3.267 m/s[/tex]
The sum of the moment about the point I is mathematically represented as
[tex]\int\limits {\sum M_I } \, dt =\Delta H_I[/tex]
From the free body diagram [tex]\int\limits {\sum M_I } \, dt = I * 0.3 = 9.8 *0.3=2.94N.m[/tex]
[tex]\Delta H_I[/tex] is the angular momentum along the horizontal axis given as
[tex]\Delta H_I[/tex] [tex]= 2mv_x(0.3)[/tex]
substituting parameters into above equation
[tex]2.94 = 2 * 1.5 * v_x * 0.3[/tex]
Making [tex]v_x[/tex] the subject
[tex]v_x = \frac{2.94}{2*1.5*0.3}[/tex]
[tex]=3.267 m/s[/tex]
the velocity of the sphere is mathematically represented as
[tex]v = \sqrt{(v_x)^2 + (v_y)^2}[/tex]
Now substituting values
[tex]v = \sqrt{3.267^2 + 3.267^2}[/tex]
[tex]v=4.62 m/s[/tex]
The physics question pertains to the concept of impulse and momentum. The velocity of each sphere after the impulse is imparted can be calculated by dividing the change in momentum by twice the mass of one sphere.
Explanation:The Physics question is relevant to impulse and momentum in the domain of mechanics. From the problem, we know that a force F acting on the sphere imparts an impulse of 9.8 N·s. Impulse, denoted as J, in physics, is the product of force and the time for which it is applied and is equivalent to the change in momentum of the body. Thus, the change in momentum, Δp is equal to the impulse imparted, which is 9.8 N·s.
Now considering the system of two spheres rigidly connected, it is an isolated system (external force F is not considered as it acts for a very short time) and therefore, the total momentum of the system remains conserved. Therefore the momentum imparted to one sphere is equally distributed to both the spheres. Therefore, the final velocity v of each sphere can be calculated by using the formula v = Δp / 2m (m being the mass of each sphere).
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At any angular speed, a uniform solid sphere of diameter D has the same rotational kinetic energy as a uniform hollow sphere of the same diameter when both are spinning about an axis through their centers. The moment of inertia of a solid sphere is :
Answer:
m = (3/5)*M
Explanation:
Given:-
- The angular speed of both hollow and solid sphere = w
- The diameter of solid & hollow sphere = D
- The mass of the solid sphere = M
- Both rotate about their common axis with similar rotational kinetic energy.
Find:-
The mass of hollow sphere (m) ?
Solution:-
- The formula for rotational kinetic energy (K.E) of any rigid body is:
K.E = 0.5*I*w^2
Where,
I : Moment of inertia of rigid body
- The rotational kinetic energies of both hollow sphere and solid sphere are same:
0.5*I_solid*w^2 = 0.5*I_shell*w^2
I_solid = I_shell
0.4*M*D^2 / 4 = (2/3)*m*D^2 / 4
(2/5)*M = (2/3)*m
m = (3/5)*M
A series RC circuit, which is made from a battery, a switch, a resistor, and a 4-μF capacitor, has a time constant of 8 ms. If an additional 7-μF is added in series to the 4-μF capacitor, what is the resulting time constant?
Answer:
The resulting time constant will be [tex]\bf{5.09~ms}[/tex].
Explanation:
Given:
the time constant of the RC circuit, [tex]\tau = 8~ms[/tex]
The value of the capacitor in the circuit, [tex]C_{1} = 4~\mu F[/tex]
The value of addition capacitor added to the circuit, [tex]C_{2} = 7~\mu F[/tex]
The value of the time constant for a series RC circuit is give by
[tex]\tau = RC[/tex]
So the value of the resistance in the circuit is
[tex]R &=& \dfrac{\tau}{C}\\&=& \dfrac{8 \times 10^{-3}~s}{4 \times 10^-6~F}\\&=& 2000~\Omega[/tex]
When the capacitor [tex]C_{2}[/tex] is added to the circuit, the net value of the capacitance in the circuit is
[tex]C &=& \dfrac{C_{1}C_{2}}{C_{1} + C_{2}}\\&=& \dfrac{4 \times 7}{4 + 7}~\mu F\\&=& \dfrac{28}{11}~\mu F[/tex]
So the new time constant will be
[tex]\tau_{n} &=& (2000~\Omega)(\dfrac{28}{11} \times 10^{-6})~s\\&=& 5.09~ms[/tex]
A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, compressing it 5.0 cm. How hard is the athlete pushing?
Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
Answer:
37.5N
Explanation:
According to Hooke's law, the load or force, F, applied on an elastic material (e.g a spring) is directly proportional to the extension or compression, e, caused by the load. i.e
F ∝ e
F = k x e -------------------------(i)
where;
k = proportionality constant known as the spring constant.
From the question;
k = 750N/m
e = 5.0cm = 0.05m
Substitute these values into equation (i) as follows;
F = 750 x 0.05
F = 37.5N
Therefore, the load applied which is a measure of how hard the athlete is pushing is 37.5N
A color television tube generates some X-rays when its electron beam strikes the screen. What is the shortest wavelength of these X-rays, in meters, if a 31 kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these X-rays from reaching viewers.)
Answer: 4.0024 x 10^ -11 m or 0.040024 nm
Explanation:
λ = h c/ΔE
λ = wave lenght
h = 6.626 x 10 ^ -34 m² kg /s = planck constant
ΔE = 31 keV potential ( 1 keV = 1.6021 x 10^-16J)
c = velocity of light = 3 x 10⁸ m/s
substitute gives
λ = 6.626 x 10 ^ -34 m² kg /s x 3 x 10⁸ m/s = 4.0024 x 10^ -11 m
31 x 1.6021x10^-16 J
A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J. Calculate the kinetic energy and speed of the ejected electron.
Explanation:
Given that,
Wavelength of the light, [tex]\lambda=4170\ A=4170\times 10^{-10}\ m[/tex]
Work function of sodium, [tex]W_o=4.41\times 10^{-19}\ J[/tex]
The kinetic energy of the ejected electron in terms of work function is given by :
[tex]KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J[/tex]
The formula of kinetic energy is given by :
[tex]KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s[/tex]
Hence, this is the required solution.
A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current later?
Complete question:
A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
Answer:
The current in the circuit 7 ms later is 0.2499 A
Explanation:
Given;
Ideal inductor, L = 45-mH
Resistor, R = 60-Ω
Ideal voltage supply, V = 15-V
Initial current at t = 0 seconds:
I₀ = V/R
I₀ = 15/60 = 0.25 A
Time constant, is given as:
T = L/R
T = (45 x 10⁻³) / (60)
T = 7.5 x 10⁻⁴ s
Change in current with respect to time, is given as;
[tex]I(t) = I_o(1-e^{-\frac{t}{T}})[/tex]
Current in the circuit after 7 ms later:
t = 7 ms = 7 x 10⁻³ s
[tex]I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A[/tex]
Therefore, the current in the circuit 7 ms later is 0.2499 A
The current later after 7 ms is 0.2499 A
The current after a time (t) is given by:
[tex]I(t)=I_o(1-e^{-\frac{t}{\tau} })\\\\I_o=initial\ current\ at\ t=0,\tau=time \ constant\\\\Given\ that\ L=45mH=45*10^{-3}H,R=60\ ohm,V = 15V,t=7\ ms\\\\\tau=\frac{L}{R}=\frac{45*10^{-3}}{60} =7.5*10^{-4} \ s\\\\I_o=\frac{V}{R}=\frac{15}{60}=0.25\ A \\\\\\I(t)=I_o(1-e^{-\frac{t}{\tau} })=0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}} })\\\\\\I(t)=0.2499\ A[/tex]
The current later after 7 ms is 0.2499 A
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If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?
P=6w
I=3A
V = ?
Answer:2 volts
Explanation:
Power=current x voltage
6=3 x voltage
Divide both sides by 3
6/3=(3 x voltage)/3
2=voltage
Voltage=2volts
If part of an electric circuit dissipates energy at 6 Watts when it draws a current of 3 Amperes, then the voltage is impressed across it would be 2 volts .
What is power ?The rate of doing work is known as power. The Si unit of power is the watt .
