Answer:
A. Using displacement =Ut + 1/2gt²
=> 0 + 1/2 (-9.8)t²
= -4.9t²
So
h(t) = 50+ displacement
= 50 - 4.9t²
B. To reach the ground
h(t) = 0
So
50-4.9t²= 0
t = √ (50/4.9)
= 3.2s
C. Using
V = u+ gt
U= 0
V= - 9.8(3.2)
= 31.4m/s
D. If u = -9m/s
Then s = ut + 1/2gt²
5t- 1/2gt²
But distance from the ground is
=.> 50-5t- 4.8t²= 0
So t solving the quadratic equation
t= 3.58s
(a) The distance of the stone above the ground level at time t is [tex]h(t) = 50 - 4.9t^2[/tex]
(b) The time taken for the stone to strike the ground is 3.19 s.
(c) The velocity of the stone when it strikes the ground is 31.4 m/s.
(d) The time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
The given parameters;
height above the ground, h₀ = 50 mThe distance of the stone above the ground level at time t is calculated as;
[tex]h(t) = h_0 - ut - \frac{1}{2} gt^2\\\\h(t) = 50 - 0 -0.5\times 9.8t^2\\\\h(t) = 50 - 4.9t^2[/tex]
The time taken for the stone to strike the ground is calculated as;
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 50}{9.8} } \\\\t = 3.19 \ s[/tex]
The velocity of the stone when it strikes the ground is calculated as;
[tex]v =u + gt\\\\v = 0 + 3.2 \times 9.8\\\\v = 31.4 \ m/s[/tex]
The time taken for the stone to reach the ground when thrown at speed of 9 m/s is calculated as;
[tex]50 = 9t + \frac{1}{2} (9.8)t^2\\\\50 = 9t + 4.9t^2\\\\4.9t^2 + 9t - 50 = 0\\\\a = 4.9 \, \ b = 9, \ \ c = -50\\\\solve \ the \ quadratic \ equation\ using \ formula \ method\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-9 \ \ + /- \ \ \sqrt{(9)^2 - 4(4.9 \times -50)} }{2(4.9)} \\\\t = 2.41 \ s \ \ or \ \ - 4.24 \ s[/tex]
Thus, the time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
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Calculate the answers to the appropriate number of significant
12.21 x 9.19 =
the charge density in an insulateed solid sphere of radius find the electric field at a distance of from the center of the solid
Answer:
Assuming the charged density in the insulated solid sphere of radius 3.1m is 8.8e-9, the electric field at 5.2 meters is 73.1256 [tex]i[/tex].
Explanation:
The electric charge linear density is equal to 8.8 x[tex]10^{-9}[/tex]
the radius of the sphere is 3.1m
The magnitude of the electric field at the radius of the sphere equal to 5.2 meters can be calculated with the formula ;
- E = λ / 4πε₀ [ r / α ( α + r ) ] [tex]i[/tex]
Solution:
E = 8.8 x[tex]10^{-9}[/tex] / 4πε₀ [ 3.1/ 5.2( 5.2 + 3.1) ] [tex]i[/tex]
= 1018.0995 [0.07183] [tex]i[/tex]
= 73.1256 [tex]i[/tex]
A sphere of radius R has charge Q. The electric field strength at distance r > R is Ei.
What is the ratio Ef /Ei of the final to initial electric field strengths if (a) Q is halved, (b) R is halved, and (c) r is halved (but is still > R)? Each part changes only one quantity; the other quantities have their initial values.
