A force of 30 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How much work is done in stretching the spring from 35 cm to 40 cm
Answer:
0.25 J
Explanation:
Given that
Force on the spring, F = 30 N
Natural length of the spring, l1 = 20 cm = 0.2 m
Final length of the spring, l2 = 35 cm = 0.35 m
Extension of the spring, x = 0.35 - 0.2 = 0.15 m
The other extension is 40 - 35 cm = 5 cm = 0.05 m
Work done = ?
Considering Hooke's Law of Elasticity.
We're told that the spring stretches through 15 cm, and thereafter asked to find the work done in stretching it through 5 cm
The question is solved in the attachment below
An automobile accelerates from zero to 30 m/s in 6 s. The wheels have a diameter of 0.4 m. What is the angular acceleration of each wheel
Answer:
12.5 rad/s²
Explanation:
Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²
From the question,
a = αr................... Equation 1
Where a = linear acceleration, α = angular acceleration, r = radius.
But,
a = (v-u)/t.............. Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
(v-u)/t = αr
make α the subject of the equation
α = (v-u)/tr................. Equation 3
Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m
Substitute into equation 3
α = (30-0)/(0.4×6)
α = 30/2.4
α = 12.5 rad/s²
A 240 m long segment of wire is hanging between 2 transmission lines. What is the total magnetic force only (ignore gravitational force) on this segment of wire if the current in the wire is 500 A and the field strength is 3e-5 T
Answer:
7.2 N
Explanation:
length of wire L = 240 m
current I = 500 A
field strength B = 3 x 10^-5 T
magnetic force on a current carrying conductor F is given as
F = BILsin∅
The wires are perpendicular with field therefore sin∅ = sin 90° = 1
therefore,
F = BIL = 3 x 10^-5 X 500 X 240 = 3.6 N
If the wire exists between this two transmission lines, then total magnetic force on the wire = 2 x 3.6 = 7.2 N
What is Dark Matter?
Explanation:
dark matter is a form of matter thought to account for approximately 85% of the matter in the universe and about a quarter of its total mass–energy density or about 2.241×10⁻²⁷ kg/m³. Its presence is implied in a variety of astrophysical observations, including gravitational effects that cannot be explained by accepted theories of gravity unless more matter is present than can be seen. For this reason, most experts think that dark matter is abundant in the universe and that it has had a strong influence on its structure and evolution. Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect.
A farmer with 600 ft of available fencing wants to enclose a rectangular are located adjacent to a farmhouse (with no fencing required along the farmhouse) and then divide it into four sections with fencing perpendicular to the side of the farmhouse. Show that the largest possible area of the four sections will be 18,000 square feet, and state the corresponding dimensions of the rectangular area.
Answer:
The corresponding dimensions will be "x = 300 & y = 60".
Explanation:
Available fencing = 600 ft
Fencing,
⇒ [tex]5y+x=600[/tex]
⇒ [tex]x=600-5y[/tex]
As we know,
⇒ [tex]Area =xy[/tex]
On substituting the given values, we get
[tex]=(600-5y)y[/tex]
[tex]=600y-5y^2[/tex] ...(equation 1)
On differentiating with respect to y, we get
⇒ [tex]\frac{dA}{dy} =600-10y[/tex]
[tex]0 = 600-10y[/tex]
[tex]600=10y[/tex]
[tex]y=\frac{600}{10}[/tex]
[tex]y=60[/tex]
On putting the values of y in equation 1, we get
⇒ [tex]600(60)-5(60)^2[/tex]
⇒ [tex]600(60)-5(3600)[/tex]
⇒ [tex]36000 - 18000[/tex]
⇒ [tex]18000[/tex]
Dimensions of the rectangular area:
x = 300
y = 60
An object 9.00 mm tall is placed 12.0 cm to the left of the vertex of a convex mirror whose radius of curvature has a magnitude of 20.0 cm.(a) calculate the position, size, orientation (erect or inverted), and nature (real or virtual) of the image. (b) draw a principal-ray diagram showing the formation of the image (how would i do that.. i know you can't draw it here)..