The power dissipated in the circuit = work / time
As given in the problem we have to find out what voltage is impressed across the circuit If part of an electric circuit dissipates energy at 6 Watts when it draws a current of 3 Amperes,
The power dissipated across the circuit = Voltage × current
6 watts = Voltage × 3 ampere
Voltage = 6 / 3
Voltage = 2 volts
Thus , the voltage across the circuit would be 2 volts .
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A local fun house incorporates a gently curved, concave, spherical mirror into its display. When a child stands 1.2 m from the mirror, her reflection is upside down and appears to float in front of the mirror. When she stands about 0.8 m from the mirror, she sees only a blur reflected in the mirror, but when she stands about 0.5 m from the mirror, her reflection is right side up and appears to be behind the mirror. The approximate focal length of the mirror is
Options:
a) more than 0.8 m .
b) equal to 0.8 m .
c) between 0.5 m and 0.8 m .
d) less than 0.5 m .
Answer:
b) equal to 0.8 m .
Explanation:
Note:
An upside down image = Inverted Image
An image that appears in front of the mirror = Real image
An image that appears behind the mirror = Virtual image
Let the object distance from the pole of the mirror be u
When the child stands 1.2 m from the mirror:
u = 1.2 m ( Real and Inverted image of the child is formed)
When the child stands about 0.8 m from the mirror:
u = 0.8 m (Virtual, erect and magnified image of the child is formed)
When the child stands about 0.5 m from the mirror:
u = 0.5 m ( Virtual and erect image of the child is formed)
Note: All objects positioned behind the focal length of a concave mirror are always real. Objects start becoming virtual when they are placed on the focal length or in front of it (Close to the pole of the mirror), although objects placed on the focus has its image formed at infinity.
Since the nature of the image formed changed from real to virtual when the child stands about 0.8 m from the mirror, then the focal length is approximately equal to 0.8 m
The approximate focal length of the mirror will be "0.8 m". A complete solution is provided below.
According to the question,
When a child stands 1.2 m, the real as well as inverted image will be formed, then
u = 1.2 mWhen a child stands 0.8 m, the virtual, erect as well as magnified image will be formed, then
u = 0.8 mWhen a child stands 0.5 m, the virtual as well as erect image will be formed, then
u = 0.5 mSince,
Whenever the image is formed from real to virtual, the focal length will become 0.8 m.
Thus the answer above is right.
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A 145 g block connected to a light spring with a force constant of k = 5 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 3 cm from equilibrium and released from rest. Find the period of its motion. (Recall that the period, T, and frequency, f, are inverses of each other.) s Determine the maximum acceleration of the block.
Answer:
T = 1.07 s
a = 1.034 m/s2
Explanation:
Metric unit conversion
m = 145 g = 0.145 kg
x = 3 cm = 0.03 m
Suppose this is a simple harmonic motion, then its period T can be calculated using the following equation
[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi\sqrt{\frac{0.145}{5}} = 1.07 s[/tex]
The maximum acceleration would occurs when spring is at maximum stretching length, aka 0.03m
The spring force at that point would be
[tex]F_s = kx = 5*0.03 = 0.15 N[/tex]
According to Newton's 2nd law, the acceleration at this point would be
[tex]a = F_s/m = 0.15 / 0.145 = 1.034 m/s^2[/tex]
Immediately after being struck by a hammer, the nail (mass of 50 g) has a velocity of 50 m/s. The total frictional force is 62.5 kM. How far does the nail move before it comes to a stop
Answer:
nail will stop after traveling 1000 m
Explanation:
We have given mass m = 50 gram = 0.05 kg
Frictional force which is used to stop the mass F = 62.5 kN
Initial velocity is given u = 50 m/sec
From newton's law force is equal to F = ma, here m is mass and a is acceleration
So [tex]a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2[/tex]
As finally nail stops so final velocity v = 0 m/sec
From third equation of motion [tex]v^2=u^2+2as[/tex]
So [tex]0^2=50^2-2\times 1.25\times 10^6\times s[/tex]
s = 1000 m
So nail will stop after traveling 1000 m
Given a PV cell with light current density of 0.035A/cm2 and reverse saturation current density of 1.5x10-10 A/cm2 . Calculate the voltage and current at the maximum power and also the fill factor under the ambient temperature of 32 o C. Use m=1 for an ideal diode. Plot current and power as a function of voltage from V=0 to V=Voc to verify that the approximation method for maximum current and voltage is a good agreement with the plotted values.
