A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
Answers
Explanation:
Given
Acceleration of the pebble is
At t=0, velocity is
considering horizontal motion
[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]
Velocity acquired during this time
[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]
Consider vertical motion
[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]
Net velocity is
[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]
Angle made is
[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]
Related Questions
Part B
What is the approximate amount of thrust you need to apply to the lander to keep its velocity roughly constant? Explain why, using Newton's first
law of motion.
Answers
Answer:
Force is zero.
Explanation:
According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.
F = m a
if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.
Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.
Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.
If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
spring? Show MATH, answer and unit.
Answers
Answer:
17.8cm
Explanation:
1.3kg --> 4cm
1kg --> 3, 1/13cm
5.8kg --> 18.8cm
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x-axis.
Answers
Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
What are stepdown transformers used for
Answers
Answer:
Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.
Explanation:
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If one lawn mower causes an 80-dB sound level at a point nearby, four lawnmowers together would cause a sound level of ____________ at that point. a.92 dB b.84 dB c.86 dB d.none of the above
Answers
Answer:
The intensity of 4 lawn movers is 86 dB.
Explanation:
Intensity of one lawnmower = 80 dB
Let the intensity is I.
Use the formula of intensity
[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]
Now the intensity of 4 lawn movers is
[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]
Four toy racecars are racing along a circular race track. The cars start at the 3-o'clock position and travel CCW along the track. Car A is constantly 2 feet from the center of the race track and travels at a constant speed. The angle Car A sweeps out increases at a constant rate of 1 radian per second.
Required:
How many radians θ does car A sweep out in t seconds?
Answers
Answer:
in t seconds, Car A sweep out t radian { i.e θ = t radian }
Explanation:
Given the data in the question;
4 toy racecars are racing along a circular race track.
They all start at 3 o'clock position and moved CCW
Car A is constantly 2 feet from the center of the race track and moves at a constant speed
so maximum distance from the center = 2 ft
The angle Car A sweeps out increases at a constant rate of 1 radian per second.
Rate of change of angle = dθ/dt = 1
Now,
since dθ/dt = 1
Hence θ = t + C
where C is the constant of integration
so at t = 0, θ = 0, the value of C will be 0.
Hence, θ = t radian
Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }
A body of mass 4kg is moving with a velocity of 108km/h . find the kenetic energy of the body.
Answers
Answer:
KE = 2800 J
Explanation:
Usually a velocity is expressed as m/s. Then the energy units are joules.
[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]
v = 30 m / sec
KE = 1/2 * 4 * (30)^2
KE =2800 kg m^2/sec^2
KE = 2800 Joules
Help me plssssssss cause I’m struggling
Answers
Answer:
I am pretty sure it is C
Explanation:
It can be found all over the universe
Thermometers and Temperature Scales
While traveling outside the United States, you feel sick. A companion gets you a thermometer, which says your temperature is 40.9. What scale is that on? What is your Fahrenheit temperature? Should you seek medical help?
Answers
Answer:
105.62°F
Explanation:
When the body temperature having fever is measured to be 40.9 on a scale then it must be a Celsius scale thermometer because 37°C is the normal temperature of a healthy human. In case of fever the given temperature is measured on a standard Celsius scale.
The relation between Fahrenheit and Celsius scale is:
[tex]\frac{C}{5}=\frac{F-32}{9}[/tex]
[tex]F=\frac{9C}{5} +32[/tex]
[tex]F=105.62^{o}F[/tex]
It is a high fever and an immediate medical help must be taken.
A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance d by a rope supplying a force F at an angle of elevation q. The surface has a frictional force acting during this motion. How much work was done by friction during this motion? The student calculates the value to be –Fd sinq. How does this value compare to the correct value?
a. It is the correct value.
b. It is too high.
c. It is too low.
d. The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Answers
Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
What is the minimum angular spread (in rad) of a 534 nm wavelength manganese vapor laser beam that is originally 1.19 mm in diameter
Answers
Answer:
Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad
Explanation:
GIven;
Wavelength of manganese vapor laser beam = 534 nm = 534 x 10⁻⁹ m
Diameter = 1.19 mm = 1.19 x 10⁻³ m
Find:
Minimum angular spread (in rad)
Computation:
Minimum angular spread (in rad) = 1.22[Wavelength / Diameter]
Minimum angular spread (in rad) = 1.222[(534 x 10⁻⁹) / (1.19 x 10⁻³)]
Minimum angular spread (in rad) = 2[448.73 x 10⁻⁶]
Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answers
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Answers
Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
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Two sinusoidal waves have the same frequency and wavelength. The wavelength is 20 cm. The two waves travel from their respective sources and reach the same point in space at the same time, resulting in interference. One wave travels a larger distance than the other. For each of the possible values of that extra distance listed below, identify whether the extra distance results in maximum constructive interference, maximum destructive interference, or something in-between.
a. 10 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
b. 15 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
c. 20 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
d. 30 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
e. 35 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
f. 40 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
Answers
Answer:
Explanation:
When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.
