Answer:
Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional. Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. the maximum extent to which a solid may be stretched without permanent alteration of size or shape, is called elastic limit
mark me brainliestt :))
There are three 20.0 Ω resistors connected in series across a 120 V generator
Answer:
That is equal to R1 + R2. If three or more unequal (or equal) resistors are connected in series then the equivalent resistance is: R1 + R2 + R3 +…, etc. One important point to remember about resistors in series networks to check that your maths is correct.
What effect does the Duck Velocity have on the waves seen by the observer?
Towards the boat:
Away from the boat:
Same as the boat:
An ideal massless spring with a spring constant of 2.00 N/m is attached to an object of 75.0 g. The system has a small amount of damping. If the amplitude of the oscillations decreases from 10.0 mm to 5.00 mm in 15.0 s, what is the magnitude of the damping constant b
Answer: 0.00693
Explanation:
Given
Spring constant [tex]k=2\ N/m[/tex]
Mass of object [tex]m=75\ g[/tex]
The amplitude of the oscillation decreases from 10 mm to 5 mm in 15 s
Equation of amplitude for the ideal spring-mass system is
[tex]\Rightarrow A=A_oe^{-\frac{bt}{2m}}\quad \quad [\text{b=damping constant}]\\\text{Insert the values}\\\\\Rightarrow 5=10e^{\frac{b\times 15}{2\times 0.075}}\\\\\Rightarrow e^{-\frac{b\times 15}{2\times 0.075}}=0.5\\\\\text{Taking natural log both sides}\\\\\Rightarrow \ln \left(e^{-\frac{b\times 15}{2\times 0.075}}\right)=\ln 0.5\\\\\Rightarrow -\dfrac{15b}{0.15}=-0.693\\\\\Rightarrow b=0.00693[/tex]
A 1460 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 18.0 m/s. They stick together. In what direction and with what speed do they move after the collision?
Answer in degrees north of east
AND
the speed after the collision in m/s
Answer:
Solution given:
North car
mass[m1]=1460kg
velocity[u1]=27 m/s
mass[m2]=2165kg
velocity [u2]=18m/s
let v be velocity after collision
we have
From the principle of conservation of linear momentum
m1u1+m2u2=(m1+m2)v
1460*27+2165*18=(1460+2165)v
v=[tex] \frac{78390}{3625} [/tex]
v=21.6m/s
the speed after the collision in 21.6 m/s.
For angle.
Tan angle =[tex] \frac{m1u1}{m2u2} [/tex]
Tan angle =[tex] \frac{1460*27}{2165*18} [/tex]
Tan angle=327.74
angle=Tan-¹(327.74)=89.82=90°
in degrees north of east is 90°
An old fashioned string of 80 Christmas lights is wired in series. Each bulb has a resistance of 2 Ohms and the entire string is plugged into a 120V outlet. What is the current passing through each of the bulbs?
The sum of the resistance = 2 ohms x 80 lights = 160 ohms.
Current = Total voltage / total resistance:
Current = 120V / 160 ohms
Current = 0.75 Amps
A closed loop conductor that forms a circle with a radius of 2.0 m is located in a uniform but changing magnetic field. If the maximum emf Induced in the loop is 5.0 V what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Answer:
The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.
Explanation:
Given;
radius of the circular loop, r = 2.0 m
maximum induced emf, E = 5.0 V
The emf induced in a magnetic field is given as;
[tex]emf = \frac{d\phi}{dt} \\\\\phi = AB\\\\emf = A\frac{dB}{dt} \\\\\frac{dB}{dt} = \frac{emf}{A} \\\\where;\\A \ is \ the \ area \ circular \ l00p = \pi r^2 = \pi (2)^2 = 4\pi \ m^2\\\\\frac{dB}{dt} = \frac{5}{4\pi} \\\\\frac{dB}{dt} = 0.398 \ T/s[/tex]
Therefore, the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.
Which of the following is form of energy:
a) Power
b) Light
C) pressure
d) None
Answer:
Explanation:
b) light
4- What force must be applied to a surface area of 0.0025m , to create a pressure ol
200.000Pa?
The atmospheric features of Neptune are easier to see than those of Uranus because A. Neptune has greater warmth and less haze. B. Neptune has more methane. C. The atmosphere of Uranus rotates differentially. D. Uranus has no significant atmosphere.
