Answer:
Explanation:
From the question we are told that
The initial velocity is [tex]u = 100 m/s[/tex]
The time taken is [tex]t = 2.0 s[/tex]
The charge to mass ratio is [tex]Q/m = 0.100 C/kg[/tex]
Generally the acceleration is mathematically evaluated as
[tex]a = \frac{u}{t }[/tex]
substituting values
[tex]a = \frac{100}{2}[/tex]
[tex]a = 50 \ m/s^2[/tex]
The electric field is mathematical represented as
[tex]E = \frac{(a+g)}{Q/m}[/tex]
substituting values
[tex]E = \frac{(50+9.8)}{0.100}[/tex]
[tex]E = 598 \ N/C[/tex]
distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the pote my jobntA total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm (b) 2ial at the following distances from the center of the sphere: (a) 48.0 cm (b) 24.0 cm (c) 12.0 cm
Answer:
(a) V = 65.625 Volts
(b) V = 131.25 Volts
(c) V = 131.25 Volts
Explanation:
Recall that:
1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:
[tex]V=k\frac{Q}{R}[/tex]
2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:
[tex]V=k\frac{Q}{r}[/tex]
where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.
Then, at a distance of:
(a) 48 cm = 0.48 m, the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]
(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:
[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
Answer:
c) a difference in electric potential
Explanation:
my insta: priscillamarquezz
WILL MARK THE BRAINLIEST!!! The diagram shows a carrier wave that is used to transmit information. Which best illustrates how the carrier wave would likely appear after pulse modulation?
Answer:
Second image in your list of possible answers
Explanation:
The second option is what you would expect from modulating a sinusoidal carrier wave of higher frequency after being modulated by a square pulse of lower frequency that allows part of the carrier signal to travel during the time the square signal is constant different from zero, and be absent (flat) during the time the square pulse signal has amplitude zero.
The second line is the best picture of a pulse-modulated carrier.
It would be easy to build a circuit where each pulse ... when it comes along ... just switches the carrier OFF for as long as the pulse lasts.
You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 220 rad/s .
Part A: What is the impedance of the circuit? ( Answer: Z = ? Ω )
Part B: What is the current amplitude? ( Answer: I = ? A )
Part C: What is the voltage amplitude across the resistor? ( Answer: VR = ? V )
Part D: What is the voltage amplitudes across the inductor? ( Answer: VL = ? V )
Part E: What is the phase angle ϕ of the source voltage with respect to the current? ( Answer: ϕ = ? degrees )
Part F: Does the source voltage lag or lead the current? ( Answer: the voltage lags the current OR the voltage leads the current )
Answer:
A. Z = 185.87Ω
B. I = 0.16A
C. V = 1mV
D. VL = 68.8V
E. Ф = 30.59°
Explanation:
A. The impedance of a RL circuit is given by the following formula:
[tex]Z=\sqrt{R^2+\omega^2L^2}[/tex] (1)
R: resistance of the circuit = 160-Ω
w: angular frequency = 220 rad/s
L: inductance of the circuit = 0.430H
You replace in the equation (1):
[tex]Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega[/tex]
The impedance of the circuit is 185.87Ω
B. The current amplitude is:
[tex]I=\frac{V}{Z}[/tex] (2)
V: voltage amplitude = 30.0V
[tex]I=\frac{30.0V}{185.87\Omega}=0.16A[/tex]
The current amplitude is 0.16A
C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:
[tex]V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV[/tex] (3)
D. The voltage across the inductor is:
[tex]V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V[/tex]
E. The phase difference is given by:
[tex]\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°[/tex]
What is the length of the shadow cast on the vertical screen by your 10.0 cm hand if it is held at an angle of θ=30.0∘ above horizontal? Express your answer in centimeters to three significant figures. View Available Hint(s)
Answer:
The length is [tex]D = 5 \ cm[/tex]
Explanation:
From the question we are told that
The length of the hand is [tex]l = 10.0 \ cm[/tex]
The angle at the hand is held is [tex]\theta = 30 ^o[/tex]
Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as
[tex]D = l * sin(\theta )[/tex]
substituting values
[tex]D = 10 * sin (30)[/tex]
[tex]D = 5 \ cm[/tex]
A 54.0 kg ice skater is moving at 3.98 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.802 m around the pole.