Answer:
A. Ef/ Ei = 1/2
B. EF/ Ei = 1
C Ef / Ei = 4
Explanation:
To solve this we apply Coulomb's law which States that
E = Kq / r^2
Where
q = charge r = straight line distance from q to the point in question and
K = Coulomb's constant
Then
Ei = K Q / r^2
So
A) If Q is halved then
Ef = K Q / (2 r^2)
Ef/Ei = 1/2
B) If R is halved, the value of the E-f
at a distance r remains unchanged. So
Ef/Ei = 1
C) if r is now r/2 then
Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2
Ef / Ei = 4
A 10-cm-long thin glass rod uniformly charged to 6.00 nC and a 10-cm-long thin plastic rod uniformly charged to - 6.00 nC are placed side by side, 4.4 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
A. Specify the electric field strength E1
B. Specify the electric field strength E2
C. Specify the electric field strength E3
Answer:
A) E(r) = 1.3957 × 10^(5) N/C
B) E(r) = 9.8864 × 10⁴ N/C
C) E(r) = 1.13 × 10^(5) N/C
Explanation:
We are given;
q = 6 nc = 6 × 10^(-9) C
L = 10 cm = 0.1 m
d = 4.4 cm = 0.044 m
r1 = 1 cm = 0.01 m
r2 = 2 cm = 0.02 m
r3 = 3 cm = 0.03 m
Formula for the electric field strength in this question is given as;
E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)
When factorized, we have;
E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]
Plugging in the relevant values for q/(2π(ε_o)L)
We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m
Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53
Thus;
E(r) = 1078.52 [1/r + 1/(d - r)]
A) E1 is at r = 1 cm = 0.01m
Thus;
E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))
E(r) = 1.3957 × 10^(5) N/C
B) E2 is at r = 2 cm = 0.02 m
Thus;
E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))
E(r) = 9.8864 × 10⁴ N/C
C) E2 is at r = 3 cm = 0.03 m
Thus;
E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))
E(r) = 1.13 × 10^(5) N/C
A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75.0 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on the
Given that,
Current = 4 A
Sides of triangle = 50.0 cm, 120 cm and 130 cm
Magnetic field = 75.0 mT
Distance = 130 cm
We need to calculate the angle α
Using cosine law
[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]
[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]
[tex]\alpha=\cos^{-1}(0.3846)[/tex]
[tex]\alpha=67.38^{\circ}[/tex]
We need to calculate the angle β
Using cosine law
[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]
[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]
[tex]\beta=\cos^{-1}(0.923)[/tex]
[tex]\beta=22.63^{\circ}[/tex]
We need to calculate the force on 130 cm side
Using formula of force
[tex]F_{130}=ILB\sin\theta[/tex]
[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]
[tex]F_{130}=0[/tex]
We need to calculate the force on 120 cm side
Using formula of force
[tex]F_{120}=ILB\sin\beta[/tex]
[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]
[tex]F_{120}=0.1385\ N[/tex]
The direction of force is out of page.
We need to calculate the force on 50 cm side
Using formula of force
[tex]F_{50}=ILB\sin\alpha[/tex]
[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]
[tex]F_{50}=0.1385\ N[/tex]
The direction of force is into page.
Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.
a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.
b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.
c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.
Given the following data:
Current = 4.00 Amperes.Magnetic field strength = 75.0 mT = [tex]7.5 \times 20^{-3}\;T[/tex]Length = 130 cm to m = 1.3 mHypotenuse = 130 cmOpposite side = 120 cmAdjacent side = 50 cmLet us assume the current is flowing in a counterclockwise direction in the right-angle triangle.
First of all, we would determine the angles by using cosine rule:
[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]
[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]
a. To the determine the magnitude of the magnetic force on the 130 cm side:
Mathematically, the force acting on a current in a magnetic field is given by the formula:
[tex]F = BILsin\theta[/tex]
Where:
B is the magnetic field strength.I is the current flowing through a conductor.L is the length of conductor.[tex]\theta[/tex] is the angle between a conductor and the magnetic field.Substituting the given parameters into the formula, we have;
[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]
b. To the determine the magnitude of the magnetic force on the 120 cm side:
[tex]F_{120}=BILsin\beta[/tex]
[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]
c. To the determine the magnitude of the magnetic force on the 50 cm side:
[tex]F_{50}=BILsin\alpha[/tex]
[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]
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What is the reason for the increase and decrease size of the moon and write down in a paragraph.
Answer:
The reason for the increase or decrease of the moon is due to the angular perception of the moon.
Explanation:
Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.
The zenith is the highest part of the sky and the horizon the lowest.
When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.
But when looking up at the sky as there is no reference point there will be a failure in the perception of size.
A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .
Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hydrogen measured on Earth.
Is this galaxy approaching or receding from the Earth, and at what speed?
Answer:
3x10^5m/s
Explanation:
See attached file
Explanation:
The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
Doppler's Effect
According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.
Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.
[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]
Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.
[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]
[tex]v = 3\times 10^5\;\rm m/s[/tex]
Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
To know more about the doppler effect, follow the link given below.
https://brainly.com/question/1330077.
