Answer:
0.69s
Explanation:
10 cm = 0.1 m
Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be
[tex]\omega = \alpha t = 2.1 t[/tex]
And so the radial acceleration is
[tex]a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2[/tex]
The tangential acceleration is always the same since angular acceleration is constant:
[tex]a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2[/tex]
For these 2 quantities to be the same
[tex]a_r = a_t[/tex]
[tex]0.441 t^2 = 0.21[/tex]
[tex]t^2 = 0.21/0.441 = 0.4762[/tex]
[tex]t = \sqrt{0.4762} = 0.69 s[/tex]
The value of t whereby the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude is; t = 0.69 s
Calculating Radial and Tangential Acceleration
Let the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude be denoted as t. Thus, angular velocity at time (t) is;
ω = αt
where;
α is angular acceleration
We are given;
radius; r = 10 cm = 0.1 mconstant angular acceleration; α = 2.1 rad/s²Thus, ω = 2.1t
Now, we can find the radial acceleration from the formula;
α_r = ω²r
Thus;
α_r = (2.1t)² × 0.1
α_r = 0.441 t² m/s²
The tangential acceleration is gotten from the formula is;
α_t = α × r
α_t = 2.1 × 0.1
α_t = 0.21 m/s²
The condition in the question implies that the tangential acceleration is equal to the radial acceleration. Thus;
α_t = α_r
0.21 = 0.441 t²
t = √(0.21/0.441)
t = 0.69 s
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A 2,537-kg truck moving at 14 m/s strikes a car waiting at a traffic light, hooking bumpers. The two continue to move together at 8 m/s. What was the mass (in kg) of the struck car
Answer:
1902.75 kg
Explanation:
From Law of conservation of momentum,
m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1
make m₂ the subject of the equation,
m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2
Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity
Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)
Substituting into equation 2.
m₂ =[2537(8) - 2537(14)]/(0-8)
m₂ = (20296-35518)/-8
m₂ = -15222/-8
m₂ = 1902.75 kg.
Thus the mass of the car = 1902.75 kg
An egg drops from a second-story window, taking 1.13 s to fall and reaching a speed of 11.1 m/s just before hitting the ground. On contact with the ground, the egg stops completely in 0.140 s. Calculate the average magnitudes of its acceleration while falling and while stopping.
m/s2 (acceleration while falling)
m/s2 (deceleration while stopping)
Answer:
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
Explanation:
Hi there!
The equation of velocity of the falling egg is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time
Let´s calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².
v = v0 + a · t
11.1 m/s = 0 m/s + a · 1.13 s (since the egg is dropped, the initial velocity is zero). Solving for "a":
11.1 m/s / 1.13 s = a
a = 9.82 m/s²
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
Now, using the same equation, let´s find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:
v = v0 + a · t
0 = 11.1 m/s + a · 0.140 s
-11.1 m/s / 0.140 s = a
a = -79.3 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
The average acceleration while the egg was falling is 9.8 m/s² and the average deceleration while the egg was stopping is -79.3 m/s².
Explanation:The subject of the question is Physics, namely kinematics in the field of Mechanics. The main concepts involved are acceleration, velocity, and time. Acceleration is the rate of change of velocity, and it can be calculated as the change in velocity divided by the change in time.
Acceleration while falling: Since the egg is falling, it's subject to gravity. On Earth the acceleration due to gravity is approximately 9.8 m/s². However if this value wasn't given, it could have been calculated using the formula: final velocity divided by the time taken. Hence, 11.1 / 1.13 = 9.8 m/s² (rounded).
Deceleration while stopping: This is calculated in the same way as acceleration, the only difference being that its direction is opposite to the direction of motion hence it's often referred as deceleration. The initial velocity, in this case the speed it was falling just before hitting the ground, will be used as the initial velocity while calculating acceleration after hitting the ground. Therefore (0 - 11.1) / 0.140 = -79.3 m/s², the negative sign indicates deceleration.
