Answer:
Explanation:
area of the coil A = .08 x .08 = 64 x 10⁻⁴ m ²
flux through the coil Φ = area of coil x no of turns x magnetic field
= 64 x 10⁻⁴ x 50 x B where B is magnetic field
emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2
= 1.6 B
current induced = emf induced / resistance
12 x 10⁻³ = 1.6 B / 15
B = 112.5 x 10⁻³ T .
g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be
In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?
Answer:
9.1Ω
Explanation:
The circuit diagram has been attached to this response.
(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e
[tex]\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]
=> [tex]R_{X} = \frac{R_1 * R_2}{R_1 + R_2}[/tex] ------------(i)
From the question;
R1 = 3Ω,
R2 = 7Ω
Substitute these values into equation (i) as follows;
[tex]R_{X} = \frac{3 * 7}{3 + 7}[/tex]
[tex]R_{X} = \frac{21}{10}[/tex]
[tex]R_{X} = 2.1[/tex]Ω
(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.
Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e
R = Rₓ + R3
Rₓ = 2.1Ω
R3 = 7Ω
=> R = 2.1Ω + 7Ω = 9.1Ω
Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω
What will the surface charge density be if the radius of the disk is doubled but its total charge remains the same
Answer:
the new surface charge density = Q/4πr²( initial surface charge density divided by 4)
Explanation:
charge density(surface) = Q/A = charge/area
let r be the initial radius of the disk
therefore, area A = πr²
charge density = Q/πr²
Now that the radius is doubled, let it be represented as R
∴ R = 2r
Recall, charge density = Q/A
A = πR = π(2r)² = 4πr²
the new surface charge density = Q/4πr²
the initial surface charge density divided by 4
At what temperature will water begin to boil and turn to steam?
212 degrees Celsius
100 degrees Fahrenheit
212 kelvins
100 degrees Celsius
Answer:
100 degrees Celsius
Explanation:
Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.
At 100 degrees Celsius water begin to boil and turns to steam.
What are the boiling point and melting point of water?The melting point for water is 0 degrees C (32 degrees F). The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower. At sea level, pure water boils at 212 °F (100°C).
Is boiling water always 212?If the temperature is much above 212°F, the water will boil. That means that it won't just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. If the water has very few dust flecks etc.
Learn more about a boiling point here https://brainly.com/question/40140
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Unpolarized light is incident upon two polarization filters in sequence. The two filters transmission axes are not aligned. If 18% of the incident light passes through this combination of filters, what is the angle between the transmission axes of the filters
Answer:
53°
Explanation:
I/Io*2= 0.18
0.18= cos²theta
Cos^-1(0.36) = 53°
A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78
Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8
Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,
Then,
Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]
Real depth = Index of refraction x apparent depth
Since the same container is used, we can make an assumption that;
real depth of fluid 1 = real depth of fluid 2
That is,
1.37 x 9 = n x 6.86
Where n = Index of refraction for the second fluid.
make n the subject of formula
n = 12.33 / 6.86
n = 1.79
Therefore, the index of refraction of the second fluid is 1.8 approximately.
Learn more about refraction here: https://brainly.com/question/10729741
The medical profession divides the ultraviolet region of the electromagnetic spectrum into three bands: UVA (320-420 nm), UVB (290-320 nm), and UVC (100-290 nm). UVA and UVB promote skin cancer and premature skin aging; UVB also causes sunburn, but helpfully fosters production of vitamin D. Ozone in Earth's atmosphere blocks most of the more dangerous UVC. Find the frequency range associated with UVB radiation.
Answer:
υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz
Explanation:
The frequency of an electromagnetic radiation can be given by the following formula:
υ = c/λ
where,
υ = frequency of electromagnetic wave = ?
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of electromagnetic wave = 290 nm to 320 nm
FOR LOWER LIMIT OF FREQUENCY:
λ = 320 nm = 3.2 x 10⁻⁷ m
Therefore,
υ = (3 x 10⁸ m/s)/(3.2 x 10⁻⁷ m)
υ = 9.375 x 10¹⁴ Hz
FOR UPPER LIMIT OF FREQUENCY:
λ = 290 nm = 3.2 x 10⁻⁷ m
Therefore,
υ = (3 x 10⁸ m/s)/(2.9 x 10⁻⁷ m)
υ = 10.34 x 10¹⁴ Hz
Therefore, the frequency range for UVB radiations is:
υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz
two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will have the electric field zero?