(a) Determine the force exerted by the horizontal rope on her arms.N
(b) What is the ratio of this force to her weight?(force from part a / her weight)
Answer:
(a) force is 1066.56N
Explanation:
(a) MV²/R
A mass of a spring vibrates in simple harmonic motion at a frequency of 8.0 Hz and an amplitude of 3.9 cm. If a timer is started with its displacement is a maximum, what is the speed of the mass when the timer reads 3 seconds
Answer:
e = 50.27 give / s
Explanation:
The expression for simple harmonic motion is
x = A cos (wt + Ф)
in this case they give us the amplitude A = 3.9 cm and frequency f = 8.0 Hz
The angular and linФear variables are related
e = 2π d
e = 2π 8
e = 50.27 give / s
let's look for the constant fi
so let's find the time to have the maximum displacement
v = dx / dt
v = -A w sin (wt +Ф)
for the point of maximum displacement the speed is I think
0 = - sin (0 + Ф)
therefore fi = 0
Let's put together the equation of motion
x = 0.039 sin (50.27 t)
v = 0.039 50.27 sin (50.27 3)
v = 1.96 50 0.01355
v = 0.0266 m / s
A square copper plate, with sides of 50 cm, has no net charge and is placed in a region where there is a uniform 80 kN / C electric field directed perpendicular to the plate. Find a) the charge density of each side of the plate and b) the total load on each side.
Answer:
a) ±7.08×10⁻⁷ C/m²
b) 1.77×10⁻⁷ C
Explanation:
For a conductor,
σ = ±Eε₀,
where σ is the charge density,
E is the electric field,
and ε₀ is the permittivity of space.
a)
σ = ±Eε₀
σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)
σ = ±7.08×10⁻⁷ C/m²
b)
σ = q/A
7.08×10⁻⁷ C/m² = q / (0.5 m)²
q = 1.77×10⁻⁷ C
The time constant of a simple RL circuit is defined as _______. We say that R is the resistance of the circuit and L is the inductance of the circuit.
Answer:
The correct answer will be "[tex]\tau =\frac{L}{R}[/tex]".
Explanation:
The time it would take again for current or electricity flows throughout the circuit including its LR modules can be connected its full steady-state condition is equal to approximately 5[tex]\tau[/tex] as well as five-time constants.
It would be calculated in seconds by:
⇒ [tex]\tau=\frac{L}{R}[/tex]
, where
R seems to be the resistor function in ohms.L seems to be the inductor function in Henries.When you release the mass, what do you observe about the energy?
Explanation:
Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.
A pendulum that has a period of 2.67000 s and that is located where the acceleration due to gravity is 9.77 m/s2 is moved to a location where it the acceleration due to gravity is 9.81 m/s2. What is its new period? (Enter your answer in seconds and to at least 5 decimal places.)
Answer:
Explanation:
Expression for time period of pendulum is given as follows
[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]
where l is length of pendulum and g is acceleration due to gravity .
Putting the given values for first place
[tex]2.67=2\pi\sqrt{\frac{l}{9.77} }[/tex]
Putting the values for second place
[tex]T=2\pi\sqrt{\frac{l}{9.81} }[/tex]
Dividing these two equation
[tex]\frac{T}{2.67} =\sqrt{\frac{9.77}{9.81} }[/tex]
T = 2.66455 s.
a beam of 1mev electrons strike a thick target. for a beam current of 100 microampere, find the power dissipated in the target
Answer:
power dissipated in the target is 100 W
Explanation:
given data
electrons = 1 mev = [tex]10^{6}[/tex] eV
1 eV = 1.6 × [tex]10^{-19}[/tex] J
current = 100 microampere = 100 × [tex]10^{-6}[/tex] A
solution
when energy of beam strike with 1 MeV so energy of electron is
E = e × v ...................1
e is charge of electron and v is voltage
so put here value and we get voltage
v = 1 ÷ 1.6 × [tex]10^{-19}[/tex]
v = [tex]10^{6}[/tex] volt
so power dissipated in target
P = voltage × current ..............2
put here value
P = [tex]10^{6}[/tex] × 100 × [tex]10^{-6}[/tex]
P = 100 W
so power dissipated in the target is 100 W
To practice Problem-Solving Strategy 15.2 Standing Waves.In lab, your instructor generates a standing wave using a thin string of length L = 2.25 m fixed at both ends. You are told that the standing wave is produced by the superposition of traveling and reflected waves, where the incident traveling waves propagate in the +x direction with an amplitude A = 2.45 mm and a speed vx = 14.5 m/s . The first antinode of the standing wave is a distance of x = 37.5 cm from the left end of the string, while a light bead is placed a distance of 18.8 cm to the right of the first antinode. What is the maximum transverse speed vy of the bead? Make sure to use consistent distance units in your calculations.To check your equation for the standing wave's transverse velocity, find the maximum transverse velocity at x = 75.0 cm .
