Answer:
1000j
Explanation:
work done = force x distance
w = 5 x 10 x 20 = 1000joules
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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A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?
Answer:
39.94m/s.Explanation:
Kinetic energy is expressed as KE = 1/2 mv² where;
m is the mass of the body
v is the velocity of the body.
For the heavier shell;
m = 5kg
KE gained = 100J
Substituting this values into the formula above to get the velocity v;
100 = 1/2 * 5 * v²
5v² = 200
v² = 200/5
v² = 40
v = √40
v = 6.32 m/s
Note that after the explosion, both body fragments will possess the same velocity.
For the lighter shell;
mass = 2.0kg and v = 6.32m/s
KE of the lighter shell = 1/2 * 2 * 6.32²
KE of the lighter shell = 6.32²
KE of the lighter shell= 39.94m/s
Hence, the lighter one gains a kinetic energy of 39.94m/s.
The gain in the kinetic energy of the smaller fragment is 249.64 J.
The given parameters;
Mass of the shell, m = 7.0 kgMass of one fragment, m₁ = 2.0 kgMass of the second fragment, m₂ = 5.0 kgKinetic energy of heavier fragment, K.E₁ = 100 JThe velocity of the heavier fragment is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\v^2 = \frac{2K.E}{m} \\\\v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 100}{5} }\\\\v = 6.32 \ m/s[/tex]
Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;
[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\-6.32(5) \ + 2u_2 = 0(7)\\\\-31.6 + 2u_2 = 0\\\\2u_2 = 31.6\\\\u_2 = \frac{31.6}{2} \\\\u_2 = 15.8 \ m/s[/tex]
The gain in the kinetic energy of the smaller fragment is calculated as follows;
[tex]K.E_2 = \frac{1}{2} mu_2^2\\\\K.E_2 = \frac{1}{2} \times 2 \times (15.8)^2\\\\K.E_2 = 249.64 \ J[/tex]
Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.
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Suppose that 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm. (a) How much work is needed to stretch the spring from 41 cm to 45 cm? (Round your answer to two decimal places.) J (b) How far beyond its natural length will a force of 15 N keep the spring stretched? (Round your answer one decimal place.)
Answer:
Explanation:
Work done on a spring is expressed as [tex]W = 1/2 ke^{2}[/tex]
k is the elastic constant
e is the extension of the material
If 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm, then;
Work done = 4J and the extension e = 47 cm - 36 cm; e = 11 cm
11cm = 0.11m
Substituting the given values into the equation above to get the elastic constant;
[tex]W = 1/2 ke^{2}\\4 = 1/2k(0.11)^{2} \\8 = 0.0121k\\k = 8/0.0121\\k = 661.16N/m[/tex]
a) In order to determine the amount of work needed work is needed to stretch the spring from 41 cm to 45 cm, wre will use the same formula as above.
[tex]W = 1/2ke^{2} \\e = 0.45 - 0.41\\e = 0.04 m\\ k = 661.16N/m[/tex]
[tex]W = 1/2 * 661.16 * 0.04^{2} \\W = 330.58*0.0016\\W = 0.53J (to\ 2d.p)[/tex]
b) According to hooke's law, F = ke where F is the applied force
We are to get the extension when a force of 15N is applied to the original length of the material.
e = F/k
e = 15/661.16
e = 0.02 m (to 1 d.p)
This means that the natural length of the spring will be stretched by 0.02 m when a force of 15N is applied to it.
5. A nail contains trillions of electrons. Given that electrons repel from each other, why do they not then fly out of the nail?
Answer:
Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.
If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.
Complete the following sentence: The term coherence relates to the phase relationship between two waves. the polarization state of two waves. the amplitude of two waves. the diffraction of two waves. the frequency of two waves.
Answer:
the phase relationship between two waves.
Explanation:
Coherence describes all properties of the correlation between physical quantities between waves. It is an ideal property of waves that determines their interference. In a situation in which there is a correlation or phase relationship between two waves. If the properties of one of the waves can be measure directly, then, some of the properties of the other wave can be calculated.