Answer:
Explanation:
a )
for convex mirror, focal length
f = + 20 / 2 = 10 cm
object distance u = 12 cm ( negative )
Using mirror formula
[tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]
Putting the values
[tex]\frac{1}{v} - \frac{1}{12} = \frac{1}{10}[/tex]
[tex]\frac{1}{v} =\frac{1}{10} + \frac{1}{12}[/tex]
[tex]v=\frac{12\times 10}{12+10}[/tex]
v = 5.45 cm
Size of image
=[tex]\frac{v}{u} \times size of object[/tex]
[tex]=\frac{5.45}{12} \times 9[/tex]
= 4.08 mm .
It will be erect because only erect image is formed in convex mirror .
It will be virtual .
b )
Draw ray from tip of object parallel to principal axis and after reflection appears to be coming from focus . Second ray falls on the pole and reflects back making equal angle with principal axis . Draw the reflected ray backwards to cut the other ray . They cut each other where image is formed .
Gravitational potential energy is greatest at the highest point of a roller coaster and least at the lowest point.
Answer:
because gravitational potential enegry is directly proportional to the height so more the height more the gravitational potential enegry. therefore gravitational potential enegry is greatest at high point than lower points.
g At 4 ft. from a point source of gamma radiation the intensity is 158 mRem/hr. How far away would you have to move to decrease the intensity to 17 mRem/hr.? [I1/I2 = (D2)2/(D1)2] I1(D1)2 = I2(D2)2
Answer:
You will have to move 12.2 ft to decrease the initial intensity to 17 mRem/h.
Explanation:
The distance can be calculated using the following equation:
[tex] \frac{I_{1}}{I_{2}} = \frac{D_{2}^{2}}{D_{1}^{2}} [/tex]
Where:
I₁ = 158 mRem/h
I₂ = 17 mRem/h
D₁ = 4 ft.
D₂=?
Hence, the distance D₂ is:
[tex] D_{2} =\sqrt{\frac{I_{1}}{I_{2}}*D_{1}^{2}} = \sqrt{\frac{158 mRem/h}{17 mRem/h}*(4 ft)^{2}} = 12.2 ft [/tex]
Therefore, you will have to move 12.2 ft to decrease the initial intensity to 17 mRem/h.
I hope it helps you!
Consider a sound wave modeled with the equation s(x, t) = 3.00 nm cos(3.50 m−1x − 1,800 s−1t). What is the maximum displacement (in nm), the wavelength (in m), the frequency (in Hz), and the speed (in m/s) of the sound wave?
Answer:
- maximum displacement = 3.00nm
- λ = 1.79m
- f = 286.47 s^-1
Explanation:
You have the following equation for a sound wave:
[tex]s(x,t)=3.00nm\ cos(3.50m^{-1}x- 1,800s^{-1} t)[/tex] (1)
The general form of the equation of a sound wave can be expressed as the following formula:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
A: amplitude of the wave = 3.00nm
k: wave number = 3.50m^-1
w: angular frequency = 1,800s^-1
- The maximum displacement of the wave is given by the amplitude of the wave, then you have:
maximum displacement = A = 3.00nm
- The wavelength is given by :
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{3.50m^{-1}}=1.79m[/tex]
The values for the wavelength is 1.79m
- The frequency is:
[tex]f=\frac{\omega}{2\pi}=\frac{1,800s^{-1}}{2\pi}=286.47s^{-1}[/tex]
The frequency is 286.47s-1
The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period
Answer:
Q = 23.49 C
Explanation:
We have,
Electric current drawn by the air conditioner is 16.2 A
Time, t = 1.45 s
It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,
[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]
So, the charge of 23.49 C is passing through the circuit during this period.
This questic concerns a tablet that
has remained in storage for 2 years
since manufacture. From the drug
monograph the decomposition
rate constant (k) was found to be 6
x 10-2 year-1. The degradation
reaction is known to proceed via
first-order kinetics. Which one of
the following is the remaining
percentage of the original drug
? concentration
Answer:
The remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
Explanation:
Recall the formula for the decay of a substance at an initial [tex]N_0[/tex] concentration at manufacture:
[tex]N(t)=N_0\,e^{-k\,\,t}[/tex]
where k is the decay rate (in our case 0.06/year), and t is the elapsed time in years. Therefore, after 2 years since manufacture we have:
[tex]N=N_0\,e^{-0.06\,\,(2)}\\N=N_0\,e^{-0.12}\\N/N_0=e^{-0.12}\\N/N_0=0.8869[/tex]
This in percent form is 88.7 %. That is, the remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
Calculate the ideal banking angle in degrees for a gentle turn of 1.88 km radius on a highway with a 136.3 km/hr speed limit, assuming everyone travels at the speed limit.