Answer:
Explanation:
Let the equation of standing wave be as follows
y = A sinωt cos kx
A = 2.45 mm
y = 2.45 cosωt sin kx
given
[tex]\frac{\omega}{k}[/tex] = velocity = 14.5
Position of first antinode = 37.5 cm
kx = π / 2
k x 37.5 = π / 2
k = π / 75
ω / k = 14.5
ω = 14.5 x π / 75
= .607 rad /s
Maximum transverse speed
= ω A
= .607 x 2.45
= 1.49 mm / s
y = A sinωt cos kx
Transverse velocity
v = dy / dt
= ω A cosωt cos kx
Maximum transverse velocity at any x = ω A .
The characteristics of standing waves allows find the results for the questions about the speed of the rope are:
The transverse oscillatory velocity is: vy = 0.298 m / sGiven parameters
The amplitude of the wave A = 2.45 mm = 2.45 10-3 m Chord length L = 2.25 m Wave velocity v = 14.5 m / d The first antinode x = 37.5 cmTo find
Maximum rope swing speed.
The movement in a string is formed by two movements, a movement in the direction of the string with constant speed and a transverse movement where the speed varies as in a simple harmonic movement.
The standing wave is formed from the sum of the incident wave and the reflected wave.
y = A cos (kx- wt)
y = A cos (kx + wt)
resulting
y = A sin wt cos kx
the speed that the wave is given by
v = w / k
They indicate the position of the first antinode at this point the cosine function must be maximum.
kx = π
k = π/x
k = [tex]\frac{\pi }{0.375}[/tex]
k = 8.38 m⁻¹
let's find the angular velocity.
w = v k
w = 14.5 8.38
w = 121.5 rad / s
The expression for displacement in simple harmonic motion is:
x = A cos wt
The speed is defined by the variation of the position with respect to time.
v = [tex]\frac{dx}{dt}[/tex] =
v = - A w sin wt
To calculate the maximum speed we make the sine equal to 1.
[tex]v_{y \ max}[/tex] = w A
[tex]v_{y \ max}[/tex] = 121.5 2.45 10⁻³
[tex]v_{y \ max}[/tex] = 0.298 m / s
For point x = 75 cm = 0.750 m
We seek the value of
kx = 8.38 0.750
kx = 6.285 = 2π
therefore this point is also an antinode and the results do not change.
In conclusion, using the characteristics of standing waves, we can find the results for the questions about the speed of the rope are:
The transverse oscillatory velocity is: [tex]v_y[/tex] = 0.298 m / s
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Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.547 T.
Answer:
Wavelength is 0.359 m
Explanation:
Given that,
Magnetic field, B = 0.547 T
We need to find the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.547 T.
The frequency of revolution of proton in the cyclotron is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
m is mass of proton
q is charge on proton
So,
[tex]f=\dfrac{1.6\times 10^{-19}\times 0.547}{2\pi \times 1.67\times 10^{-27}}\\\\f=8.34\times 10^6\ Hz[/tex]
We know that,
Speed of light, [tex]c=f\lambda[/tex]
[tex]\lambda[/tex] = wavelength
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{834\times 10^6}\\\\\lambda=0.359\ m[/tex]
So, the wavelength of the radiation produced by a proton is 0.359 m.
Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2 . A constant horizontal force F to the right is applied to m1 . What is the horizontal force acting on m2?