A tennis ball is thrown from ground level with velocity v0 directed 30 degrees above the horizontal. If it takes the ball 1.0s to reach the top of its trajectory, what is the magnitude of the initial velocity?
Answer:
vi = 19.6 m/s
Explanation:
Given:
final velocity vf = 0
gravity a = -9.8
time t = 1
Initial velocity vi = vf - at
vi = 0 + 9.8 (1.0)
vi = 9.8 m/s
the y component of velocity is the initial velocity.
therefore v sin 30 = 9.8
vi/2 = 9.8
vi = 19.6 m/s
A tennis ball is thrown from ground level with velocity v0 directed 30 degrees above the horizontal. If it takes the ball 1.0s to reach the top of its trajectory, the magnitude of the initial velocity vi = 19.62 m/s.
Given data to find the initial velocity,
final velocity vf = 0
gravity a = -9.8
time t = 1
What is deceleration?The motion of the tennis ball on the vertical axis is an uniformly accelerated motion, with deceleration of (gravitational acceleration).
The component of the velocity on the y-axis is given by the following law:
Initial velocity vi = vf - at
At the time t=0.5 s, the ball reaches its maximum height, and when this happens, the vertical velocity is zero (because it is a parabolic motion), Substituting into the previous equation, we find the initial value of the vertical component of the velocity:
vi = 0 + 9.8 (1.0)
vi = 9.8 m/s
the y component of velocity is the initial velocity.
However, this is not the final answer. In fact, the ball starts its trajectory with an angle of 30°. This means that the vertical component of the initial velocity is,
therefore, v sin 30° = 9.8
We found before the value of y component, so we can substitute to find the initial speed of the ball:
vi/2 = 9.8
vi = 19.6 m/s
Thus, the initial velocity can be found as 19.62 m/s.
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A long solenoid (1500 turns/m) carries a current of 20 mA and has an inside diameter of 4.0 cm. A long wire carries a current of 2.0 A along the axis of the solenoid. What is the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire
Answer:
The magnitude of the magnetic field is 55μT
Explanation:
Given;
number of turns of the solenoid per length, n = N/L = 1500 turns/m
current in the solenoid, I = 20 mA = 20 x 10⁻³ A
diameter of the solenoid, d = 4 cm = 0.04 m
The magnetic field at a point that is inside the solenoid;
B₁ = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Given;
current in the wire, I = 2 A
distance of magnetic field from the wire, r = 1 cm = 0.01 m
The magnetic field at 1.0 cm from the wire;
[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]
The magnitude of the magnetic field;
[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]
Therefore, the magnitude of the magnetic field is 55μT
The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
Given the following parameters from the question
Number of turns of the solenoid per length, n = N/L = 1500 turns/m current in the solenoid, I = 20 mA = 20 x 10⁻³ A Diameter of the solenoid, d = 4 cm = 0.04 mThe magnetic field at a point that is inside the solenoid is expressed according to the formula;
B₁ = μ₀nIWhere;
μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Next is to get the magnetic field strength in the second wire.
Current in the wire, I = 2 A Distance of magnetic field from the wire, r = 1 cm = 0.01 mThe magnetic field at 1.0 cm from the wireSubstitute into the formula:
[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]
Get the resultant magnetic field:
[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]
Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
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Cables supporting a suspension bridge have a linear mass density of 3700 kg/m; the tension is 1.7 x 10 8 N. What would be the transverse wave speed in such a cable?
Answer:
The speed is [tex]v = 214.35 \ m/s[/tex]
Explanation:
From the question we are told that
The linear mass density is [tex]\mu = 3700 \ kg/m[/tex]
The tension is [tex]T = 1.7*10^8 \ N[/tex]
The transverse wave speed is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
substituting values
[tex]v = \sqrt{\frac{1.7 *10^{8}}{3700} }[/tex]
[tex]v = 214.35 \ m/s[/tex]
soaring birds and glider pilots can remain aloft for hours without expending power. Discuss why this is so.