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The second-order rate constants for the reaction of oxygen atoms ·with aromatic hydrocarbons have been measured (R. Atkinson and J.N . Pitts, J. Phys. Chem. 79, 295 (1975)). In the reaction with benzene the rate constants are 1.44x 107dm' mor ' s ' at 300.3 K, 3.03 x 107 dm' mol"'s"' at 34 1.2K, and 6.9 x 107 dm' mor's ' at 392.2K. Find the frequency factor and activation energy of the reaction.
Answer: Frequency factor A = 8 × 10⁹
activation energy Ea = 15.5 KJ/Mol
Explanation: to begin, let us first define the parameters given;
K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹
K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹
K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹
also T₁ = 300.3 K
T₂ = 341.2 K
T₃ = 392.2 K
we know that;
㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]
where R is given as 8.314 J/mol-k
Ea = activation energy
K₁, K₂ = rate constant
T₁, T₂ = Temperature
therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]
this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol
∴ Ea = 15.5 KJ/ Mol
also given that K = A e⁻∧Ea/RT
here A = frequency factor
∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)
A = 7.99 × 10⁹ = 8 × 10⁹
Final answer:
The Arrhenius equation can be used to determine the frequency factor and activation energy of a reaction.
Explanation:
The given question requires determining the frequency factor and activation energy for a reaction. The relationship between the rate constant and temperature is described by the Arrhenius equation:
k = Ae-Ea/RT
To find the frequency factor (A), we can rearrange the equation and use the rate constant (k) at a specific temperature and the activation energy (Ea) to solve for A. Similarly, to find the activation energy, we can rearrange the equation and use the rate constants at two different temperatures to solve for Ea.
What is the distance between the motion sensor and the object?
The motion sensor acts as a transmitter, when it sends out pulses, and as a receiver, when it listens for echoes. The delay t between sending a pulse and receiving the echo is equal to 8.8 ms (=0.0088 s). The air temperature id 20 degrees Celsius.
Answer:
1.5092 m
Explanation:
Time taken by the wave for the round trip is 8.8 ms
Velocity of sound at 20 degrees Celsius is 343 m/s
The distance to the object will be given by the one way trip
Time taken for one way trip is
[tex]\dfrac{8.8}{2}=4.4\ ms[/tex]
Distance is given by
[tex]Distance=Speed\times time\\\Rightarrow Distance=343\times 4.4\times 10^{-3}\\\Rightarrow Distance=1.5092\ m[/tex]
The distance between the motion sensor and the object is 1.5092 m
The distance between the motion sensor and the object is 1.5092 m
What is a motion sensor?The motion sensor serves as a transmitter for pulses when it sends out pulses, and as a receiver, when it listens for echoes.
From the parameters given:
The time delay = 0.0088 sThe speed of sound (s) = 343 m/sThe temperature of air = 20° CThe distance traveled by the signal is:
d = s × Δ t
d = 343 m/s × 0.0088s
d = 3.0184 m
Provided that the distance is the circular trip between the motion sensor and the object,
Therefore, we can conclude that the distance between the motion sensor and the object will be equal to half the distance traveled by the signal.
i.e.
= 3.0184 m/2
= 1.5092 m
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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/s2 .A. How much distance is between you and the deer when you come to a stop?
B. What is the maximum speed you could have and still not hit the deer?
Answer:
given,
speed of the car = 20 m/s
final speed of car = 0 m/s
distance between car and the deer = 38 m
reaction time, t = 0.5 s
deceleration of the car = 10 m/s².
a) distance between deer and car
distance travel in the reaction time
d₁ = v x t
d₁ = 20 x 0.5 = 10 m
distance travel after you apply brake
using equation of motion
v² = u² + 2 a s
0 = 20² - 2 x 10 x s
s = 20 m
total distance traveled by the car
D = d₁ + d₂
D = 20 + 10 = 30 m
distance between car and the deer = 38 m - 30 m
= 8 m
b) now, maximum speed car.