Answer:
Option B is the correct answer (Neptune has more methane)Explanation:
From the options given,
The atmospheric features of Neptune are easier to see than those of Uranus because Neptune has more methane
Neptune has small amount of methane and water which gives it blue colour and white patches which distinguish it from uranus
For more information, visit
http://abyss.uoregon.edu/~js/ast121/lectures/lec20.html
paano matutugunan o matutulungan ng pamahalaan at ng mga guro yubg mga estudyanteng nakararanas nag stress at anxiety.
Answer: how the government and teachers can address or help students experiencing stress and anxiety.
Explanation:
What is the speed of a wave if it has a wavelength of
42 m and a frequency of 7 hertz?
Answer:
♕ [tex]\large{ \red{ \tt{Step - By - Step \: Explanation}}}[/tex]
☃ [tex] \underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} : [/tex]
Frequency ( f ) = 7 HertzWavelength ( λ ) = 42m♨ [tex] \underline {\underline{ \orange{ \large{ \tt{T \: O \: \: F \: I \: N\: D}}}} }: [/tex]
Wave velocity ( v )☄ [tex]\underline{ \underline{ \large{ \pink{ \tt{S\: O \: L \: U \: T\: I \: O \: N}}}}}: [/tex]
✧ [tex] \red{ \boxed{ \large{ \purple{ \sf{Wave \: velocity(v) = Frequency(f) \times Wavelength(λ)}}}}}[/tex]
~Plug the known values and then multiply!
↦ [tex] \large{ \tt{7 \times 42}}[/tex]
↦ [tex] \boxed{ \boxed{ \large{ \bold{ \tt{294 \: m {s}^ {- 1} }}}}}[/tex]
☥ [tex] \large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}} [/tex]
---------------------------------------------------------------
❁ [tex] \underline{ \large{ \red{ \tt{D\: E\: T \: A \: I \: L\: E \: D \: \: I\: N \: F \: O}}}} : [/tex]
Frequency ( f ) : The number of complete waves , set up in a medium in one second is called frequency of the wave. The SI unit of frequency is Hertz ( Hz ). For example : if a sound wave completes 15 compressions and 15 rarefactions in one second , it's frequency is 15 Hz.Wavelength ( λ ) : The distance between two consecutive troughs or crests in a transverse wave or the distance between two consecutive compressions or rarefactions in a longitudinal wave us called wavelength. It is the distance travelled by a wave in a time equal to it's time period. It's SI unit is metre ( m ).Wave velocity ( v ) : The velocity with which a wave propagates in a medium is called wave velocity. It's SI unit is m/s.# KILL : Excuses
KISS : Opportunities
MARRY : Goals
♪ Hope I helped! ♡
☂ Have a wonderful day / night ! ツ
✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✔
▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁
An irrigation canal has a rectangular cross section. At one point where the canal is 18.2 m wide and the water is 3.55 m deep, the water flows at 2.55 cm/s . At a second point downstream, but on the same level, the canal is 16.3 m wide, but the water flows at 11.6 cm/s . How deep is the water at this point
Answer:
Explanation:
Rate of volume flow at two points will be same at two points .
A₁ V₁ = A₂V₂
A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .
A₁ = 18.2 x 3.55 = 64.61 m²
V₁ = 2 .55 x 10⁻² m/s
A₂ = 16.3 x d = 16.3 d m²
d is depth at second point .
V₂ = 11.6 x 10⁻² m/s
64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d m² x 11.6 x 10⁻² m/s
d = .87 m
so canal is .87 m deep.
Can someone help me
An athlete is performing squats in the weight room. The knee is going from anatomical position to 92 degrees and then back to anatomical position each squat. The athlete performs a total of 10 squats. This is done over a time period of 30 seconds. What is the angular acceleration (rad/sec 2) of the knee
Answer:
α = 0.357 ras / s²
Explanation:
This is a rotational kinematics exercise, it tells us that it performs 10 squats in 30 s, for which it performs one squat at t = 3 s, also indicates that the angle of the squat is θ = 92º
θ = θ₀ + w₀ t + ½ α t²
the athlete starts from rest, whereby w₀ = 0 and the initial angle in the vertical position is zero (θ₀=0)
θ = ½ α t²
α = 2 θ /t²
let's reduce the magnitudes to the SI system
θ = 92º (π rad /180º) = 0.511π rad
let's calculate
α = 2 0.5111π /3²
α = 0.1136π rad / s²
α = 0.357 ras / s²
Why is it important for equipment for sport to be strong? To protect us
Answer:
To protect us.