Answer:
Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection
Explanation:
A nano-satellite has the shape of a disk of radius 0.80 m and mass 8.50 kg.
The satellite has four navigation rockets equally spaced along its edge. Two
navigation rockets on opposite sides of the disk fire in opposite directions
to spin up the satellite from zero angular velocity to 14.5 radians/s in 30.0
seconds. If the rockets each exert their force tangent to the edge of the
satellite (the angle theta between the force and the radial line is 90
degrees), what was is the force of EACH rocket, assuming they exert the
same magnitude force on the satellite?
Answer:
Explanation:
moment of inertia of satellite I = 1/2 m R²
m is mass and R is radius of the disc
I = 0.5 x 8.5 x 0.8²
= 2.72 kg m²
angular acceleration α = change in angular velocity / time
α = (14.5 - 0) / 30
α = .48333
Let force of each rocket = F
torque created by one rocket = F x R
= F x .8
Torque created by 4 rockets = 4 x .8 F = 3.2 F
3.2 F = I x α
3.2 F = 2.72 x .48333
F = 0 .41 N
A circular loop of wire of radius 10 cm carries a current of 6.0 A. What is the magnitude of the magnetic field at the center of the loop
Answer:
3.77x10^-5T
Explanation:
Magnetic field at center of the loop is given as
B=uo*I/2r =(4pi*10-7)*6/2*0.1
B=3.77*10-5Tor 37.7 uTi
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 .
Which of the following statements is not true?
1) The average power supplied to an inductor in an AC circuit is proportional to the angular frequency of the power source.
2) By stepping up AC voltage with a transformer, we can transport electricity across large distances with minimal power loss.
3) Voltage and current are in phase across a resistor connected to an AC power source.
4) In AC circuits, RMS stands for Root Mean Square.
Answer:
Explanation:
1 ) Average power supplied to an inductor is zero because the phase difference of potential and current is π / 2 .
So it is a wrong statement .
2 ) Step up transformer increases the voltage . At high voltage , lesser current is required to transport electrical energy . When current is reduced , the loss of energy due to heating effect is reduced .
3 ) voltage and current are in phase in resistance in ac .
3 ) RMS stands for Root Mean Square .
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Given data :
Landing speed of Jet = 200 km/h
Distance = 425 m
Total mass of aircraft = 140 Mg with mass center at G
Determine the reaction N under the nose of wheel B First step : calculate the value of the Jet accelerationJet acceleration = 1 / (2 *425) * (200² - 60² ) * 1 / (3.6)²
= 3.3 m/s²
Next step : determine the reaction N under the nose of WheelReaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140). ----- ( 1 )
∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )
Hence Reaction N = 257 KN
We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
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A wagon wheel consists of 8 spokes of uniform diameter, each of mass ms and length L cm. The outer ring has a mass mring. What is the moment of inertia of the wheel
Answer:
The moment of inertial of the wheel, [tex]I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2[/tex]
Explanation:
Given;
8 spokes of uniform diameter
mass of each spoke, = [tex]M_s[/tex]
length of each spoke, = L
mass of outer ring, = [tex]M_r[/tex]
The moment of inertial of the wheel will be calculated as;
[tex]I = 8I_{spoke} + I_{ring}[/tex]
where;
[tex]I_{spoke[/tex] is the moment of inertia of each spoke
[tex]I_{ring[/tex] is the moment of inertia of the rim
Moment of inertia of each spoke [tex]=\frac{1}{3}M_sL^2[/tex]
Moment of inertial of the wheel
[tex]I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2[/tex]
Rays that pass through a lens very close to the principle axis are more sharply focused than those that are very far from the axis. This spherical aberration helps us understand why:_______
Answer: it is easier to read in bright light than dim light.
Explanation:
The ray of light is the direction that is used by light in travelling through a medium. Rays that pass through a lens very close to the principle axis are more sharply focused than those that are very far from the axis.
Because of the fact that the rays are close to the principle axis, the spherical aberration helps us to understand the reason why it is easier for people to read in bright light than readin iin dim light.