Answer:
Explanation:
find the solution below
Final answer:
Calculating voltage, current at maximum power, and fill factor for a PV cell involves applying the Shockley diode equation under illumination. These calculations help in identifying the Maximum Power Point (MPP), crucial for optimizing the cell's performance. Detailed calculations were not provided, but the importance of plotting the I-V characteristics to validate the MPP was emphasized.
Explanation:
The task requires calculating the voltage and current at maximum power and the fill factor for a photovoltaic (PV) cell, given its light current density and reverse saturation current density, at an ambient temperature of 32°C. This involves understanding the behavior of PV cells under varying conditions and applying theoretical models to determine the operating characteristics that yield maximum power output. The fill factor (FF), the maximum power point (MPP), and the open-circuit voltage (Voc) are crucial aspects in assessing the performance and efficiency of PV cells. To derive an accurate answer, one would typically employ the Shockley diode equation adjusted for light conditions and solve for the power at different voltages to find the maximum. However, the calculation details are beyond this response scope due to the required complexity and dependency on specific model parameters not provided. Generally, these calculations involve detailed semiconductor physics and numerical methods to solve nonlinear equations reflective of the cell's I-V characteristic in illuminated conditions. Plotting current and power as functions of voltage from V=0 to Voc, helps validate the maximum power point approximation. This plot demonstrates the balance between voltage and current that a PV cell must achieve to optimize power output, usually found using methods like the perturb and observe algorithm in practical MPPT (Maximum Power Point Tracking) systems.
A 62.0 kgkg skier is moving at 6.90 m/sm/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 mm long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 mm high.How fast is the skier moving when she gets to the bottom of the hill?
How much internal energy was generated in crossing the rough patch?
Answer:
the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.
the internal energy generated in crossing the rough patch is 820.26 J
Explanation:
Given that,
Mass, m = 62 kg,
Initial speed, = 6.90 m/s
Length of rough patch, L = 4.50 m,
coefficient of friction, = 0.3
Height of inclined plane, h = 2.50 m
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
According to energy conservation equation,
Part (b)
The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.
The magnitude of the friction force is:
[tex]F_f=\mu mg[/tex]
Therefore, the work done by friction is:
[tex]W=-F_f d =-\mu mg d[/tex]
The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier
Substituting,
[tex]W=-(0.300)(62.0)(9.8)(4.50)=-820.26 J[/tex]
So, the internal energy generated in crossing the rough patch is 820.26 J
Part (a)
If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:
[tex]E=K_i +U_i = \frac{1}{2}mu^2 + mgh[/tex]
After crossing the rough patch, the new mechanical energy is
[tex]E'=E+W[/tex]
where
W = -820.26 J is the work done by friction
At the bottom of the hill, the final energy is just kinetic energy,
[tex]E' = K_f = \frac{1}{2}mv^2[/tex]
where v is the final speed.
According to the law of conservation of energy, we can write:
[tex]E+W=E'[/tex]
So we find v:
[tex]\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\\\v=\sqrt{u^2+2gh+\frac{2W}{m}}\\\\=\sqrt{6.9^2+2(9.8)(2.50)+\frac{2(-820.26)}{62.0}}\\\\=8.38 m/s[/tex]
Thus, the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.
The question repeated the units attached the the values
So in order of their appearance in the question, the correct units are;
62kg, 6.9m/s, 4.5m, 0.30, 2.5m
Answer:
A) Speed at bottom of hill = 8.38 m/s
B) Internal energy generated in crossing the patch = 820.26J
Explanation:
From the question, it's clear that there are 2 distinct stages of of her skis. The first stage is when she is on the rough patch while the second stage is when she skis down the hill.
A) For the first stage when she is on the rough patch;
If we apply Newton's second law of motion to the skis along the vertical direction, we obtain;
ΣF_y = N - mg = 0
Thus,N = mg
m = 62kg,thus,N = 62 x 9.8 = 607.6N
Now, let's find the kinetic friction. It's given by; f_k = μN
Where μ is coefficient of kinetic friction
Thus, f_k = 0.3 x 607.6 = 182.28N
Now, let's calculate the work done by this frictional force,
So, W_fk = f_k x distance = 182.28 x
4.5 = 820.26J
This work is done in a direction opposite to the displacement and it will have a negative sign. Thus,
W_fk = - 820.26J
Now, since the skier skies horizontally and perpendicular to the gravitational force, the work done due to gravity is zero. Thus,
W_grav = 0
Let's calculate the kinetic energy at the beginning of the rough patch.
