Answer:
4.41 m/s^2
Explanation:
(v_f)^2 - (v_i)^2 = 2a * change in distance
(21)^2 - (0)^2 = 2a * 50
a = (21^2)/(2*50)
a = 4.41 m/s^2
Current folw in which dirction
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle so that each side has a length of b. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.
Answer:
Mb²/2
Explanation:
Pls see attached file
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
In a shipping yard, a crane operator attaches a cable to a 1,390 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 33 m. Determine the amount of work done (in J) by each of the following.
a) the tension in the cable.
b) the force of gravity.
Answer:
a) A = 449526 J, b) 449526 J
Explanation:
In this exercise they do not ask for the work of different elements.
Note that as the box rises at constant speed, the sum of forces is chorus, therefore
T-W = 0
T = W
T = m g
T = 1,390 9.8
T = 13622 N
Now that we have the strength we can use the definition of work
W = F .d
W = f d cos tea
a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel
A = A x
A = 13622 33
A = 449526 J
b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180
W = T x cos 180
W = - 13622 33
W = - 449526 J
Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiver pass through
The number of maxima of the standing wave pattern is two.
Maxima problem:At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.
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A 900 kg roller coaster car starts from rest at point A. rolls down the track, goes
around a loop (points B and C) and then flies off the inclined part of the track (point D),
Figure 2.
The dimensions are: H =80 m.
r= 15m, h=10m and theta =9.30°
Calculate the
(a) gravitational potential energy at point A.
(b) velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J.
c) distance of the car land (in the horizontal direction) from point D if given the
velocity at point D is 37.06 m/s
I
Answer:
gravitational potential energy at point A.
A) The gravitational potential energy at point A is; 705600 J
B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s
A) Formula for gravitational potential energy is;
PE = mgh
At point A;
mass; m = 900 kg
height; h = 80 m
Thus;
PE = 900 × 9.8 × 80
PE = 705600 J
B) Kinetic energy of the roller coaster at point C is given as;
KE = PE - W
We are given Workdone; W = 264870 J
Thus;
KE = 705600 - 264870
KE = 440730 J
Thus, velocity at point C is gotten from the formula of kinetic energy;
KE = ½mv²
v = √(2KE/m)
v = √(2 × 440730/900)
v = 31.295 m/s
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Two particles, of charges q1 and q2, are separated by a distance d on the x-axis with q1 at the origin and q2 in the positive direction. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x-axis (other than infinity) at which the electric potential due to the two particles is zero.
Answer:
No point on the x-axis
Pls see attached file
A square coil of wire with side 8.0 cm and 50 turns sits in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is pulled quickly out of the magnetic field in 0.2 s. If the resistance of the coil is 15 ohm and a current of 12 mA is induced in the coil, calculate the value of the magnetic field.
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 .
If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B
Answer:
0.9m/s²
Explanation:
See attached files
1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.
Answer:
The answer is CExplanation:
Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.
A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10
Answer:
It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water
Explanation:
An air conditioner connected to a 103 V rms AC line is equivalent to a 20 resistance and a 1.68 inductive reactance in series. a) What is the impedance of the air conditioner
Answer:
20.07ohms
Explanation:
Impedance is defined as the opposition to the flow of current through the elements of the circuit.
Impedance for R-L AC circuit is expressed as Z = √R²+XL²
R is the resistance
XL is the inductive reactance.
Given resistance of the air condition = 20 ohms
Inductive reactance XL = 1.68 ohms
Z = √20²+1.68²
Z = √400+2.8224
Z = √402.8224
Z = 20.07 ohms
Hence the impedance of the air conditioner is 20.07ohms
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
E = 1.24MeV
Explanation:
The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.
Its energy can be determined by;
E = hf
= (hc) ÷ λ
where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.
Therefore,
E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]
= 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]
E = 1.9878 × [tex]10^{-13}[/tex] J
But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;
E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]
= 1242375 eV
∴ E = 1.24MeV
The energy of the photon is 1.24MeV.
A string passing over a pulley has a 3.85-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.5 cm and mass 0.79 kg .
A. If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
B. In fact, it is found that if the heavier mass is given a downward speed of 0.20 m/s , it comes to rest in 6.4 s . What is the average frictional torque acting on the pulley?