What is the wavelength λλlambda of the wave described in the problem introduction? Express the wavelength in terms of the other given variables and constants
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda= \frac{2 \pi }{k}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]
The magnetic field is [tex]\= B = B_0 sin (kx -wt) \r k[/tex]
From the above equation
and k is the wave number which is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]\lambda= \frac{2 \pi }{k}[/tex]
Where [tex]\lambda[/tex] is the wavelength
Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in opposite directions.
(d) Calculate the change in kinetic energy of the system, in joules, in this case.
Answer:
a) 6.57 m/s
b) 53.75 J
c) 6.37 m/s
d) -98.297 J
Explanation:
mass of player = [tex]m_{p}[/tex] = 117.5 kg
speed of player = [tex]v_{p}[/tex] = 6.5 m/s
mass of ball = [tex]m_{b}[/tex] = 0.43 kg
velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s
Recall that momentum of a body = mass x velocity = mv
initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s
initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s
initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] = 2482.187 J
a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.
for this first case that they travel in the same direction, their momenta carry the same sign
[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
where v is the final velocity of the player.
inserting calculated momenta of ball and player from above, we have
763.75 + 11.395 = (117.5 + 0.43)v
775.145 = 117.93v
v = 775.145/117.93 = 6.57 m/s
b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J
change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained
c) if they travel in opposite direction, equation becomes
[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
763.75 - 11.395 = (117.5 + 0.43)v
752.355 = 117.93v
v = 752.355/117.93 = 6.37 m/s
d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex] = 2383.89 J
change in kinetic energy = 2383.89 - 2482.187 = -98.297 J
that is 98.297 J lost
Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3900 N on the car for 0.55 s. Use the initial direction of the cars motion as the positive direction.
What impulse, in kilogram meters per second, is imparted to the car by this force?
Find the horizontal components of the final velocity of the bumper car, in meters per second, if its initial velocity was 2.95 m/s and the car plus driver have a mass of 190 kg. You may neglect friction between the car and floor.
Find the horizontal components of the final velocity of the bumper car, in meters per second, if its initial velocity was 2.95 m/s and the car plus driver have a mass of 190 kg. You may neglect friction between the car and floor.
Answer:
The impulse is 2145 kg-m/s
The final velocity is -8.34 m/s or 8.34 m/s in he opposite direction.
Explanation:
Force on the rail = 3900 N
Elapsed time of impact = 0.55 s
Impulse is the product of force and the time elapsed on impact
I = Ft
I is the impulse
F is force
t is time
For this case,
Impulse = 3900 x 0.55 = 2145 kg-m/s
If the initial velocity was 2.95 m/s
and mass of car plus driver is 190 kg
neglecting friction, the initial momentum of the car is given as
P = mv1
where P is the momentum
m is the mass of the car and driver
v1 is the initial velocity of the car
initial momentum of the car P = 2.95 x 190 = 560.5 kg-m/s
We know that impulse is equal to the change of momentum, and
change of momentum is initial momentum minus final momentum.
The final momentum = mv2
where v2 is the final momentum of the car.
The problem translates into the equation below
I = mv1 - mv2
imputing values, we have
2145 = 560.5 - 190v2
solving, we have
2145 - 560.5 = -190v2
1584.5 = -190v2
v2 = -1584.5/190 = -8.34 m/s
A particle moving along the x axis has a position given by x = 54t - 2.0t3 m. At the time t = 3.0 s, the speed of the particle is zero. Which statement is correct?
Answer:
v=54-6t^2, 54-6 (9)=0
Suppose a particle moves back and forth along a straight line with velocity v(t), measured in feet per second, and acceleration a(t). What is the meaning of ^120∫60 |v(t)| dt?
Answer:
The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Explanation:
We are told that the particle moves back and forth along a straight line with velocity v(t).
Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.
Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
A box of mass 115 kg sits on an inclined surface with an angle of 59. What is the component of the weight of the box along the surface?