K1 = (1/2)m(v1)²
K1 = (1/2) x 62 x (6.9)²= 1475.91J
Also,let's calculate the kinetic energy at end of rough patch.
K2 = (1/2)m(v2)²
where v2 is final velocity at end of rough patch
K2 = (1/2)(62)(v2)² = 31(v2)²
Now, the total work done when other forces other than gravity do work, is given by ;
W_total = W_other + W_grav = K2 - K1
In this case, W_other is W_fk
Thus,
- 820.26J + 0 = 31(v2)² - 1475.91J
31(v2)² = 1475.91J - 820.26J
31(v2)² = 655.65
(v2)² = 655.65/31
v2 =√21.15 = 4.6 m/s
Now, for the second stage when she skis down the hill;
In this case the only force acting is gravity, thus, W_other = 0
Work done by gravity = mgh
W_grav = 62 x 9.8 x 2.5 = 1519 J
Now K2 will be kinetic energy at top of hill while K3 will be kinetic energy at bottom of hill.
Thus,
K2 = (1/2)m(v2)²
K2 = (1/2)(62)(4.6)²
K2 = 655.96J
Similarly,
K3 = (1/2)m(v3)²
Where v3 is velocity at bottom of hill.
Thus,
K3 = (1/2)(62)(v3)²
K3 = 31(v3)²
Again, the total work done when other forces other than gravity do work, is given by ;
W_total = W_other + W_grav = K3 - K1
So,
W_other is zero.
Thus,
1519J = 31(v3)² - 655.96J
31(v3)² = 1519J + 655.96J
31(v3)² = 2174.96J
(v3)² = 2174.96/31
v3 = √70.16
v3 = 8.38 m/s
B) Workdone by non-conservative forces manifests itself as changes in the internal energy of bodies. Now, frictional force experienced in the first stage in crossing the patch is a non - conservative force. Thus,
Internal energy = W_fk = 820.26J
An object is electrically charged if the amounts of positive and negative charge it contains are not _______.
equal
negative
positive
increased
Answer:oops
Explanation:
Which are examples of short-term environmental change? Check all that apply.
tsunamis
El Niño
large asteroid and comet impacts
volcanic eruptions
global warming
i just took the test it's: tsunamis, El Nino, and volcanic eruptions.
Answer: tsunamis,El Nino,and volcanic eruptions
Explanation:
the correct answer for e d g e n u i t y
The examples of short-term environmental change are Tsunamis, El Niño, and Volcanic eruptions.
Tsunamis are huge ocean waves that are triggered by undersea disturbances like earthquakes or landslides. They can inflict major damage to coastal regions, however they usually happen in a short period of time.
El Nio is a climatic trend characterised by higher-than-normal equatorial Pacific ocean temperatures.
It can cause changes in weather patterns, such as greater rainfall in some areas and droughts in others, although the impacts are usually transient and endure for a few months to a couple of years.
Volcanic eruptions spew lava, ash, and gases into the atmosphere, which can have direct effects on the environment, including as changes in air quality and local weather patterns.
Thus, these effects, however, are often short-lived and localized.
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g An object spins in place with no unbalanced forces or torques acting upon it, what do we expect this object to do? The object’s spin will slow and eventually reverse direction. The object will continue spinning as it has been. The object will spin faster and faster. The object will begin to roll. None of these answers. The object will slowly stop spinning.
Answer:
The object will continue spinning as it has been.
Explanation:
When an object is spinning in a closed system with no unbalance
forces and no external force or torques are applied to it, it will have no change in angular momentum.
According to the law of conservation of angular momentum which states that when no external torque acts on an object, no change of angular momentum will occur
If the change in angular momentum is zero, then the angular momentum is constant; therefore, the object will continue spinning as it has been.
The same scenario is applicable to the rotation and the spinning of the earth.
Answer: The object will continue spinning as it has been.