Answer:
On that line segment between the two charges, at approximately [tex]0.7\; \rm m[/tex] away from the smaller charge (the one with a magnitude of [tex]5 \times 10^{-19}\; \rm C[/tex],) and approximately [tex]1.3\; \rm m[/tex] from the larger charge (the one with a magnitude of [tex]20 \times 10^{-19}\; \rm C[/tex].)
Explanation:
Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.
Let [tex]k[/tex] denote the Coulomb constant, and let [tex]q[/tex] denote the size of a point charge. At a distance of [tex]r[/tex] away from the charge, the electric field due to this point charge will be:
[tex]\displaystyle E = \frac{k\, q}{r^2}[/tex].
At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.
Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.
When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.
On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.
Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.
Let [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] and [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex]. Assume that the electric field is zero at [tex]r[/tex] meters to the right of the [tex]5\times 10^{-19}\; \rm C[/tex] point charge. That would be [tex](2 - r)[/tex] meters to the left of the [tex]20 \times 10^{-19}\; \rm C[/tex] point charge. (Since this point should be between the two point charges, [tex]0 < r < 2[/tex].)
The electric field due to [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] would have a magnitude of:
[tex]\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}[/tex].
The electric field due to [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex] would have a magnitude of:
[tex]\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}[/tex].
Note that at all point in this section, the two electric fields [tex]E_1[/tex] and [tex]E_2[/tex] will be acting in opposite directions. At the point where the two electric fields balance each other precisely, [tex]| E_1 | = | E_2 |[/tex]. That's where the actual electric field is zero.
[tex]| E_1 | = | E_2 |[/tex] means that [tex]\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}[/tex].
Simplify this expression and solve for [tex]r[/tex]:
[tex]\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0[/tex].
[tex]\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0[/tex].
Either [tex]r = -2[/tex] or [tex]\displaystyle r = \frac{2}{3}\approx 0.67[/tex] will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, [tex]0 < r < 2[/tex]. Therefore, [tex](-2)[/tex] isn't a valid value for [tex]r[/tex] in this context.
As a result, the electric field is zero at the point approximately [tex]0.67\; \rm m[/tex] away the [tex]5\times 10^{-19}\; \rm C[/tex] charge, and approximately [tex]2 - 0.67 \approx 1.3\; \rm m[/tex] away from the [tex]20 \times 10^{-19}\; \rm C[/tex] charge.
An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you measure its length to be 12.40cm .
If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=
Answer
0.2067m or 0.2067m
Explanation;
Let lenght of spring= Lo= 11cm=0.110m
It is hang from a mass of
3.05-kg having a length of L1= 12.40cm= 0.124m
Force required to stretch the spring= Fkx
But weight of mass mg= kx then K= Mg/x
K= 3.05-kg× 9.8)/(0.124m-.110m)
K=2135N
But potential Energy U= 0.5Kx
X=√ 2U/k
√(2*10)/2135
X=0.0967m
The required new length= L2= L0 ±x
=
.110m ± 0.0967m
X= 0.2067m or 0.2067m hence the total lenghth
A millionairess was told in 1992 that she had exactly 15 years to live. However, if she immediately takes off, travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
The expected year is 2017.