Answer:
Explanation:
Let the acceleration be a of the system
T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass
for motion of 3.85 kg , applying newton's law
3.85g - T₁ = 3.85 a
for motion of 2.6 kg
T₂ - 2.6g = 2.6 a
T₂ - T₁ + 1.25 g = 6.45 a
T₁ - T₂ = 1.25 g - 6.45 a
for motion of pulley
(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration
(T₁ - T₂ ) x R = 1 /2 m R² x a / R
(T₁ - T₂ ) = m x a / 2 = .79 x a / 2 = . 395 a
1.25 g - 6.45 a = .395 a
1.25 g = 6.845 a
a = 1.79 m /s²
B )
When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s
Average deceleration = .2 / 6.4 = .03125 m /s²
Total deceleration created by frictional torque = 1.79 + .03125
= 1.82125 m /s²
If R be the average frictional torque acting on the pulley
angular deceleration of pulley = a / R
= 1.82125 / .045
= 40.47 rad /s²
Now R = I x 40.47 , I is moment of inertia of pulley
= 1 /2 x .79 x .045² x 40.47
= .0323 N.m
Torque created = .0323 Nm
The acceleration of the two masses hanging from ends of the pulley is 31 m/s².
The average frictional torque acting on the pulley is 0.55 Nm.
The given parameters;
mass of the pulley, = M = 0.79 kgfirst mass, m₁ = 3.85 kgsecond mass, m₂ = 2.6 kgradius, R = 4.5 cm = 0.045 mThe acceleration of the two masses is determined by taking net torque acting on the pulley;
[tex]\tau _{net} = I \alpha[/tex]
[tex]T_1R - T_2R = I \alpha\\\\[/tex]
where;
T is the tension on both stings suspending the masses = mgI is the moment of inertia of the pulley [tex]= \frac{MR^2}{2}[/tex]α is the angular acceleration[tex]R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)[/tex]
Substitute the given parameters, to solve for the acceleration of the masses;
[tex]a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2[/tex]
The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.
[tex]a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm[/tex]
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•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?
Answer:
Explanation:
For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.
Therefore the potential on the ferric surface is
V = k Q / r
where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest
a) On the surface the potential
V = 9 10⁹ Q / 0.5
V = 18 10⁹ Q
Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V
b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials
for V = 1300V let's find the radius
r = k Q / V
r = 9 109 1 10-7 / 1300
r = 0.69 m
other values are shown in the following table
V (V) r (m)
1800 0.5
1300 0.69
800 1,125
300 3.0
In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V
C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape
E = k Q / r²
A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed
Answer:
The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.
Explanation:
This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 6.3 m from two double slits 0.49 mm apart illuminated by 739-nm light. (give answer in millimeters)
Answer:
Explanation:
distance of third dark fringe
= 2.5 x λ D / d
where λ is wavelength of light , D is screen distance and d is slit separation
putting the given values
required distance = 2.5 x 739 x 10⁻⁹ x 6.3 / .49 x 10⁻³
= 23753.57 x 10⁻⁶
= 23.754 x 10⁻³ m
= 23.754 mm .
A flashlight is held at the edge of a swimming pool at a height h = 1.6 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 1.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.
Required:
What is the horizontal distance from the edge of the pool to the bottom of the pool where the light strikes? Write your answer in meters.
one of the answers that i found was 5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.
A block of mass m is suspended by a vertically oriented spring. If the mass of a block is increased to 4m, how does the frequency of oscillation change, if at all
Answer:
The frequency will be reduced by a factor of √2/2
Explanation:
Pls see attached file
The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Let the initial mass of block be m.
And new mass is, 4m.
The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,
[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
Here,
k is the spring constant.
If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,
[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]
Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
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Rick spends four hours researching on the internet and does 1090 J of work. In the process, his internal energy decreases by 2190 J. Determine the value of Q, including the algebraic sign.
Answer:
Q = -3280J
Explanation:
From the First Law of Thermodynamics, energy cannot be created nor destroyed but it can be converted from one form to another with the interaction of heat. Mathematically, this can be expressed as:
ΔU = Q + W ----------(i)
Where;
ΔU = total change in internal energy of a system.
Q = heat exchanged between the system and the surrounding
W = work done by or on the system.