Answer:
Ф = 4.4°Explanation:
given:
radius (r) = 1.88 km
velocity (v) = 136.3 km/hr
required:
banking angle ∡ ?
first:
convert 1.88 km to m = 1.88km * 1000m / 1km
r = 1880 m
convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)
v = 37.86 m/s
now.. calculate the angle
Ф = inv tan (v² / r * g) we know that gravity = 9.8 m/s²
Ф = inv tan (37.86² / (1880 * 9.8))
Ф = 4.4°
What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)
Answer: Use this F=Ma.
Explanation: So your answer will be
F=1 Kg+9.8 ms-2
So the answer will be
F=9.8N
How'd I do this?
I just used Newton's second law of motion.
I'll also put the derivation just in case.
Applied force α (Not its alpha, proportionality symbol) change in momentum
Δp α p final- p initial
Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)
or then
F α m(v-u)/t
So, as we know v=final velocity & u= initial velocity and v-u/t =a.
So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.
So, F=kma (where k is the proportionality constant)
k is 1 so you can ignore it.
So, our equation becomes F=ma
A bicycle tire pump has a piston with area 0.43 In2. If a person exerts a force of 16 lb on the piston while Inflating a tire, what pressure does this produce on the air in the pump?
Answer:
The pressure produced on the air in the pump is 37.209 pounds per square inch.
Explanation:
By definition, the pressure is the force exerted on the piston divided by its area. Given that, force is distributed uniformly on the piston area, the formula to determine the pressure is:
[tex]p = \frac{F}{A}[/tex]
Where:
[tex]p[/tex] - Pressure, measured in pounds per square inch.
[tex]F[/tex] - Force exerted on the piston, measured in pounds.
[tex]A[/tex] - Piston area, measured in square inches.
If [tex]F = 16\,lb[/tex] and [tex]A = 0.43\,in^{2}[/tex], the pressure produced on the air in the pump is:
[tex]p = \frac{16\,lb}{0.43\,in^{2}}[/tex]
[tex]p = 37.209\,psi[/tex]
The pressure produced on the air in the pump is 37.209 pounds per square inch.
The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?
Answer:
The potential at the center of the sphere is -924 V
Explanation:
Given;
radius of the sphere, R = 0.22 m
electric field at the surface of the sphere, E = 4200 N/C
Since the electric field is directed towards the center of the sphere, the charge is negative.
The Potential is the same at every point in the sphere, and it is given as;
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)
The electric field on the sphere is also given as;
[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]
[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]
Substitute in the value of q in equation (1)
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]
Therefore, the potential at the center of the sphere is -924 V
According to Newton, when the distance between two interacting objects doubles, the gravitational force is
Answer:
1/4 of its original value
Explanation:
Newton's law of universal gravitation states that when two bodies of masses M₁ and M₂ interact, the force of attraction (F) between these bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between these bodies. i.e
F ∝ [tex]\frac{M_1 M_2}{r^2}[/tex] ------------(i)
From the equation above, it can be deduced that;
F ∝ [tex]\frac{1}{r^2}[/tex]
=> F = G [tex]\frac{1}{r^2}[/tex] -----------(ii)
Where;
G = constant of proportionality called the gravitational constant
Equation (ii) can be re-written as
Fr² = G
=> F₁r₁² = F₂r₂² -----------(iii)
Where;
F₁ and r₁ are the initial values of the force and distance respectively
F₂ and r₂ are the final values of the force and distance respectively
From the question, if the distance doubles i.e;
r₂ = 2r₁,
Then the final value of the gravitational force F₂ is calculated as follows;
Substitute the value of r₂ = 2r₁ into equation (iii) as follows;
F₁r₁² = F₂(2r₁)²
F₁r₁² = 4F₂r₁² [Divide through by r₁²]
F₁ = 4F₂ [Make F₂ subject of the formula]
F₂ = F₁ / 4 [Re-write this]
F₂ = [tex]\frac{1}{4} F_1[/tex]
Therefore the gravitational force will be 1/4 of its original value when the distance between the bodies doubles.
A bucket of water with total mass 23 kg is attached to a rope, which in turn is wound around a 0.050-m radius pulley at the top of a well. The bucket is raised to the top of the well and released from rest. The bucket is falling for 2 s and has a speed of 8.0 m/s upon hitting the water surface in the well. What is the moment of inertia of the pulley?