Explanation: Unbalanced forces refers a force that changes the position, speed or direction of the object to which it is applied. While torque refers to otational or twisting effect of a force; a moment of force, defined for measurement purposes as an equivalent straight line force multiplied by the distance from the axis of rotation.
Any object that spins in place with no unbalanced forces or torques acting upon it will continue spinning as it has been. An example can be seen in the rotation of planets.
A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is the velocity of the ball of mass 2m after the collision?
Answer:
The velocity of mass 2m is [tex]v_B = 0.67 m/s[/tex]
Explanation:
From the question w are told that
The mass of the billiard ball A is =m
The initial speed of the billiard ball A = [tex]v_1[/tex] =1 m/s
The mass of the billiard ball B is = 2 m
The initial speed of the billiard ball B = 0
Let the final speed of the billiard ball A = [tex]v_A[/tex]
Let The finial speed of the billiard ball B = [tex]v_B[/tex]
According to the law of conservation of Energy
[tex]\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2[/tex]
Substituting values
[tex]\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2[/tex]
Multiplying through by [tex]\frac{1}{2}m[/tex]
[tex]1 =v_A^2 + 2 v_B ^2 ---(1)[/tex]
According to the law of conservation of Momentum
[tex]mv_1 + 2m(0) = mv_A + 2m v_B[/tex]
Substituting values
[tex]m(1) = mv_A + 2mv_B[/tex]
Multiplying through by [tex]m[/tex]
[tex]1 = v_A + 2v_B ---(2)[/tex]
making [tex]v_A[/tex] subject of the equation 2
[tex]v_A = 1 - 2v_B[/tex]
Substituting this into equation 1
[tex](1 -2v_B)^2 + 2v_B^2 = 1[/tex]
[tex]1 - 4v_B + 4v_B^2 + 2v_B^2 =1[/tex]
[tex]6v_B^2 -4v_B +1 =1[/tex]
[tex]6v_B^2 -4v_B =0[/tex]
Multiplying through by [tex]\frac{1}{v_B}[/tex]
[tex]6v_B -4 = 0[/tex]
[tex]v_B = \frac{4}{6}[/tex]
[tex]v_B = 0.67 m/s[/tex]
The stationary billiard ball of mass 2m will have a final velocity of 1/3 m/s after the head-on collision with a ball of mass m moving at 1 m/s due to conservation of momentum.
Explanation:In this problem, we are dealing with a one-dimensional collision. The principle of conservation of momentum is key in solving this problem. Momentum before the collision is the same as momentum after the collision. The initial momentum is given by mass * velocity of moving ball = m * v1, while the stationary ball has 0 momentum.
So, we can write this as m * 1 = (m + 2m) * v', where v' is the final velocity of the stationary ball after the collision. Solving for v', we get v' = m / 3m = 1/3 m/s. Hence, the stationary ball (of mass 2m) has a final velocity of 1/3 m/s after the collision.
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An 80 kg hockey player is skating at 4.1 m/s when an 85 kg hockey player from
another team, moving in the same direction at 5.5 m/s, hits him from behind. The
two becoming tangled up together and continue in a straight line. Find the
velocity of the two hockey players after they become attached.
Answer: 4.821m/s
Explanation:
initial momentum = final momentum
(80 x 4.1) + (85 x 5.5) = (80 + 85)u
328 + 467.5 = 165u = 795.5 / 165 = 4.821m/s
Identify two everyday phenomena that exhibit diffraction of sound and explain how diffraction of sound applies. Identify two everyday phenomena that exhibit diffraction of light and explain how diffraction of light applies. Comparing your examples, what are some significant differences between diffraction of light and diffraction of sound, if any?
Answer:
Explanation:
Diffraction is the term used to describe the bending of a wave around an obstacle. It is one of the general properties of waves.
1. Diffraction of sound is the bending of sound waves around an obstacle which propagates from source to a listener. Two of the daily phenomena that exhibit diffraction of sound are:
i. The voices of people talking outside a building can be heard by those inside.
ii. The sound from the horn of a car can be heard by people at certain distances away.
When sound waves are produced, the surrounding air molecules are required for its transmission. This is because sound wave is a mechanical wave which requires material medium for its propagation. When a source produces a sound, the sound waves bend around obstacles on its path to reach listeners.