Answer:
966 N
Explanation:
For computing the component of the weight we first need to compute the object weight which is shown below:
Weight = (115 kg)(9.8 m/s²)
= 1,127 N
Now the weight of the box along the surface with an angle of 59 is
= (1,127 N) (sin 59 degree)
= 966 N
Hence, the weight component of the box along the surface is 966 N
A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string, what will be the wave's speed? A second wave is sent in the string, what is the new speed of each of the two waves?
Answer:
a
The speed of wave is [tex]v_1 = 129.1 \ m/s[/tex]
b
The new speed of the two waves is [tex]v = 129.1 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the string is [tex]m = 60 \ g = 60 *10^{-3} \ kg[/tex]
The length is [tex]l = 2.0 \ m[/tex]
The tension is [tex]T = 500 \ N[/tex]
Now the velocity of the first wave is mathematically represented as
[tex]v_1 = \sqrt{ \frac{T}{\mu} }[/tex]
Where [tex]\mu[/tex] is the linear density which is mathematically represented as
[tex]\mu = \frac{m}{l}[/tex]
substituting values
[tex]\mu = \frac{ 60 *10^{-3}}{2.0 }[/tex]
[tex]\mu = 0.03\ kg/m[/tex]
So
[tex]v_1 = \sqrt{ \frac{500}{0.03} }[/tex]
[tex]v_1 = 129.1 \ m/s[/tex]
Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )
A sound wave of frequency 162 Hz has an intensity of 3.41 μW/m2. What is the amplitude of the air oscillations caused by this wave? (Take the speed of sound to be 343 m/s, and the density of air to be 1.21 kg/m3.)
Answer:
I believe it is 91
Explanation:
A conventional current of 3 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.093 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
The current in the small radius loop must be 0.9677 A
Explanation:
Recall that the formula for the magnetic field at the center of a loop of radius R which runs a current I, is given by:
[tex]B=\mu_0\,\frac{I}{2\,R}[/tex]
therefore for the first loop in the problem, that magnetic field strength is:
[tex]B=\mu_0\,\frac{I}{2\,R} =\mu_0\,\frac{3}{2\,(0.093)} =16.129\,\mu_{0}\,[/tex]
with the direction of the magnetic field towards the plane.
For the second smaller loop of wire, since the current goes counterclockwise, the magnetic field will be pointing coming out of the plane, and will subtract from the othe field. In order to the addition of these two magnetic fields to be zero, the magnitudes of them have to be equal, that is:
[tex]16.129\,\,\mu_{0}=\mu_0\,\frac{I'}{2\,R'} =\mu_{0}\,\frac{I'}{2\,(0.03)} \\I'=16.129\,(2)\,(0.03)=0.9677\,\,Amps[/tex]
A horizontal uniform meter stick is supported at the 50.0 cm mark. It has a mass of 0.52 kg, hanging from it at the 20.0 cm mark and a mass of 0.31 kg mass hanging from the 60.0 cm mark. Determine the position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance. Group of answer choices
Answer: 70.5 cm
Explanation:
The position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance will be at the side of 0.31kg.
You will use the moment techniques.
That is,
Sum of the clockwise moment = sum of anticlockwise moments
Please find the attached file for the remaining explanation and solution.
At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a second point in the line, 18.5 m lower than the first, if the pipe diameter at the second point is twice that at the first. Remember that the density of water is 1000 kg/m3. Please give your answer in units of kPa.