Complete each of the statements
A. Lines of force are lines used to represent ________ an ________ electric field
B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge
C. The electric field is uniform if at any point in the field its _________ and ________ is the same
D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation
Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
HELP!!! 35 point question. answer at least 3 correctly. please include equations and how you did it
Answer:
9. (B) ¼ Mv²
10. (A) √(3gL)
11. 20 N
12. 5 m/s²
Explanation:
9. The rotational kinetic energy is:
RE = ½ Iω²
RE = ½ (½ MR²) (v/R)²
RE = ¼ Mv²
10. Energy is conserved.
Initial potential energy = rotational energy
mgh = ½ Iω²
Mg(L/2) = ½ (⅓ ML²) ω²
g(L/2) = ½ (⅓ L²) ω²
gL = ⅓ L² ω²
g = ⅓ L ω²
ω² = 3g / L
ω = √(3g / L)
The velocity of the top end is:
v = ωL
v = √(3gL)
11. Sum of torques about the hinge:
∑τ = Iα
-(Mg) (L/2) + (T) (r) = 0
T = MgL / (2r)
T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)
T = 20 N
12. Sum of forces on the block in the -y direction:
∑F = ma
mg − T = ma
Sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a / R)
T = ½ Ma
Substitute:
mg − ½ Ma = ma
mg = (m + ½ M) a
a = mg / (m + ½ M)
Plug in values:
a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))
a = 5 m/s²
A 0.100-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.550 m apart and 2.0 m long,
(a) If the coefficient of kinetic friction between the rod and rails is 0.120, what vertical magnetic field is required to keep the rod moving at a constant speed?
(b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).
Answer:
The speed of the rod is 2.169 m/s.
Explanation:
Given that,
Mass = 0.100 kg
Current = 15.0 A
Distance = 2 m
Length = 0.550 m
Kinetic friction = 0.120
(a). We need to calculate the magnetic field
Using relation of frictional force and magnetic force
[tex]F_{f}=F_{B}[/tex]
[tex]\mu mg=Bli[/tex]
[tex]B=\dfrac{\mu mg}{li}[/tex]
Where, l = length
i = current
m = mass
Put the value into the formula
[tex]B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}[/tex]
[tex]B=0.01425\ T[/tex]
[tex]B=1.425\times10^{-2}\ T[/tex]
(b). If the friction between the rod and rail is reduced zero.
So, [tex]f_{f}=0[/tex]
We need to calculate the acceleration
Using formula of force
[tex]F_{net}=f_{f}+F_{B}[/tex]
[tex]F_{net}=0+Bil[/tex]
[tex]ma=Bil[/tex]
[tex]a=\dfrac{Bil}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}[/tex]
[tex]a=1.176\ m/s^2[/tex]
We need to calculate the speed of the rod
Using equation of motion
[tex]v^2=u^2+2as[/tex]
Put the value into the formula
[tex]v^2=0+2\times1.176\times2[/tex]
[tex]v^2=\sqrt{4.704}\ m/s[/tex]
[tex]v=2.169\ m/s[/tex]
Hence, The speed of the rod is 2.169 m/s.
Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 500 nm on the slits
Answer:
The values is [tex]m_{max} = 8001 \ bright \ spots[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]
At the first half of the screen from the central maxima
The number of bright spot according to the condition for constructive interference is
[tex]n = \frac{d * sin (\theta )}{\lambda}[/tex]
For maximum number of spot [tex]\theta = 90^o[/tex]
So
[tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]
[tex]n =4000[/tex]
Now for the both sides plus the central maxima we have
[tex]m_{max} = 2 * n + 1[/tex]
substituting values
[tex]m_{max} = 2 * 4000 + 1[/tex]
[tex]m_{max} = 8001 \ bright \ spots[/tex]
What happens to the magnetic field when you reverse the direction of current by sliding the battery voltage bar past 0 volts
Answer:
The polarity of the magnetic field changes
Explanation:
This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)
The work function of a certain metal is φ = 3.55 eV. Determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (Planck's constant is: h = 4.1357×10-15 eVs.)
Answer:
Explanation:
Let f₀ be the frequency .
energy of photons having frequency of f₀
= hf₀ where h is plank's constant
for electron to get ejected , work function should be equal to energy of photon
hf₀ = 3.55
4.1357 x 10⁻¹⁵ x f₀ = 3.55
f₀ = 8.58 x 10¹⁴ Hz .