For constructive interference , path diff = n λ
For destructive interference path diff = ( 2n+ 1 ) λ /2
where λ is wave length of wave , n is an integer.
a )
path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.
b )
path diff = 15 cm which is neither half the wavelength nor full wavelength , so in between is the right option.
c )
path diff = 20 cm which is equal to the wavelength , so maximum constructive interference will occur.
d)
path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.
e)
path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.
f )
path diff = 40 cm which is 2 times the wavelength , so maximum constructive interference will occur
Two pendulums have the same dimensions (length {L}) and total mass (m). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar.
1. Find the period of pendulum A for small oscillations.
2. Find the period of pendulum B for small oscillations.
Answers
Answer:
1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].
Explanation:
1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:
[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)
Where:
[tex]l[/tex] - Length of the massless bar.
[tex]g[/tex] - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)
Where:
[tex]m[/tex] - Total mass of the pendulum.
[tex]g[/tex] - Gravitational acceleration.
[tex]l[/tex] - Length of the uniform bar.
[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]
[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)
By applying (3) in (2) we get the following expression:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]
[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]
[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]
1. The period of pendulum A for small oscillations is
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
2. The period of pendulum B for small oscillations.
[tex]T_B=1.137.T_A[/tex]
What is simple harmonic motion?Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.
1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
Where:
l - Length of the massless bar.
g - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2
Where:
m - Total mass of the pendulum.
g - Gravitational acceleration.
l - Length of the uniform bar.
Io- Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]
[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3
By applying (3) in (2) we get the following expression:
[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]
[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]
[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]
[tex]TB=1.137.T_A[/tex]
Thus to know more about Simple harmomnic motion follow
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A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east
Answers
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex] Filling in:
[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
(b) Name the devices used to measure the volume of liquid.
Answers
Answer:
Liquid volume is usually measured using either a graduated cylinder or a buret. As the name implies, a graduated cylinder is a cylindrical glass or plastic tube sealed at one end, with a calibrated scale etched (or marked) on the outside wall.
A system is acted on by its surroundings in such a way that it receives 50 J of heat while simultaneously doing 20 J of work. What is its net change in internal energy
Answers
Answer:
30J
Explanation:
Given data
The total quantity of heat recieved= 50J
Quantity of heat used to do work= 20J
Hence the net change is
ΔU= Total Heat - Net work
ΔU= 50-20
ΔU= 30J
Hence the change in the internal energy is 30J
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath. The temperature of the water bath is monitored, and it is determined from this data that 133.0 kJ of heat flows into the system during the reaction. The position of the piston is also monitored, and it is determined from the data that the piston does 241.0 kJ of work on the system during the reaction.
a. Does the temperature of the water bath go up or down?
b. Does the piston move in or out?
c. Does heat flow into or out of the gaseous mixture?
d. How much heat flows?
Answers
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answers
Answer:
Average Velocity = 3.63 m/s
Explanation:
First, we will calculate the total displacement of the quarterback, taking forward direction as positive:
Total Displacement = 15 m - 3 m + 24 m = 36 m
Now, we will calculate the total time taken for this displacement:
Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s
Therefore, the average velocity will be:
[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]
Average Velocity = 3.63 m/s
Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then the power transferred to the turbine will be: a) 100 KW b) 10 KW c) 1 KW d) 100 W
Answers
Answer:
Natae Si Jordan Kaya Sya Napaihe
Explanation:
haha
A 9.0 V battery is connected across two resistors in series. If the resistors have resistances of and what is the voltage drop across the resistor?
Select one:
A. 4.6 V B. 9.4 V C. 8.6 V D. 4.4 V
Answers
Answer:
the answer to the question is known as D
3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)
Answers
Answer:
150 N
Explanation:
Given that,
Acceleration (a) = 3 m/s²Mass of the bike (m) = 50 kgWe are asked to calculate force required.
[tex]\longrightarrow[/tex] F = ma
[tex]\longrightarrow[/tex] F = (50 × 3) N
[tex]\longrightarrow[/tex] F = 150 N
An electron has an initial speed of 8.06 x10^6 m/s in a uniform 5.60 x 10^5 N/C strength electic field.The field accelerates the electron in the direction opposite to its initial velocity.