Explanation:
For ex. your dunking on a basketball hoop if that wasn't strong you would fall on your back and get injured.
Help me please I don’t understand
Answer:
a. hydroelectric power plant
5. Stopping a fast-moving object is harder than stopping a slow-moving
one.
True
False
Look at the Position vs. Time and Velocity vs. Time plots. What is the person's velocity when his position is at its maximum value (around 6 m )
Answer:
The person's velocity is zero.
Explanation:
Are surface currents warm or cold?
A:war m
B:cold
Answer:
Cold
Explanation:
Im pretty sure im sorry if I am wrong
A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What is the longest wavelength that forms an intensity maximum in the fifth order
Answer:
377 nm
Explanation:
Number of lines per meter is, [tex]N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}[/tex]
Grating element is, [tex]d=\frac{1}{N}[/tex]
[tex]=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda[/tex]
[tex]\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\ &=0.377 \times 10^{-6} \mathrm{~m} \\ &=377 \mathrm{~nm}[/tex]
An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose the electron is released from rest a distance 0.050 m from the ring center. It then oscillates through the ring center. Calculate its period. (The electron is always much closer to the ring center than a radius.)
Answer:
T = 1.12 10⁻⁷ s
Explanation:
This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.
All the charge dq is at a distance r
dE = k dq / r²
Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry
cos θ =[tex]\frac{dE_x}{dE}[/tex]
dEₓ = dE cos θ
cos θ = x / r
substituting
dEₓ = [tex]k \frac{dq}{r^2 } \ \frac{x}{r}[/tex]
DEₓ = k dq x / r³
let's use the Pythagorean theorem to find the distance r
r² = x² + a²
where a is the radius of the ring
we substitute
dEₓ = [tex]k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq[/tex]
we integrate
∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} } ∫ dq
Eₓ = [tex]k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}[/tex]
In the exercise indicate that the electron is very central to the center of the ring
x << a
Eₓ = [tex]k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}[/tex]
if we expand in a series
[tex](\ 1+ (x/a)^2 \ )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2[/tex]
we keep the first term if x<<a
Eₓ = [tex]\frac{ k Q}{a^3} \ x[/tex]
the force is
F = q E
F = [tex]- \frac{kQ }{a^3} \ x[/tex]
this is a restoring force proportional to the displacement so the movement is simple harmonic,
F = m a
[tex]- \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }[/tex]
[tex]\frac{d^2 x}{dt^2} = \frac{keQ}{ma^3} \ x[/tex]
the solution is of type
x = A cos (wt + Ф)
with angular velocity
w² = [tex]\frac{keQ}{m a^3}[/tex]
angular velocity and period are related
w = 2π/ T
we substitute
4π² / T² = \frac{keQ}{m a^3}
T = 2π [tex]\sqrt{\frac{m a^3 }{keQ} }[/tex]
let's calculate
T = 2π [tex]\sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19} \ 0.021 \ 10^{-3} } }[/tex]
T = 2π pi [tex]\sqrt{320.426 \ 10^{-18} }[/tex]
T = 2π 17.9 10⁻⁹ s
T = 1.12 10⁻⁷ s
A 5.0-m radius playground merry-go-round with a moment of inertia of 1,630 kg m2 is rotating freely with an angular speed of 1.6 rad/s. Two people, each having a mass of 69.5 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on
Answer:
The right solution is "0.511".
Explanation:
Given:
Initial moment of inertia,
= 1630 kg.m²
Radius,
= 5 m
Angular speed,
= 1.6 rad/s
Now,
The moment of inertia after stepping on will be:
= [tex]1630+2\times (69.5\times (5)^2)[/tex]
= [tex]1630+2\times (69.5\times 25)[/tex]
= [tex]5105 \ Kg.m^2[/tex]
hence,
As per the question, the angular speed is conserved, then
⇒ [tex]1630\times 1.6=5105\times \omega'[/tex]
[tex]2608=5105\times \omega'[/tex]
[tex]\omega'=\frac{2608}{5105}[/tex]
[tex]=0.511[/tex]
How are soil and air similar?