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a point r < R1 is:
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
What is the threshold velocity vthreshold(ethanol) for creating Cherenkov light from a charged particle as it travels through ethanol (which has an index of refraction of n
Explanation:
The velocity of light in a medium of refractive index [tex]n[tex] is given by,
[tex]v=\frac{c}{n}[/tex]
[tex]v \text { is the velocity of light in the medium }[/tex]
[tex]c \text { is speed of light in vacuum }[/tex]
The exact value of speed of light in vacuum is [tex]299792458 \mathrm{m} / \mathrm{s}[/tex].
For Cherenkov radiation to be emitted, the velocity of the charged particle traversing the medium must be greater than this velocity. Thus, the threshold velocity of for creating Cherenkov radiation is,
[tex]v_{\text {Cherenkov }} \geq \frac{c}{n}[/tex]
[tex]v_{\text {threshod }}=\frac{c}{n}[/tex]
For water [tex]n=1.33,[tex] thus the threshold velocity for producing Cherenkov radiation in water is,
[tex]v_{\text {threatold }(\text { water })} &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.33}[/tex]
[tex]=225407863 \mathrm{m} / \mathrm{s}[/tex]
[tex]=2.254 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]
For ethanol [tex]n=1.36[tex], thus the threshold velocity for producing Cherenkov radiation in water is,
[tex]v_{\text {threstold }( \text { ettanol) } } &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.36}[/tex]
[tex]=220435630 \mathrm{m} / \mathrm{s}[/tex]
[tex]=2.204 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]
Answer:
The answer is "2.2 × [tex]\bold{10^8}[/tex]".
Explanation:
In the given question the value of n is missing which can be defined as follows:
n= 1.36
The velocity value of the threshold(ethanol) for a generation the Cerenkov light from the charged particle by travel through ethanol as:
know we will have to use an equation as follows:
Formula:
(ethanol) or the vthreshold = [tex]\frac{c}{n}[/tex]
[tex]= \frac{3\times 10^8} {1.36} \\\\= 2.2 \times 10^8[/tex]
The water in vthreshold:
[tex]= 2.2 \times 10^8 \ \ \frac{m}{ s} \\\\[/tex]
Express the value in c, that is multiple, so, the value of vthreshold(water) is:
=(0.735) c
Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin. Part A What is the maximum wavelength of electromagnetic radiation that can cause a transition
Answer:
The maximum wavelength of the e-m wave is 6.9 x 10^-7 m
Explanation:
Energy required to trigger a response = 1.8 eV
we convert to energy in Joules.
1 eV = 1.602 x 10^-19 J
1.8 eV = [tex]x[/tex] J
[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J
The energy of an electromagnetic wave is gotten as
E = hf
where
h is the Planck's constant = 6.63 x 10^-34 J-s
and f is the frequency of the wave.
substituting values, we have
2.88 x 10^-19 = 6.63 x 10^-34 x f
f = (2.88 x 10^-19)/(6.63 x 10^-34)
f = 4.34 x 10^14 Hz
We know that the frequency of an e-m wave is given as
f = c/λ
where
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength of the e-m wave
From this we can say that
λ = c/f
λ = (3 x 10^8)/(4.34 x 10^14)
λ = 6.9 x 10^-7 m
Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 1.50-mm-diameter superconducting wire.
What current is needed?
Answer:
The current needed is 1790.26 A
Explanation:
Given;
magnitude of magnetic field, B = 1.5 T
length of the solenoid, L = 1.8 m
diameter of the solenoid, d = 75 cm = 0.75 m
The magnetic field is given by;
[tex]B = \frac{\mu_o NI }{L}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is current in the solenoid
N is the number of turns, calculated as;
[tex]N = \frac{Length \ of\ solenoid}{diameter \ of \ wire} \\\\N = \frac{1.8}{1.5*10^{-3}} =1200 \ turns[/tex]
The current needed is calculated as;
[tex]I = \frac{BL}{\mu_o N} \\\\I = \frac{1.5 *1.8}{4\pi *10^{-7} *1200} \\\\I = 1790.26 \ A[/tex]
Therefore, the current needed is 1790.26 A.
Answer:
I = 1790.5 A
Explanation:
The magnetic field due to a solenoid is given by the following formula:
B = μ₀NI/L
where,
B = Magnetic Field Required = 1.5 T
μ₀ = 4π x 10⁻⁷ T/A.m
L = length of Solenoid = 1.8 m
I = Current needed = ?