Answer:
The pressure at point 2 is [tex]P_2 = 254.01 kPa[/tex]
Explanation:
From the question we are told that
The speed at point 1 is [tex]v_1 = 3.57 \ m/s[/tex]
The gauge pressure at point 1 is [tex]P_1 = 68.7kPa = 68.7*10^{3}\ Pa[/tex]
The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]
Let the height at point 1 be [tex]h_1[/tex] then the height at point two will be
[tex]h_2 = h_1 - 18.5[/tex]
Let the diameter at point 1 be [tex]d_1[/tex] then the diameter at point two will be
[tex]d_2 = 2 * d_1[/tex]
Now the continuity equation is mathematically represented as
[tex]A_1 v_1 = A_2 v_2[/tex]
Here [tex]A_1 , A_2[/tex] are the area at point 1 and 2
Now given that the are is directly proportional to the square of the diameter [i.e [tex]A= \frac{\pi d^2}{4}[/tex]]
which can represent as
[tex]A \ \ \alpha \ \ d^2[/tex]
=> [tex]A = c d^2[/tex]
where c is a constant
so [tex]\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}[/tex]
=> [tex]\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}[/tex]
=> [tex]A_2 = 4 A_1[/tex]
Now from the continuity equation
[tex]A_1 v_1 = 4 A_1 v_2[/tex]
=> [tex]v_2 = \frac{v_1}{4}[/tex]
=> [tex]v_2 = \frac{3.57}{4}[/tex]
[tex]v_2 = 0.893 \ m/s[/tex]
Generally the Bernoulli equation is mathematically represented as
[tex]P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2[/tex]
So
[tex]P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )[/tex]
=> [tex]P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )[/tex]
substituting values
[tex]P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )[/tex]
[tex]P_2 = 254.01 kPa[/tex]
A tungsten filament used in a flashlight bulb operates at 0.20 A and 3.0 V. If its resistance at 20°C is 1.5Ω, what is the temperature of the filament when the flashlight is on?
Answer:
The temperature of the filament when the flashlight is on is 2020 °C.
Explanation:
The resistivity varies linearly with temperature:
[tex] R = R_{0}[1 + \alpha*(T-T_{0})] [/tex] (1)
Where:
T: is the temperature of the filament when the flashlight is on=?
T₀: is the initial temperature = 20 °C
α: is the temperature coefficient of resistance = 0.0045 °C⁻¹
R₀: is the resistance at T₀ = 1.5 Ω
When V = 3.0 V, R is:
[tex]R = \frac{V}{I} = \frac{3.0 V}{0.20 A} = 15 \Omega[/tex]
By solving equation (1) for T we have:
[tex]T = \frac{R-R_{0}}{\alpha*R_{0}} + T_{0} = \frac{15-1.5}{0.0045*1.5} + 20 = 2020 ^{\circ} C[/tex]
Therefore, the temperature of the filament when the flashlight is on is 2020 °C.
I hope it helps you!
Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank positive forces as larger than negative forces.
1. q1=+1nC
q2=-1nC
q3 =-1nC
2. q1= -1nC
+ q2 = + 1nC
q3= +1nC
3. q1= +1nC
q2= +1nC
q3= +1nC
4. q1= +1nC
q2= + 1nC
q3= -1nC
5. q1= -1nC
q2= - 1nC
q3= -1nC
6. q1=+1nC
q2=-1nC
q3 =+1nC
Answer:
Plss see attached file
Explanation:
A "590-W" electric heater is designed to operate from 120-V lines.
A)What is its operating resistance?
b)What current does it draw?
c)If the line voltage drops to 110 V, what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.)
d)The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in the previous part?
a. It will be smaller. The resistance will be smaller so the current drawn will increase, decreasing the power.
b. It will be smaller. The resistance will be smaller so the current drawn will decrease, decreasing the power.
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
d. It will be larger. The resistance will be smaller so the current drawn will decrease, increasing the power.
Answer:
a) 24.4 Ω
b) 4.92 A
c) 495.9 W
d)
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
Explanation:
b)
The formula for power is:
P = IV
where,
P = Power of heater = 590 W
V = Voltage it takes = 120 V
I = Current Drawn = ?
Therefore,
590 W = (I)(120 V)
I = 590 W/120 V
I = 4.92 A
a)
From Ohm's Law:
V = IR
R = V/I
Therefore,
R = 120 V/4.92 A
R = 24.4 Ω
c)
For constant resistance and 110 V the power becomes:
P = V²/R
Therefore,
P = (110 V)²/24.4 Ω
P = 495.9 W
d)
If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.