Answers
Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
Related Questions
The source of sound moves away from the listener.The listener has the impression that the source is lower in pitch. Why?
Answers
When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
Mention the importance of occupation??
Answers
Answer:
ln the contemporary time , farming can be considered as comparitively important occupation as it can feed the population , So agriculture is having a greater importance than any other occupation.
Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with her? Why or why not?
Answers
Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2
Answers
The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶ W/m².
The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.
a) Intensity of the electromagnetic wave for the electromagnetic field.
The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:
[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex] (1)
Where:
c: is the speed of light = 3.00*10⁸ m/s
E: is the magnitude of the electric field = 1.5 V/m
ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²
Hence, the intensity of the electromagnetic wave (eq 1) is:
[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]
b) Intensity of the electromagnetic wave for the magnetic field
We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:
[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex] (2)
Where:
B: is the magnitude of the magnetic field = 10⁻¹⁰ T
μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A
Therefore, the intensity of the electromagnetic wave (eq 2) is:
[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]
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These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor
Answers
Answer:
Following are the solution to the given question:
Explanation:
For charging plates that are connected in a similar manner:
Calculating the total charge:
[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]
Calculating the common potential:
[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]
Calculating the charge after redistribution:
[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]
[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Answers
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answers
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answers
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answers
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
Answers
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
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cyclist always bends when moving the direction opposite to the wind. Give reasons
Answers
An object undergoing simple harmonic motion takes 0.40 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.
Answers
Answer:
a) [tex]P=0.80[/tex]
b) [tex]1.25Hz[/tex]
c) [tex]A=25cm[/tex]
Explanation:
From the question we are told that:
Travel Time [tex]T=0.40s[/tex]
Distance [tex]d=50cm[/tex]
a)
Period
Time taken to complete one oscillation
Therefore
[tex]P=2*T\\\\P=2*0.40[/tex]
[tex]P=0.80[/tex]
b)
Frequency is
[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]
[tex]1.25Hz[/tex]
c)
Amplitude:the distance between the mean and extreme position
[tex]A=\frac{50}{2}[/tex]
[tex]A=25cm[/tex]
What is an internal resistance?
Answers
Explanation:
some thing inside a resistor
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answers
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.
Answers
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
[tex]\phi = BAcos \theta[/tex]
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]
Hence the magnetic flux Φ through the loop is 0.5849Weber
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
Answers
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced
Answers
Answer:
The total surface area is "90 km²".
Explanation:
Given:
Power from solar installations,
= 27 GW
Other natural installations,
= 740 GW
Intensity,
[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]
%n,
= 30%
Now,
⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]
then,
⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]
[tex]=90 \ GW[/tex]
As we know,
⇒ [tex]I=\frac{P}{A}[/tex]
by substituting the values, we get
[tex]1000=\frac{90\times 10^9}{A}[/tex]
[tex]A = \frac{90\times 10^9}{10^3}[/tex]
[tex]=90\times 10^6[/tex]
[tex]=90 \ km^2[/tex]
define nortons theorem
Answers
Answer:
In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.
The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answers
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
Answers
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.
Answers
Answer:
Florida is tropical, with a significant rainy seson
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
Answers
Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3
Answers
Answer:
d = 30kg/cm³
Explanation:
d = m/v
d = 1080kg/(3cm*4cm*3cm)
d = 30kg/cm³
Identify the correct descriptions of alpha particles. Select one or more: Alpha particles are more massive than beta particles. An alpha particle is a helium nucleus. An alpha particle has a negative charge. An alpha particle is a form of electromagnetic radiatio
Answers
Answer:
Alpha particles are more massive than beta particles.
Explanation:
The alpha particles are also called double-positive Heilum Nuclei because they have a charge of "+2" and a mass of 4 a.m.u. The properties of the alpha particles are as follows:
1. It possesses high energy due to high velocity. It is 7.7 MeV for most energetic from Rac (i.e: Bismuth-214)
2. It has a very high ionizing power. A 7.7 MeV particle produces about 0.2 x 10⁶ ions.
3. The range of alpha particles is very small. It is about 7 x 10⁻² m and only 4 x 10⁻⁵ m in aluminum for 7.7 MeV alpha-particle.
4. Alpha particles produce fluorescence on striking certain substances, such as zinc sulphide and bariumplatinocynide.
The beta particles are fast-moving electrons, which have a negligible mass.
Hence, the correct option is:
Alpha particles are more massive than beta particles.
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)
Answers
Answer:
[tex]E=35921.96N/C[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.321mm[/tex]
Charge Density [tex]\mu=0.100[/tex]
Distance [tex]d= 5.00 cm[/tex]
Generally the equation for electric field is mathematically given by
[tex]E=\frac{mu}{2\pi E_0r}[/tex]
[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]
[tex]E=35921.96N/C[/tex]
PLEASE ANSWER IF YOU CAN AND NOT FOR THE SAKE OF GAINING POINTS!
Answers
The cells lie odjacent to the sieve tubes
Answers
Answer:
Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.
Answers
Answer:
PlROCA
Explanation:
Two long, straight wires are separated by 0.120 m. The wires carry currents of 11 A in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field.
Answers
Answer:
The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.
Explanation:
The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.
In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.
The magnetic field (B) between the given wires can be determined by:
B = [tex]\frac{U_{o}I }{2\pi r}[/tex]
where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.
But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A
So that;
B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]
= 1.8333 x [tex]10^{-5}[/tex]
B = 1.83 x [tex]10^{-5}[/tex] T
a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm
Answers
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