Answer:
[tex]I = 0.083 kg m^2[/tex]
Explanation:
Mass of the bucket, m = 23 kg
Radius of the pulley, r = 0.050 m
The bucket is released from rest, u = 0 m/s
The time taken to fall, t = 2 s
Speed, v = 8.0 m/s
Moment of Inertia of the pulley, I = ?
Using the equation of motion:
v = u + at
8 = 0 + 2a
a = 8/2
a = 4 m/s²
The relationship between the linear and angular accelerations is given by the equation:
[tex]a = \alpha r[/tex]
Angular acceleration, [tex]\alpha = a/r[/tex]
[tex]\alpha = 4/0.050\\\alpha = 80 rad/s^2[/tex]
Since the bucket is falling, it can be modeled by the equation:
mg - T = ma
T = mg - ma = m(g-a)
T = 23(9.8 - 4)
The tension, T = 133.4 N
The equation for the pulley can be modeled by:
[tex]T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2[/tex]
find the value of k for which the given pair of vectors are not equal
2ki +3j and 8i + 4kj
Answer:
5
Explanation:
Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .
Answer:
Explanation:
v= u + at
v is final velocity , u is initial velocity . a is acceleration and t is time
Initial velocity u = 0 . Putting the given values in the equation
v = 0 + g sin 18 x 3.5
= 10.6 m /s
For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.
What are Velocity and Acceleration?The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.
There are several types of velocity :
Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform VelocityThe pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.
According to the question, the given values are :
Time, t = 3.50 sec
Initial Velocity, u = 0 m/s
Use equation of motion :
v = u+at
v = 0+ g sin 18 × 3.5
v = 10.6 m/s.
So, the final velocity will be 10.6 m/s.
To get more information about Velocity and Acceleration :
https://brainly.com/question/14683118
#SPJ2
6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.
Answer:
same direction = 80 (n)
opposite direction = 20 (n) going one direction
Explanation:
same direction means they are added to each other
and opposite means acting on eachother
What is the potential difference across one wire of a 70 m extension cord made of 16 gauge copper wire carrying a current of 4 A at room temperature (20degrees Celsius)
Answer:
V = 3.6 volts
Explanation:
From Ohm's Law, we know that:
V = IR
but,
R = ρL/A
Therefore,
V = IρL/A
where,
V = Potential Difference = ?
I = Current = 4 A
ρ = resistivity of copper = 1.68 x 10⁻⁸ Ω.m
L = Length = 70 m
A = Cross-sectional Area = πd²/4 = π(1.29 x 10⁻³ m)²/4 [16 gauge wire has a diameter of 1.29 mm]
A = 1.31 x 10⁻⁶ m²
V = (4 A)(1.68 x 10⁻⁸ Ω.m)(70 m)/(1.31 x 10⁻⁶ m²)
V = 3.6 volts
The potential difference across the given length of copper wire is obtained to be 3.6 V.
Ohm's LawThe resistance of the wire depends on the length, area of cross-section of the wire and resistivity of the material.
Given the length of the wire, [tex]L = 70\,m[/tex].
The diameter of a 16 gauge copper wire is given by, [tex]d = 1.29\times 10^{-3}\,m[/tex].
Therefore, the radius of the cross-section is;
[tex]r=\frac{d}{2} = \frac{1.29\times 10^{-3}\,m}{2} =6.45\times 10^{-4}\,m[/tex]
So, the resistance of the given length of wire can be written as,
[tex]R= \rho\, \frac{L}{A} =(1.68 \times 10^{-8} \, \Omega .m)\times \frac{70\,m}{\pi \times (6.45\times 10^{-4}m)^2} = 0.9\, \Omega[/tex]
According to Ohm's Law, we can write;
[tex]V=IR[/tex]
Substituting the known values, we get the voltage drop is;
[tex]V = 4A \times 0.9\Omega = 3.6\,V[/tex]
Learn more about Ohm's law here:
https://brainly.com/question/3656434
How much electricity is in human body?
Answer:
100 on average
Explanation:
The human body is a physical substance from where a human can breathe, drink, eat, walk, etc, and consists of various cells. A human being cannot live with their functions and when functions stop working that means a human being is dead.
Therefore, in the human body, there is also electricity that contains 100 watts on average.
So, According to the given situation, the correct answer is 100 on average.