Explanation:
Total years that the millionaire to live = 15 years
Travel away from the earth at = 0.8 c
This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:
[tex]T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years[/tex]
Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017
A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficient of kinetic friction between the rod and rails is 0.18 and the kinetic friction force is 0.212 N , what vertical magnetic field is required to keep the rod moving at a constant speed of 5.1 m/s
Answer:
The magnetic field is [tex]B = 8.20 *10^{-3} \ T[/tex]
Explanation:
From the question we are told that
The mass of the metal rod is [tex]m = 0.12 \ kg[/tex]
The current on the rod is [tex]I = 4.1 \ A[/tex]
The distance of separation(equivalent to length of the rod ) is [tex]L = 6.3 \ m[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.18[/tex]
The kinetic frictional force is [tex]F_k = 0.212 \ N[/tex]
The constant speed is [tex]v = 5.1 \ m/s[/tex]
Generally the magnetic force on the rod is mathematically represented as
[tex]F = B * I * L[/tex]
For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so
[tex]F_ k = B* I * L[/tex]
=> [tex]B = \frac{F_k}{L * I }[/tex]
=> [tex]B = \frac{0.212}{ 6.3 * 4.1 }[/tex]
=> [tex]B = 8.20 *10^{-3} \ T[/tex]
Two copper wires have the same volume, but wire 2 is 10% longer than wire 1. The ratio of the resistances of the two wires R2/R1 is Group of answer choices 1.2. 1.1. 0.82. 0.91. 1.0.
Answer:
Explanation:
volume is same
π r₁² L₁ =π r₂²L₂
L₁ / L₂ = r₂² / r₁²
For resistance the formula is
R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area
R₁ = ρ L₁ / S₁
R₂ = ρ L₂ / S₂
Dividing
R₁ / R₂ = L₁ / L₂ x S₂ / S₁
= L₁ / L₂ x r₂² / r²₁
= L₁ / L₂ x L₁ / L₂
= L₁²/ L₂²
L₂ = 1.1 L₁ ( GIVEN )
= L₁²/ (1.1L₁)²
1 / 1.21
R₂ / R₁ = 1.21 .
= 1.2
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?
Answer:
t = 2.95 min
Explanation:
Given that,
The diameter of flywheeel, d = 1.5 m
Mass of flywheel, m = 250 kg
Initial angular velocity is 0
Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]
We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.
Firstly, we will find the angular acceleration of the flywheel.
The moment of inertia of the flywheel,
[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]
Now,
Let the torque is 50 N-m. So,
[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]
So,
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]
or
t = 2.95 min
Wiley Coyote has missed the elusive road runner once again. This time, he leaves the edge of the cliff at 52.4 m/s with a purely horizontal initial trajectory. If the canyon is 141 m deep, how far from the base of the cliff does the coyote land
Answer:
280.86 mExplanation:
Range is defined as the distance covered in the horizontal direction. In projectile, range is expressed as x = vt where;
x is the range
v is the velocity of the runner
t is the time taken
Before we can get the range though, we need to find the time taken t using the relationship S = ut + 1/2gt²
if u = 0
S = 1/2gt²
2S = gt²
t² = 2S/g
t = √2S/g
t = √2(141)/9.8
t = √282/9.8
t = 5.36secs
The range x = 52.4*5.36
x =280.86 m
Hence, the coyort lies approximately 280.86 m from the base of the cliff
Two 2.0 g plastic buttons each with + 65 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on either side of a 5.0 g button charged to +250 nC. All three are released simultaneously.
a. How many interactions are there that have a potential energy?
b. What is the final speed of the left 2.0 g plastic button?
c. What is the final speed of the right 2.0 g plastic button?
d. What is the final speed of the 5.0 g plastic button?
Answer:
A) 3 interactions
B) 2.78 m/s
C) 2.78 m/s
D) 0 m/s
Explanation:
Given Data :
plastic buttons ( 2 ) = 2.0 g each
charge on each plastic buttons = +65 nC
A) The interactions that have potential energy can be expressed as
V net = v1 + v2 + v3
V net = 0 + [tex]\frac{kq1q2}{r12} + \frac{kq1q3}{r13} + \frac{kq2q3}{r23}[/tex]
hence the total interactions is (3)
B) The final speed of the left 2.0 g plastic button
U = [tex]\frac{9*10^9* 10^-18}{0.02} [ 65*250 + 65*250 + (65*65)/2 ][/tex]
U = 0.0155 J
also U = [tex]2( 1/2 mv^2 )[/tex]
U = mv^2 = 0.0155 J
therefore ; v = [tex]\sqrt{0.0155/(2*10^-3)}[/tex] = 2.78 m/s
C) final speed of the right 2.0 g plastic button = Final speed of the right 2.0 g plastic button
v = 2.78 m/s
D) The final speed of the 5.0 g plastic button
= 0 m/s
A cylindrical rod 120 mm long and having a diameter of 15 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2x10-2 mm. Of the materials listed below, which are possible candidates? Justify your choice(s).