Answer:
The horizontal force acting on m2 is F + 9.8m1
Explanation:
Given;
Block m1 on left of block m2
Make a sketch of this problem;
F →→→→→→→→→→→-------m1--------m2
Apply Newton's second law of motion;
F = ma
where;
m is the total mass of the body
a is the acceleration of the body
The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1
F₂ = F + W1
F₂ = F + m1g
F₂ = F + 9.8m1
Therefore, the horizontal force acting on m2 is F + 9.8m1
The force acting on the block of mass m₂ is [tex]\frac{m_2F}{m_1+m_2}[/tex]
Force acting on the block:Given that there are two blocks of mass m₁ and m₂.
m₁ is on the left of block m₂. They are in contact with each other.
A force F is applied on m₁ to the right.
According to Newton's laws of motion:
The equation of motion of the blocks can be written as:
F = (m₁ + m₂)a
here, a is the acceleration.
so, acceleration:
a = F / (m₁ + m₂)
Now, the force acting on the block of mass m₂ is:
f = m₂a
[tex]f = \frac{m_2F}{m_1+m_2}[/tex]
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When the center of a bicycle wheel has linear velocity relative to the ground, the velocity relative to the ground of point P' at the top of the wheel is
Answer:
The velocity relative to the ground at the top of the wheel is twice the velocity relative to the ground at the center of the wheel.
Explanation:
The velocity relative to the ground is directly proportional to the distance from the point of contact between the ground and the bottom of the wheel, which represents the center of rotation. Hence, the velocity relative to the ground at the top of the wheel is:
[tex]\frac{v_{top}}{v_{center}} = \frac{2\cdot R}{R}[/tex]
Where [tex]R[/tex] is the radius of the wheel.
[tex]v_{top} = 2\cdot v_{center}[/tex]
The velocity relative to the ground at the top of the wheel is twice the velocity relative to the ground at the center of the wheel.
n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnetic field.
Answer:
The corresponding magnetic field is
Explanation:
From the question we are told that
The electric field amplitude is [tex]E_o = 611\ V/m[/tex]
Generally the magnetic field amplitude is mathematically represented as
[tex]B_o = \frac{E_o }{c }[/tex]
Where c is the speed of light with a constant value
[tex]c = 3.0 *0^{8} \ m/s[/tex]
So
[tex]B_o = \frac{611 }{3.0*10^{8}}[/tex]
[tex]B_o = 2.0 4 *10^{-6} \ Vm^{-2} s[/tex]
Since 1 T is equivalent to [tex]V m^{-2} \cdot s[/tex]
[tex]B_o = 2.0 4 *10^{-6} \ T[/tex]
A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, then the final particle energy is proportional to which of the following?
A. 1/RB. RC. R^2D. R^3E. R^4
Answer:
C. R^2
Explanation:
A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.
R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.
A hair dryer draws 1200 W, a curling iron draws 800 W, and an electric light fixture draws 500 W. If all three of these appliances are operating in parallel on a 120-V circuit, what is the total current drawn
Answer:
The Total current drawn is 20.83 Ampere.
Explanation:
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A 0.20-kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of 40 N/m. The block is initially displaced 4.0 cm from the equilibrium point and then released to set up a simple harmonic motion. A frictional force of 0.3 N exists between the block and surface. What is the speed of the block when it passes through the equilibrium point after being released from the 4.0-cm displacement point
Answer:
Approximately [tex]0.45\; \rm m \cdot s^{-1}[/tex].
Explanation:
The mechanical energy of an object is the sum of its potential energy and kinetic energy. Consider this question from the energy point of view:
Mechanical energy of the block [tex]0.04\; \rm m[/tex] away from the equilibrium position:
Elastic potential energy: [tex]\displaystyle \frac{1}{2} \, k\, x^2 = \frac{1}{2}\times \left(0.04\; \rm m\right)^2 \times 40\; \rm N \cdot m^{-1} = 0.032\; \rm J[/tex].Kinetic energy: [tex]0\; \rm J[/tex].While the block moves back to the equilibrium position, it keeps losing (mechanical) energy due to friction:
[tex]\begin{aligned}& \text{Work done by friction} = (-0.3\; \rm N) \times (0.04 \; \rm m) = -0.012\; \rm J\end{aligned}[/tex].
The opposite ([tex]0.012\; \rm N[/tex]) of that value would be the amount of energy lost to friction. Since there's no other form of energy loss, the mechanical energy of the block at the equilibrium position would be [tex]0.032\; \rm N - 0.012\; \rm N = 0.020\; \rm N[/tex].