Answer:
The air in the soil is similar in composition to that in the atmosphere with the exception of oxygen, carbon dioxide, and water vapor. In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.036%.
hope this helps
have a good day :)
Explanation:
A retired bank president can easily read the fine print of the financial page when the newspaper is held no closer than arm's length, 59.1 cm from the eye. What should be the focal length of an eyeglass lens that will allow her to read at the more comfortable distance of
Answer:
Explanation:
comfortable distance is 25 cm .
He must be using convex lens . In that case rays coming from object placed at 25 cm appears to be coming from 59.1 cm due to converging nature of convex lens.
object distance u = -25 cm
image distance v = -59.1 cm
Lens formula
1 / v - 1 /u = 1 /f
-1 / 59.1 + 1 / 25 = 1/f
- .0169 + .04 = 1 / f
.0231 = 1 / f
f = 43.3 m
Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulley without slipping. The mass of block A is 2.1 kg and the mass of block B is 4.1 kg. Block A is also connected to a horizontally-mounted spring with a spring constant of 358 J/m2. What is the angular frequency (in rad/s) of oscillations of this system
Answer:
[tex]F=1.159[/tex]
Explanation:
From the question we are told that:
Mass of pulley [tex]M=1kg[/tex]
Radius [tex]r=12cm[/tex]
Mass of block A [tex]M_a=2.1kg[/tex]
Mass of block B [tex]m_b=4.1kg[/tex]
Spring constant[tex]\mu= 358 J/m2[/tex]
Generally the equation for Torque is mathematically given by
Since [tex]\sumF=ma[/tex]
At mass A
[tex]T_2-f_3=2.1a[/tex]
At mass B
[tex]4.8-T_1=4.1a[/tex]
At Pulley
[tex]R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}[/tex]
[tex]R(T_1-T_2)=0.55a[/tex]
Therefore the equation for total force F
At mass A+At mass B+At Pulley
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a[/tex]
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a[/tex]
[tex]-f_3+4.1=6.75a[/tex]
[tex]-f_3=6.75a+4.8[/tex]
Since From above equation
[tex]M_{eff}=6.7kg[/tex]
Therefore
[tex]T=2\pi \sqrt{{\frac{M_{eff}}{k}}[/tex]
[tex]T=2\pi \sqrt{{\frac{6.75}{\mu}}[/tex]
[tex]T=0.862s[/tex]
Generally the equation for frequency is mathematically given by
[tex]F=\frac{1}{T} \\F=\frac{1}{0.862}[/tex]
[tex]F=1.159[/tex]
how dose an exam question outed from text book
Answer:
In which school you are???
Explanation:
3 نقطة (نقاط)
السؤال 2
A block slides on a rough horizontal surface from paint A to point B. A force (magnitude P -
2.0 N) acts on the block between A and B, as shown. If the kinetic energies of the block at A
and B are 5.0 J and 4.0 J, respectively and the work done on the block by the force of friction
as the block moves from A to B is -4.5 J what is the distance between point A and B?
40°
B
The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is
[tex]W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J[/tex]
Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.
The only other force acting on the block as it moves is the force P. Let [tex]W_P[/tex] be the work done by the force P. Then the total work done on the block is
[tex]W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}[/tex]
A sound wave with a frequency of 700 Hz and a wavelength of 5 m travels through a liquid. How fast does sound travel through the liquid?
A.
140 m/s
B.
0.007 m/s
C.
3500 mHz
D.
3500 m/s
BRAINLIST A wave travels at a constant speed. How does the wavelength change if the
frequency is reduced by a factor of 3? Assume the speed of the wave remains
unchanged.
A. The wavelength does not change.
B. The wavelength increases by a factor of 3.
C. The wavelength decreases by a factor of 3.
D. The wavelength increases by a factor of 9.
24. A anvil with a mass of 60 kg falls from a height of 9.5 m. How fast is it going right
before it hits the ground?
V= I*R
V = voltage (measured in volts) V
I = current (measured in amperes) A
R = resistance (measured in Ohms) Ω
So they give us this
V=IR
V= 1.8
I=0.4
R=?
So we insert the thing that we know.
1.8=0.4*R
We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.
So we have this.
Lastly we solve.
R=4.5ohms
The formula to find R is V=IR
V/I=R
So the resistance will be the Voltage divided by the Current