Material Modulus of Elasticity (GPa) Yield Strength (MPa) Poisson's ratio
Titanium alloy 70 250 0.33
Steel alloy 105 850 0.36
Magnesium alloy 205 550 0.27
Aluminium alloy 45 170 0.20
Answer:
The only suitable candidate is steel.
Explanation:
We are given;
Initial length;L_o = 120 mm = 0.12 m
Diameter;d_o = 15 mm = 0.015 m
Radius;r = 0.015/2 m = 0.0075 m
Applied force;F = 35000 N
Allowable Diameter reduction;Δd = 1.2 x 10^(-2) mm = 0.012 mm
Formula for Poisson’s ratio is;
v = ε_x/ε_z
Where;
ε_x is lateral strain = = Δd/d_o
ε_z is axial strain = σ/E
Thus;
v = (Δd/d_o)/(σ/E)
Making Δd the subject, we have;
Δd = (vσd_o)/E
Where σ is force per unit area
σ = F/A = 35000/πr² = 35000/(π0.0075)² = 198 × 10^(6) Pa
Now, let's find Δd for each of the metals given;
For Titanium alloy, E = 70 GPa = 70 × 10^(9) Pa and Poisson’s ratio (v) = 0.33
Thus;
Δd_ti = (0.33 × 198 × 10^(6) × 0.015)/70 × 10^(9) = 0.000014m = 0.014 mm
Similarly, for steel;
Δd_st = (0.36 × 198 × 10^(6) × 0.015)/(105 × 10^(9)) = 0.0000102 m = 0.0102 mm
For magnesium;
Δd_mn = (0.27 × 198 × 10^(6) × 0.015)/(205 × 10^(9)) = 0.0000039 m = 0.0039 mm
For aluminum;
Δd_al = (0.2 × 198 × 10^(6) × 0.015)/(45 × 10^(9)) = 0.0000132 m = 0.0132 mm
Since they must not experience a diameter reduction of more than 0.012 mm, then the only alloy that has a reduction diameter less than 0.012 is steel which has 0.0102.
Thus,the only possible candidate is stee.
Tom is climbing a 3.0-m-long ladder that leans against a vertical wall, contacting the wall 2.5 m above the ground. His weight of 680 N is a vector pointing vertically downward. (Weight is measured in newtons, abbreviated N).
A) What is the magnitude of the component of Tom's weight parallel to the ladder?
B) What is the magnitude of the component of Tom's weight perpendicular to the ladder?
Answer: A) [tex]P_{x}[/tex] = 564.4 N
B) [tex]P_{y}[/tex] = 374 N
Explanation: The ladder forms with the wall a right triangle, with one unknown side. To find it, use Pythagorean Theorem:
[tex]hypotenuse^{2} = side^{2} + side^{2}[/tex]
[tex]side = \sqrt{hypotenuse^{2} - side^{2}}[/tex]
side = [tex]\sqrt{3^{2} - 2.5^{2}}[/tex]
side = 1.65
Tom's weight is a vector pointing downwards. Since he is at an angle to the floor, the gravitational force has two components: one that is parallel to the floor ([tex]P_{x}[/tex]) and othe that is perpendicular ([tex]P_{y}[/tex]). These two vectors and weight, which is gravitational force, forms a right triangle with the same angle the ladder creates with the floor.
The image in the attachment illustrates the described above.
A) [tex]P_{x}[/tex] = P sen θ
[tex]P_{x} = P.\frac{oppositeside}{hypotenuse}[/tex]
[tex]P_{x}[/tex] = 680.[tex]\frac{1.65}{3}[/tex]
[tex]P_{x}[/tex] = 564.4 N
B) [tex]P_{y}[/tex] = P cos θ
[tex]P_{y} = P.\frac{adjacentside}{hypotenuse}[/tex]
[tex]P_{y}[/tex] = 680. [tex]\frac{1.65}{3}[/tex]
[tex]P_{y}[/tex] = 374 N
The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to:__________.
a) 4 Q.
b) 2 Q.
c) Q.
d) Q/2.
e) Q/4.