The elastic potential energy of the block at the equilibrium position is zero. As a result, all that [tex]0.020\; \rm N[/tex] of mechanical energy would all be in the form of the kinetic energy of that block.
Elastic potential energy: [tex]0\; \rm J[/tex].Kinetic energy: [tex]0.020\; \rm J[/tex].Given that the mass of this block is [tex]0.020\; \rm kg[/tex], calculate its speed:
[tex]\begin{aligned}v &= \sqrt{\frac{2\, \mathrm{KE}}{m}} \\ &= \sqrt{\frac{2 \times 0.020\; \rm J}{0.20\; \rm kg}} \approx 0.45\; \rm m\cdot s^{-1}\end{aligned}[/tex].
A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 4.40 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
Answer:
F = umg where u is coefficient of dynamic friction
Explanation:
F = 0.4 x 80 x 9.81 = 313.92 N
Which force does not operate at a distance of 1 m?
O A. Strong nuclear
B. Electric
O C. Gravitational
O D. Magnetic
SUBMI
Answer: A. Strong nuclear
The max effective range of strong nuclear force is about 1.2 femtometers ( which is 1.2*10^(-15) meters). This is well below 1 meter. Strong nuclear forces are the forces that hold together a nucleus. Specifically it holds together the protons that would otherwise repel one another due to similar charge.
Which characteristic gives the most information about what kind of element an atom is ?
Answer:
The atomic number
Explanation:
A student applies a constant horizontal 20 N force to a 12 kg box that is initially at rest. The student moves the box a distance of 3.0 m. What is the speed of the box at the end of the motion
Answer:
u = 10.02m/s
Explanation:
a = f/m
a = 20/12 = 1.67m/s²
U =2aS
u = 2 x 1.67 x 3
U = 10.02m/s
The process by which energy is realized is known as?
Answer:
did you mean released?
Explanation:
If so the process is called respiration
A parallel-plate capacitor has a plate separation of 1.5 mm and is charged to 450 V. 1) If an electron leaves the negative plate, starting from rest, how fast is it going when it hits the positive plate
Answer:
Explanation:
this is the answer to your question
The electron is going with a velocity of 1.25 × 10⁷ m/s when it hits the positive plate.
What is law of the conservation of mechanical energy?According to the law of the conservation of mechanical energy, the total mechanical energy is always conserved by an electron. We can say that the sum of potential energy (U) and kinetic energy (K) is always constant.
K + U = E
Given, the distance between the two parallel plates = 1.5 mm
The potential difference between the plates, V = 450V
The charge on an electron, q = [tex]-1.6\times 10^{-19} C[/tex]
The mass of an electron, m = 9.1× 10⁻³¹ Kg
The change in the potential energy of the charge moving through the potential difference of 450V.
ΔU = qΔV = (-1.6× 10⁻¹⁹)(450) = -7.2 × 10⁻¹⁷J
From the law of the conservation of mechanical energy, we can write:
K + U = E
ΔK + ΔU = 0
ΔK = -ΔU
1/2mv² = -ΔU
v² = -2ΔU/m
[tex]v^2 =\frac{-2\times (-7.2\times 10{-17})}{9.1\times 10^{-31}}[/tex]
[tex]v=\sqrt{1.58\times 10^{14}}[/tex]
v = 1.25 × 10⁷ m/s
Therefore, the electron is going with the speed of 1.25 × 10⁷ m/s when it hits the positive plate.
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A girl weighing 600 N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm under her weight. Find the spring constant and the total work done on it during the compression.
Answer:
The spring constant is 60,000 N
The total work done on it during the compression is 3 J
Explanation:
Given;
weight of the girl, W = 600 N
compression of the spring, x = 1 cm = 0.01 m
To determine the spring constant, we apply hook's law;
F = kx
where;
F is applied force or weight on the spring
k is the spring constant
x is the compression of the spring
k = F / x
k = 600 / 0.01
k = 60,000 N
The total work done on the spring = elastic potential energy of the spring, U;
U = ¹/₂kx²
U = ¹/₂(60000)(0.01)²
U = 3 J
Thus, the total work done on it during the compression is 3 J
A whistle is often used in dog-‐‑training exercises and is an integral part of field-‐‑marking competitions, where signals and commands are relayed to the dog via whistle. Suppose the whistle produces a sound wave with a frequency of 25,100 Hz (outside the range of human hearing) and the propagation speed of sound in air is 343 m/s.