2. Light waves are electromagnetic waves which can undergo diffraction. Diffraction of light is the bending of the rays of light around an obstacle. Two of the daily phenomena that exhibit diffraction of light are:
i. The shadow of objects which has the umbra and penumbra regions.
ii. The apparent color of the sky.
A ray of light is the path taken by light, and the combination of two or more rays is called a beam. A ray or beam of light travels in a straight line, so any obstacle on its path would subject the light to bending around it during propagation. These are major applications in pin-hole cameras, shadows, rings of light around the sun etc.
Some significant differences between diffraction of light and that of the sound are:
i. Diffraction of light is not as common as that of sound.
ii. Sound propagates through a wider region than light waves.
iii. Sounds are longitudinal waves, while lights are transverse waves.
The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit. The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit. twice / twice twice / half half / half the same as / half half / twice half / the same as
Answer:
Current is half
Voltage is half
Explanation:
according to Ohms law V = IR
Case 1
I₁ = V / R
I₂ = V / (R+R) = V / 2R
∴ I₁ / I₂ = (V/R) / (V/2R)
=> I₁ / I₂ = (V/R) * (2R/V)
=> I₁ / I₂ = (2VR / RV)
I₁ / I₂ = 2
I₂ = I₁ / 2
The current through each resistor in the two-resistor circuit is half the current through the resistor in the one-resistor circuit (the circuit in Part A).
The voltage across each resistor in the two-resistor circuit is half the voltage across the resistor in the one-resistor circuit.
The current through each resistor in a two-resistor series circuit is the same as the current through a one-resistor circuit. However, the voltage across each resistor in the two-resistor circuit is half of the voltage across the single resistor due to how voltages distribute in series circuits.
Explanation:In a series circuit, such as the two-resistor circuit you have mentioned, the current through each resistor is the same as the current through the single resistor in the one-resistor circuit. This is due to Kirchhoff's Current Law which states that the current entering a junction or a node must equal the current leaving it. However, the voltage accross each resistor in the two-resistor circuit is half of the voltage across the single resistor in the one-resistor circuit. This is because in a series circuit, the total voltage is the sum of the individual voltage drops across each resistor, according to Ohm’s law.
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At the instant when the speed of the loop is 3.00 m/sm/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
Complete question:
A rectangular loop of wire with dimensions 2.0 cm by 10.0 cm and resistance 1.0 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude of 2.0 T and is directed into the plane. At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
Check the image uploaded
Answer:
The magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N
Explanation:
Given;
resistance of the wire; R = 1.0 Ω
magnitude of magnetic field strength, B = 2.0 T
speed of the loop, v = 3.00 m/s
Induced emf is given as;
ε = IR
[tex]I = \frac{emf}{R} = \frac{VBL}{R}[/tex]
magnitude of the force that the magnetic field exerts on the loop:
F = BIL
Substitute in the value of I
[tex]F = \frac{VB^2L^2}{R}[/tex]
where;
L is the displacement vector between the initial and final end of the portion of the wire inside the field = 2.0 cm
Substitute the given values and solve for F
[tex]F = \frac{3*2^2*(2*10^{-2})^2}{1} \\\\F = 4.8 *10^{-3} \ N[/tex]
Therefore, the magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N
Complete Question:
A rectangular loop of wire with dimensions 1.50 cm by 8.00 cm and resistance 0.700 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.20 T and is directed into the plane of.
At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
Answer:
F = 0.133 N
Explanation:
Magnitude of the magnetic field, B = 2.20 T
Length of the loop = 1.5 cm = 0.015 m
The speed of the loop, v = 3.00 m/s
The emf induced in the loop , e = Blv
e = 2.20 * 0.015 * 3
e = 0.099 V
Current induced in the loop, I = e/R
I = 0.099/0.7
I = 0.1414 A
The magnitude of the force is given by, F = I *l *B sin90
F = 0.1414 * 0.015 * 2.20
F = 0.00467 N
Which of the following energy sources is no way derived from the sun is it nuclear energy title energy when energy or petroleum
Answer:nuclear energy is not derived from the sun
Explanation:
Nuclear energy is not derived from the sun.nuclear energy comes from the energy released when atoms are split apart and some mass is converted to energy