Answer:
D. Q/2
Explanation:
See attached file
How do you stay hydrated during warm-up and scheduled activity?
Answer:
In order to stay hydrated during warm-up(s) drink 8oz of water 20-30 mintues before you start exercising or during your warm-up(s), make sure you drink 7 to 10 oz of water every 10 to 20 minutes during exercise, and drink 8oz of water no more than 30 minutes after you exercise.
In order to stay hydrated during scheduled activity(s) drink 17 to 20 oz of water 2 to 3 hours before you start to exercise, like said before drink 8 oz of water 20 to 30 minutes before you start exercising or during your warm-up(s), drink 7 to 10 oz of water every 10 to 20 minutes during exercise, also said before drink 8 ounces of water no more than 30 minutes after you exercise.
Answer: My scheduled activity was one hour of softball practice. I play catcher, so my thighs and knees take a lot of abuse from kneeling and standing. The lunges were excellent at preparing my thighs for softball. The high knees exercise and arm pumping didn’t feed into softball too well. I suppose that they might help me with base running.
Explanation: EDMENTUM
A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.60 s. Find the force constant of the spring.
Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.
N/m
Explanation:
In the figure, suppose the length L of the uniform bar is 3.2 m and its weight is 220 N. Also, let the block's weight W = 270 N and the angle θ = 45˚. The wire can withstand a maximum tension of 450 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?
Answer:
a) x = 2.46 m
b) 318.2 N
c) 177.8 N
Explanation:
Need to resolve the tension of the string at end say B.
The vertical upward force at B due to tension is 450 sin 45°.
Using Principle of Moments, with the pivot at A,
Anti clockwise moments = Clockwise moments
450 sin 45° X 3.2 = 220 X (3.2/2) + (270 X x)
x = 2.46 m
(b) The horizontal force is only due to the wire's tension, so it is
450 cos 45° = 318.2 N
(c) total downward forces = 270 + 220 = 496 N
Total upward forces = 450 sin 45° (at B) + upForce (at A)
Equating, upForce = 496 - 318.2
= 177.8 N
What magnification in multiples is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being viewed?
Answer:
51.6Explanation:
A microscope is made up of a convex lens and the nature of the image formed by the object viewed from it is a virtual image. Since the image is virtual, the image distance will be negative and the focal length will be positive (for convex lenses).
Using the lens formula to first calculate the image distance from the lens;
1/f = 1/u + 1/v
f is the focal length = 0.150 cm,
u is the object distance = 0.155 cm
v is the image distance.
Since the image distance is negative,
1/f = 1/u - 1/v
1/0.150 = 1/0.155 - 1/v
1/v = 1/0.155 - 1/0.15
1/v = 6.542 - 6.667
1/v = -0.125
v = 1/-0.125
v = -8 cm
Magnification = image distance/object distance
Mag = 8/0.155
Mag = 51.6
Magnification produces is 51.6
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle of 10 o and released. If the spring has a torsion constant of 370 N-m/rad, what is the frequency of the motion
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin(kx−ωt)j^,where j^ is the unit vector in the y direction.
What is the Poynting vector S⃗ (x,t), that is, the power per unit area associated with the electromagnetic wave described in the problem introduction?
Given that,
The electric field is given by,
[tex]\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}[/tex]
Suppose, B is the amplitude of magnetic field vector.
We need to find the complete expression for the magnetic field vector of the wave
Using formula of magnetic field
Direction of [tex](\vec{E}\times\vec{B})[/tex] vector is the direction of propagation of the wave .
Direction of magnetic field = [tex]\hat{j}[/tex]
[tex]B=B_{0}\sin(kx-\omega t)\hat{k}[/tex]
We need to calculate the poynting vector
Using formula of poynting
[tex]\vec{S}=\dfrac{E\times B}{\mu_{0}}[/tex]
Put the value into the formula
[tex]\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}[/tex]
[tex]\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]
Hence, The poynting vector is [tex]\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]
Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the following function.
h=1/2t2+1/2t
(a) What is the height of the balloon at the end of 40 sec?