If heat is lost into the surrounding, then Q = -ve, else Q = +ve
If work is done on the system, then W = -ve, else W = -ve
=> From the question, Rick is the system and does a work of
W = +1090J [since Rick does the work, W = +ve]
=>Also, the internal energy decreases by 2190J, therefore,
ΔU = -2190J [since there is a decrease in internal energy]
Substitute the values of W and ΔU into equation (i) as follows;
-2190 = Q + 1090
=> Q = -2190 - 1090
=> Q = -3280J
Therefore, the value of Q = -3280J
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.
Answer:
206.67NExplanation:
The sum of force along both components x and y is expressed as;
[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]
The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
[tex]a_x = \frac{d^2 x }{dt^2}[/tex]
[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]
Similarly,
[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]
[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]
[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-axis is the electric potential zero?
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, [tex]q_1=-3\ nC[/tex]
It is placed at a distance of 9 cm at x axis
Charge, [tex]q_2=+4\ nC[/tex]
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,
[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]
Here,
[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]
So,
[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]
Squaring both sides,
[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]
So, at a distance of 10.2 m on the y axis the electric potential equals 0.
According to the question,
Charge,
[tex]q_1 = -3 \ nC[/tex] (9 cm at x-axis)[tex]q_2 = +4 \ nC[/tex] (16 cm at x-axis)Now,
→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]
or,
→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]
→ [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]
here,
[tex]r_1 = \sqrt{y^2+81}[/tex]
[tex]r^2 = \sqrt{y^2+225}[/tex]
By substituting the values,
→ [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]
By applying cross-multiplication,
[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]
By squaring both sides, we get
→ [tex]9(y^2+225) = 16(y^2+81)[/tex]
[tex]9y^2+2025 = 16 y^2+1296[/tex]
[tex]2025-1296=7y^2[/tex]
[tex]7y^2=729[/tex]
[tex]y = 10.2 \ m[/tex]
Thus the solution above is correct.
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A solenoid 26.0 cm long and with a cross-sectional area of 0.580 cm^2 contains 490 turns of wire and carries a current of 90.0 A.
Calculate:
(a) the magnetic field in the solenoid;
(b) the energy density in the magnetic field if the solenoid is filled with air;
(c) the total energy contained in the coil’s magnetic field (assume the field is uniform);
(d) the inductance of the solenoid.
Answer:
A.21.3T
B.1.8x 10^6J/m^3
C.0.27x10^2J
D.6.6x10^-3H
Explanation:
Pls see attached file
At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s
Answer:
5.81 X 10^3 Ns
Explanation:
Given that
F = At² and F at t = 1.25 s is 781.25 N ?
A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²
d(Impulse) = Fdt
Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s
Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s
Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns
5.81 X 10^3 N.s
If the speed of a "cheetah" is 150 m / s. How long does it take to cover 800 m?
Answer:
5.33333... seconds
Explanation:
800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....
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:
Suppose a space vehicle with a rest mass of 150 000 kg travels past the International Space Station at a constant speed of 2.6 x 108 m/s with respect to the I.S.S. When an observer on the I.S.S. measures the moving vehicle, her measurement of the space vehicle length is 25.0 m. Determine the relativistic mass of the space vehicle. Determine the length of the space vehicle as measured by an astronaut on the space vehicle.
Answer:
m = 300668.9 kg
L₀ = 12.47 m
Explanation:
The relativistic mass of the space vehicle is given by the following formula:
[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]
where,
m = relativistic mass = ?
m₀ = rest mass = 150000 kg
v = relative speed = 2.6 x 10⁸ m/s
c = speed of light = 3 x 10⁸ m/s
Therefore
[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]
m = 300668.9 kg
Now, for rest length of vehicle:
L = L₀√(1 - v²/c²)
where,
L = Relative Length of Vehicle = 25 m
L₀ = Rest Length of Vehicle = ?
Therefore,
25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]
L₀ = (25 m)(0.499)
L₀ = 12.47 m
Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a central bright spot?
A. Young
B. Fresnel
C. Poisson
D. Arago
Answer:
Your answer is( D) - Arago
A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.
Answer:
Explanation:
Total mass m = 18 kg .
Spring are parallel to each other so total spring constant
= 4 x 24 = 96 N/cm = 9600 N/m
Time period of vibration
[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]
Putting the given values
[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]
= .27 s .