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated
Answer:
It is calculated as Force × perpendicular distance.
Explanation:
Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.
The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.
But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.
A soccer ball is kicked straight upwards with an initial vertical speed of 8.0\,\dfrac{\text m}{\text s}8.0 s m 8, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. How long does it take the ball to have a downwards speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
Answer:
1.2 s
Explanation:
Given:
v₀ = 8.0 m/s
v = -4.0 m/s
a = -10 m/s²
Find: t
v = at + v₀
(-4.0 m/s) = (-10 m/s²) t + (8.0 m/s)
t = 1.2 s
A 12.5-µF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.
(a) How much energy is stored in the capacitor before and after the dielectric is inserted?
(b) By how much did the energy change during the insertion? Did it increase or decrease?
Answer:
a) 0.0036 J & 0.0135 J
b) 0.0099 J
Explanation:
Given that
Capacitance of a capacitor, C(i) = 12.5*10^-6 F
Potential difference across the capacitor, V = 24 V
Dielectric constant, k = 3.75
Energy in a capacitor is given by the formula
U = ½ CV²
Now, applying this to our solution, we have.
a, the energy stored before the dielectric was inserted is 0.0036. After inserting the dielectric, we then take cognizance into it, essentially multiplying by the constant of the dielectric, we got 0.0135 J
b) Now, Change in energy is gotten by subtracting the two energies, and we have
ΔU = U - U(1)
ΔU = 0.0036 - 0.0135
ΔU = 00099 J
See attachment for calculation
An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?
Answer:
(a) max. height = 3.641 m
(b) flight time = 1.723 s
(c) horizontal range = 31.235 m
(d) impact velocity = 20 m/s
Above values have been given to third decimal. Adjust significant figures to suit accuracy required.
Explanation:
This problem requires the use of kinematics equations
v1^2-v0^2=2aS .............(1)
v1.t + at^2/2 = S ............(2)
where
v0=initial velocity
v1=final velocity
a=acceleration
S=distance travelled
SI units and degrees will be used throughout
Let
theta = angle of elevation = 25 degrees above horizontal
v=initial velocity at 25 degrees elevation in m/s
a = g = -9.81 = acceleration due to gravity (downwards)
(a) Maximum height
Consider vertical direction,
v0 = v sin(theta) = 8.452 m/s
To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,
solve for S in equation (1)
v1^2 - v0^2 = 2aS
S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s
(b) total flight time
We solve for the time t when the vertical height of the object is AGAIN = 0.
Using equation (2) for vertical direction,
v0*t + at^2/2 = S substitute values
8.452*t + (-9.81)t^2 = 3.641
Solve for t in the above quadratic equation to get t=0, or t=1.723 s.
So time for the flight = 1.723 s
(c) Horiontal range
We know the horizontal velocity is constant (neglect air resistance) at
vh = v*cos(theta) = 25*cos(25) = 18.126 m/s
Time of flight = 1.723 s
Horizontal range = 18.126 m/s * 1.723 s = 31.235 m
(d) Magnitude of object on hitting ground, Vfinal
By symmetry of the trajectory, Vfinal = v = 20, or
Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s
Two children are balanced on a seesaw, but one child weighs twice as much as the other child. The heavier child is sitting half as far from the pivot as is the lighter child. Since the seesaw is balanced, the heavier child is exerting on the seesaw:_______.
a. a force that is less than the force the lighter child is exerting.
b. a force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Answer:
B. A force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Explanation:
If they are sitting at the same distance away from the pivot yet the seesaw is balanced, the only conclusion is the heavier child is exerting a lower force. This causes the pivot exertion and balances to be equal. The equilibrium of the pivot-seesaw is not affected by the weight because of force exertion.
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the initial current (at t = 0), and τ is a constant having dimensions of time. Consider a fixed observation point within the conductor.
Required:
a. How much charge passes this point between t = 0 and t = τ?
b. How much charge passes this point between t = 0 and t = 10 τ?
c. What If ? How much charge passes this point between t = 0 and t = [infinity]?
Answer:
Pls see attached file
Explanation:
(9) A dancer spins at 72rpm about an axis through the center with arms outstretched. The dancer mass=90kg, model the head as a uniform sphere r=8.0cm Model the trunk and legs as uniform solid cylinder r=12.0cm, arms as slender rods L=60.0cm. On the average the head=7%, both arms together=13% and trunk and legs=80% of body mass. Find the K.E. of the dancer note: arms are I around an axis for slender rod: 1/3ML^2 each arm note2: Moments of Inertia are summable, i.e. I total = I1 + I2 + I 3 +...