(b) What is the average velocity of the balloon between t = 0 and t = 30?
ft/sec
(c) What is the velocity of the balloon at the end of 30 sec?
ft/sec
Answer:
Explanation:
Given the height reached by a balloon after t sec modeled by the equation
h=1/2t²+1/2t
a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t
If h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
b) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec
c) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec
A) The height of the balloon at the end of 40 sec is 820 feet.
B) The average velocity of the balloon is 30 ft/sec.
C) The velocity of the balloon at the end of 30 sec is
VelocityGiven :
h=1/2t²+1/2tPart A)
The height of the balloon after 40 secs is :
h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
Part B)
The average velocity of the balloon is :
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0 sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
When at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon = 30 ft/sec
The average velocity of the balloon is 30 ft/sec.
Part C)
The velocity of the balloon at the end of 30 sec is :
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec.
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The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?
Answer:
Torque = 8.38Nm
Explanation:
Time= 8.00s
angular speed (w) =400 rpm
Moment of inertia (I)= 1.60kg.m2 about its rotation axis
We need to convert the angular speed from rpm to rad/ sec for consistency
2PI/60*n = 0.1047*409 = 41.8876 rad/sec
What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?
Then we need to use the formula below for our torque calculation
from basic equation T = J*dω/dt ...we get
Where : t= time in seconds
W= angular velocity
T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm
Therefore, constant torque that is required is 8.38 Nm
Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.
Given-
Inertia of the flywheel is 1.60 kg m squared.
Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,
[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]
[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]
[tex]\omega = 41.89[/tex]
Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.
When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,
[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]
[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]
[tex]\tau=8.38[/tex]
Hence, the required constant torque is 8.38 N-m.
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A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb
Answer:
931.00ft-lb
Explanation:
Pls see attached file
The work done in moving the object from x = 1 ft to x = 18 ft is 935 ft-lb.
What is work?
Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.
Given that force = 6x - 2 pounds.
So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]
= [ 3x² - 2x]¹⁸₁
= 3(18² - 1² ) - 2(18-1) ft-lb
= 935 ft-lb.
Hence, the work done is 935 ft-lb.
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An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plates. All the geometric parameters of the capacitor (plate diameter andplate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, howmuch energy does it now store?
Answer:
U_f = (U_o)/2)
Explanation:
The capacitance of a given capacitor is given by the formula;
C = ε_o•A/d
While energy stored in plates capacitor is given as; U_o = Q²/2C
Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;
C' = ε_o•4A/2d
Hence, C' = 2C
so capacitance is now doubled
Thus, the final energy stored between the plates of capacitor is given as;
U_f = Q²/2C'
From earlier, we saw that C' = 2C.
Thus;
U_f = Q²/2(2C)
U_f = Q²/4C
Rearranging, we have;
U_f = (1/2)(Q²/2C)
From earlier, U_o = Q²/2C
Hence,
U_f = (1/2)(U_o)
Or
U_f = (U_o/2)
Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)
Answer:
Power=50.17dioptre
Power=50.17D
Explanation:
P=1/f = 1/d₀ + 1/d₁
Where d₀ = the eye's lens and the object distance= 5.70m=
d₁= the eye's lens and the image distance= 0.02m
f= focal length of the lense of the eye
We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m
Therefore, we can calculate the power using above formula
P= 1/5.70 + 1/0.02
Power=50.17dioptre
Therefore, the power the eye's is using to see the object from distance is 5.70D
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time
Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]
The induced emf in the shorter coil is calculated as;
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]
Induced EMF:The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:
[tex]E=-\frac{\delta\phi}{\delta t}[/tex]
where E is the induced emf,
[tex]\phi[/tex] is the magnetic flux through the coil.
The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:
[tex]\phi = \frac{\mu_oNIA}{L}[/tex]
so;
[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]
where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.
[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]
The emf produced in the coil at the center of the solenoid which has 14 turns will be:
[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]
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