1) What is the wavelength and wave number of this sound from the whistle?
2) Write the wave function for the longitudinal sound wave described above, assuming the amplitude of the sound wave is Smax = 1.57 x 10-‐‑6 m and it is moving to the right (in the positive x direction).
Answer:
The wavelength is [tex]\lambda = 0.01367 m[/tex]
The wave number is [tex]N = 73.18\ m^{-1}[/tex]
The wave function is [tex]y= 1.57 *10^{-6} sin 2 \pi ( 73.178 x -25100t)[/tex]
Explanation:
From the question we are told that
The frequency of the sound wave is [tex]f = 25,100 Hz[/tex]
The speed of the wave is [tex]v = 343 m/s[/tex]
The wavelength of the wave is mathematically evaluated as
[tex]\lambda = \frac{v}{f}[/tex]
substituting values
[tex]\lambda = \frac{343}{25100}[/tex]
[tex]\lambda = 0.01367 m[/tex]
The wave number is mathematically represented as
[tex]N= \frac{1}{\lambda }[/tex]
substituting values
[tex]N = \frac{1}{ 0.01367 }[/tex]
[tex]N = 73.18\ m^{-1}[/tex]
The general form of wave function is
[tex]y= A sin (kx -wt)[/tex]
given that the amplitude is [tex]A = 1.57*10^{-6} \ m[/tex]
While [tex]w[/tex] which is the angular velocity is represented as [tex]w = 2 \pi f[/tex]
and k which is the angular wave number is mathematically represented as [tex]k = \frac{2 \pi }{\lambda }[/tex]
The wave function becomes
[tex]y= A sin 2 \pi (\frac{1}{\lambda} x -ft)[/tex]
substituting values
[tex]y= 1.57 *10^{-6} sin 2 \pi ( 73.178 x -25100t)[/tex]
A horizontal spring with spring constant 290 N/m is compressed by 10 cm and then used to launch a 300 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?
Answer:
Explanation:
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In an experiment to measure the acceleration due to gravity, g two values, 9.96 m/s2 and 9.72 m/s2 , are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value: g 5 9.80 m/s2 .)
Answer:
(1) Percent Difference = 2.47%
(2) Percent Error (9.96 m/s²) = 1.63 %
Percent Error (9.72 m/s²) = 0.82 %
(3) Percent Error (Mean) = 0.41 %
Explanation:
(1)
Percent Difference = [(9.96 m/s² - 9.72 m/s²)/(9.72 m/s²)]*100 %
Percent Difference = 2.47%
(2)
Percent Error = (|Measured Value - Original Value|/Original Value)*100%
Therefore,
Percent Error (9.96 m/s²) = (|9.96 m/s² - 9.8 m/s²|/9.8 m/s²)*100%
Percent Error (9.96 m/s²) = 1.63 %
Now,
Percent Error (9.72 m/s²) = (|9.72 m/s² - 9.8 m/s²|/9.8 m/s²)*100%
Percent Error (9.72 m/s²) = 0.82 %
(3)
First we need to find the mean of values:
Mean = (9.96 m/s² + 9.72 m/s²)/2
Mean = 9.84 m/s²
Therefore,
Percent Error (Mean) = (|9.84 m/s² - 9.8 m/s²|/9.8 m/s²)*100%
Percent Error (Mean) = 0.41 %
The specific heat of a certain type of cooking oil is 1.75J/g°C. How much heat energy is needed to raise the temperature of 2.63kg of this oil from g
Complete question:
The specific heat of a certain type of cooking oil is 1.75J/g°C. How much heat energy is needed to raise the temperature of 2.63 kg of this oil from 23 °C to 191 °C?
Answer:
The heat energy required to raise the temperature of this cooking oil is 773220 J
Explanation:
Given;
mass of the cooking oil, m = 2.63 kg = 2630 g
specific heat capacity of the cooking oil, C = 1.75J/g°C
initial temperature of the cooking oil, T₁ = 23° C
final temperature of the cooking oil, T₂ = 191° C
The heat energy required to raise the temperature of this cooking oil;
Q = mcΔT
Q = 2630 x 1.75 x (191 - 23)
Q = 773220 J
Therefore, the heat energy required to raise the temperature of this cooking oil is 773220 J