(a) What is the direction of the electric field?
i. opposite
ii. direction to the electron's initial velocity
iii. same direction as the electron's initial velocity
iv. not enough information to decide
(b) How far does the electron travel before coming to rest? m
(c) How long does it take the electron to come to rest? s
(d) What is the electron's speed when it returns to its starting point?
Answers
Answer:
Explanation:
a)
The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .
Hence direction of electric field must be in the same in which electrons travels.
Hence option iii is correct.
b )
deceleration a = force / mass
= qE / m
= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹
= .98 x 10²⁰ m /s²
v² = u² - 2 a s
0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s
s = 64.96 x 10¹² / 1.96 x 10²⁰
= 33.14 x 10⁻⁸ m
c ) time required
= 8.06 x 10⁶ / .98 x 10²⁰
= 8.22 x 10⁻¹² s .
d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .
Answer:
(a) Option (i)
(b) 6.6 x 10^-4 m
(c) 8.2 x 10^-11 s
Explanation:
initial velocity, u = 8 .06 x 10^6 m/s
Electric field, E = 5.6 x 10^5 N/C
(a) The direction of field is opposite.
Option (i).
(b) Let the distance is s.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times \frac{qE}{m}\times s\\\\8.06\times 10^6\times 8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} s\\\\s = 6.6\times 10^{-4} m[/tex]
(c) Let the time is t.
Use first equation of motion.
[tex]v = u + a t \\\\0 = u - \times \frac{qE}{m}\times t\\\\8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} t\\\\t = 8.2\times 10^{-11} s[/tex]
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answers
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Three 30 g metal balls, one of aluminum, copper and lead, are placed in a large beaker of hot water for a few minutes. [The specific heats of aluminum, copper, and lead are 903, 385, and 130 J / (kg ° C), respectively].
to. Which of the balls, if any, will reach the highest temperature? Explain.
b. Which of the balls, if any, will have the most heat energy? Explain.
Answers
Answer:
The answer is below
Explanation:
Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.
a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.
Hence lead will have the highest temperature since it has the least specific heat capacity.
b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.
Hence aluminum will have the most heat since it has the highest specific heat capacity.
An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC
Answers
Explanation:
Given:
L = 0.02 H
C = [tex]2\:\mu \text{F}[/tex]
f = 200 Hz
The general form of the impedance Z is given by
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to
[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]
[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]
A seesaw made of a plank of mass 10.0 kg and length 3.00 m is balanced on a fulcrum 1.00 m from one end of the plank. A 20.0-kg mass rests on the end of the plank nearest the fulcrum. What mass must be on the other end if the plank remains balanced?
Answers
Answer:
7.5 kg
Explanation:
We are given that
[tex]m_1=10 kg[/tex]
Length of plank, l=3 m
Distance of fulcrum from one end of the plank=1 m
[tex]m_2=20 kg[/tex]
We have to find the mass must be on the other end if the plank remains balanced.
Let m be the mass must be on the other end if the plank remains balanced.
In balance condition
[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]
[tex]20=10(0.5)+2m[/tex]
[tex]20=5+2m[/tex]
[tex]2m=20-5=15[/tex]
[tex]\implies m=\frac{15}{2}[/tex]
[tex]m=7.5 kg[/tex]
Hence, mass 7.5 kg must be on the other end if the plank remains balanced.
Answer:
The mass at the other end is 7.5 kg.
Explanation:
Let the mass is m.
Take the moments about the fulcrum.
20 x 1 = 10 x 0.5 + m x 2
20 = 5 + 2 m
2 m = 15
m = 7.5 kg
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0 m and Ay 450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0, but the magnitude (450.0 m) and direction of vector remain unchanged, as in drawing b. What are the scalar components, Ax and Ay, of the vector in the rotated x, y coordinate system
Answers
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answers
Complete question:
A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answer:
The tension on the rope is 1.95 N
Explanation:
The general equation of a progressive wave is given as;
[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]
Compare the given equation with the general equation of wave, the following parameters will be deduced.
A = 0.75 cm
k = 0.400π cm⁻¹
ω = 250π s⁻¹
The frequency of the wave is calculated as;
ω = 2πf
2πf = 250π
2f = 250
f = 250/2
f = 125 Hz
The wavelength of the wave is calculated as;
[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]
The velocity of the wave is calculated as;
v = fλ
v = 125 x 0.05
v = 6.25 m/s
The tension on the rope is calculated as;
[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]
Therefore, the tension on the rope is 1.95 N