Answer:
Explanation:
ω = angular velocity = 2π n = 2π x 72 / 60
= 7.536 rad /s
mass of head = 90 x .07 = 6.3 kg
moment of inertia of head = 2 /5 m R²
= .4 x 6.3 x .08²
= .016128 kg m²
moment of inertia of trunk + legs
= 1/2 m R²
= .5 x .8 x 90 x .12²
= .5184 kg m²
moment of inertia of arms
= 1/3 m L²
= 1 / 3 x .13 x 90 x .60²
= 1.404 kg m²
Total moment of inertia
I = 1.938 kg m²
kinetic energy = 1/ 2 I ω ²
where I is moment of inertia and ω is angular velocity
= .5 x 1.9338 x 7.536²
=55 J approx .
The definition of kinetic energy and moment of inertia allows finding the result for the kinetic energy of the dancer with arms extended is:
The kinetic energy is: KE = 56.3J
Given parameters
Frequency is: f = 72 rad / min (1 min / 60s) = 1.2 rad / s The mass is m = 90 k Head radius is r₁ = 8.0 cm = 0.08 m The leg and trunk are cylinders of radius r₂ = 12.0 cm = 0.12 m The length of the arms, approximated as rods L = 60.0 cm = 0.600 m The dancer spins with outstretched arms The percentage of the mass is:* Head 7%
* Arms 13%
* Trunk and legs 80%
To find.
Kinetic energy
The kinetic energy is the energy due to the movement, in this case the movement is rotational, therefore the expression is:
KE = ½ I w²
The angular velocity is related to the frequency.
w = 2π f
w = 2π 1.2
w = 7.540 s
The moment of inertia is a scalar, therefore it is a quantity that can be added, the total moment of inertia of the dancer is the sum of the moments of inertia of each part with respect to the axis of rotation of the person.
[tex]I_{toal}= I_{head}+ I_{trunk}+I_{arms}[/tex]
The moments of inertia with respect to the centers of mass are tabulated.
Sphere I = 2/5 mr²
Cylinder I = ½ m r²
Rod I = ⅓ m r²
The axis of rotation of the head and the trunk are in the axis of rotation of the person, therefore their moment of inertia is those corresponding to the center of mass.
At the end of the arms it is at a distance of D = [tex]\frac{r_2}{2}[/tex] from the axis of rotation of the dancer, therefore to find the moment of inertia we must use the theorem of parallel axes, see attached.
[tex]I_{arms} = I_{cm} + M D^2[/tex]
[tex]I_{arms}[/tex] = ⅓ m L² + m D²
[tex]I_{arms} = m_{arms} ( \frac{L^2}{3} + D^2)[/tex]
The masses of each part of the body are:
[tex]m_{head}[/tex] = m 0.07
[tex]m_{trunk}[/tex] = m 0.80
[tex]m_{arms}[/tex] = m 0.13
Let's find the total moment of inertia.
[tex]I_{total} = \frac{2}{5} \ 0.07m \ r_1^2 + 0.13m\ ( \frac{L^2}{3} + D^2) + \frac{1}{2} \ 0.80m \ r_2^2[/tex]
[tex]I_{total} = m ( 0.028 \ r_1^2 + 0.13 (\frac{L^2}{3} + D^2) + 0.40 \ r_2^2}[/tex]
Let's calculate.
[tex]I_{total} = 90\ ( 0.028 \ 0.08^2 + 0.13\ ( \frac{0.6^2}{3} + 0.06^2) + 0.40 \ 0.12^2 )[/tex]
[tex]I_{total}= 90 \ 0.0220 \\ \\I_{total} = 1.9806 \ Kg m^2[/tex]
we substitute in the kinetic energy
KE = ½ 1.9806 7,540²
KE = 56.3 J
In conclusion using the definition of kinetic energy and moment of inertia we can find the result for the kinetic energy of the dancer with the extended arms is:
The kinetic energy is: KE = 56.3 J
Learn more here: brainly.com/question/15246709
what is the difference between a good conductor and a good insulator?
Answer:
Explanation:
In a conductor, electric current can flow freely, in an insulator it cannot.
Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.
Most atoms hold on to their electrons tightly and are insulators.