N = No. of turns = L/diameter of wire = 1.8 m/1.5 x 10⁻³ m = 1200
Therefore,
1.5 T = (4π x 10⁻⁷ T/A.m)(1200)(I)/1.8 m
I = (1.5 T)(1.8 m)/(1200)(4π x 10⁻⁷ T/A.m)
I = 1790.5 A
an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas
Answer:
Explanation:
Charge on an electron (q) = 1.6 * 10 ^ -19 C
Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec
We know that, Force exerted on moving particle moving through a magnetic field :
[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]
1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B
B = 0.08573 T
An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
What is the average velocity of the airplane? Round your answer to the nearest whole number.
Answer:
433
Explanation:
Compare the value for the inductor when the current was increasing vs decreasing. Which statement matches the expected results. The inductance should be the same regardless of whether the current is increasing or decreasing. The inductance should be greater while the current is increasing. The inductance should be greater while the current is decreasing.
Answer:
see that the inductance depends on the variation with respect to time of the current, therefore it is independent, increase decreases,
Explanation:
The express for inductance is
[tex]E_{L}[/tex]= L dI / dt
L = E_{L} (di / dt)⁻¹
where L is the inductance, E_{L} the induced electromotive force, di/dt the variation of the current as a function of time.
When analyzing this equation we see that the inductance depends on the variation with respect to time of the current, therefore it is independent, increase decreases,
Correct answer the inductance must be the same regardless of whether the current increases or decreases.
Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.
A small insect viewed through a convex lens is 1.5 cmcm from the lens and appears 2.5 times larger than its actual size. Part A What is the focal length of the lens
Answer:
The focal length of the lens is 2.5 cm
Explanation:
Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;
[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]
Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:
[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]
then, solving for the focal distance knowing that the object distance is 1.5 cm:
[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/tex]
A car starts from rest and accelerates with a constant acceleration of 5 m/s2 for 4 s. The car continues for 18 s at constant velocity. How far has the car traveled from its starting point
To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.68 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.
Answer:
1.8/0.61 =2.95 ft
Hope it helped u if yes mark me BRAINLIEST!
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The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as seen from the left)?
Answer:
The induced current is clockwise
A5 kg box slides 3 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3 m / s?
Answer:
Coefficient of kinetic friction (Cof. KE) = 0.153
Explanation:
Given:
Mass of box (M) = 5 kg
Distance = 3 m
Initial speed (v) = 3 m/s
Find:
Coefficient of kinetic friction (Cof. KE)
Computation:
v² = u² + 2as
a = v² / 2s
a = 9 / 2(3)
a = 1.5 m/s²
Coefficient of kinetic friction (Cof. KE) = a / g
Coefficient of kinetic friction (Cof. KE) = 1.5 / 9.8
Coefficient of kinetic friction (Cof. KE) = 0.153
What do we call a substance in
which two or more elements are
chemically bonded
Answer:
A compound
Explanation:
A compound is a substance formed when two or more elements are chemically joined
Answer:
Compound
Explanation:
A compound is a substance derived from the chemical combination of two or more elements
e.g Water ;
= [tex]H_2O\\Hydrogen\:and\:Oxygen[/tex]
Salt ;
[tex]NaCl\\Sodium\:and\: Chlorine[/tex]
how many electrons do calcium have in their outer shell
Answer:
Calcium has two electrons in its outer shell.
Explanation:
Calcium is defined as a metal due to its physical and chemical traits. The two outer electrons are very reactive. Calcium has a valence of 2.
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Required:
How far away did the impact occur?
Answer:
The distance is [tex]d = 193.6 \ m[/tex]
Explanation:
From the question we are told that
The time interval between the sounds is k[tex]t_1 = k + t_2[/tex] = 0.50 s
The speed of sound in air is [tex]v_s = 343 \ m/s[/tex]
The speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]
Generally the distance where the collision occurred is mathematically represented as
[tex]d = v * t[/tex]
Now from the question we see that d is the same for both sound waves
So
[tex]v_c t = v_s * t_1[/tex]
Now
So [tex]t_1 = k + t[/tex]
[tex]v_c t = v_s * (t+ k)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]t = 0.0645 \ s[/tex]
So
[tex]d = 3000 * 0.0645[/tex]
[tex]d = 193.6 \ m[/tex]