distance travel in reaction time
d₁ = s x t
d₁ = 0.5 V
distance left between them
d₂ = 38 - d₁
d₂ = 38 - 0.5 V
distance travel after you apply brake
using equation of motion
v² = u² + 2 a d₂
0 = (V)² - 2 x 10 x (38 - 0.5 V)
V² + 10 V - 760 = 0
now, solving the quadratic equation
[tex]x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}[/tex]
V = 23.01 , -33.01
rejecting the negative term.
hence, maximum speed of the car could be V = 23.01 m/s
Ideal Meters Problem. What internal resistance is ideal for a voltmeter? What internal resistance is ideal for an ammeter?
A. Infinity for both
B. Zero for both
C. Zero for ammeter; infinity for voltmeter
D. Infinity for ammeter; zero for voltmeter
E. None of these
Answer:
C. Zero for ammeter; infinity for voltmeter
Explanation:
Ammeter: A device that is used to measure the current through a circuit.
Voltmeter: A device that is used to measure the electric potential difference between two points in a circuit.
The relation among voltage (v), current (i), and resistance (r) is given as,
[tex]v = i\times r[/tex]
In ideal cases an ammeter must have zero resistance so that there is no voltage drop across it and accurate value of current will be measured.
In case of an ideal voltmeter, the resistance should be infinite so that the current across it is zero.
The ideal internal resistance for a voltmeter is infinity, to prevent it from altering the circuit by drawing current. For an ammeter, it's zero to avoid introducing additional resistance into the circuit. Thus, the correct answer is C: Zero for ammeter; infinity for voltmeter.
Explanation:The ideal internal resistance for a voltmeter and an ammeter varies due to their different roles in electrical measurements. A voltmeter is used to measure the voltage across two points in a circuit and ideally should have an infinite internal resistance to prevent it from drawing any significant current from the circuit.
This is because a voltmeter is connected in parallel with the circuit component, and a high resistance ensures that it does not affect the circuit's operation by introducing a significant parallel path for current. On the other hand, an ammeter is used to measure the current flowing through a circuit and is placed in series with the circuit elements. Therefore, an ammeter should have a zero internal resistance to ensure that it does not introduce any additional resistance into the circuit, which could alter the current it is meant to measure.
If a car is moving with a velocity v, the distance the car takes to come to a halt is ___________ the distance that same car would take if it was starting with a velocity 2v.
Answer:
one forth.
Explanation:
car is moving at the speed of v
car stops, final speed = s
distance to stop the car = ?
using equation of motion
u_i² = u_f² + 2 a s
0² = v² - 2 a s
-ve sign is used because the car is decelerating.
[tex]s = \dfrac{v^2}{2a}[/tex]
now, if the velocity of the car is 2v distance to stop
[tex]s' = \dfrac{(2v)^2}{2a}[/tex]
[tex]s' = 4\dfrac{v^2}{2a}[/tex]
[tex]s' = 4 s[/tex]
[tex]s = \dfrac{s'}{4}[/tex]
now, the distance is one forth.
so, car with speed v has to cover one forth of the distance cover by car with speed 2 v.
Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic meters per second, liters per minute and cubic feet per second.
Answer
given,
discharge rate from pipe = 1000 gallons/minutes
now,
flow rate in cubic meters per second
1 gallon = 0.00378541 m³
1 min = 60 s
Q = [tex]1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}[/tex]
Q = 0.063 m³/s
flow rate in liters per minute
1 gallon = 3.78541 L
Q = [tex]1000\times \dfrac{3.78541\ m^3}{1\ gallon}[/tex]
Q = 3785.41 m³/min
flow rate in cubic feet per second
1 gallon = 0.133681 ft³
1 min = 60 s
Q = [tex]1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}[/tex]
Q = 2.23 ft³/s
A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its centre and perpendicular to its length. A mass m1 = 5.40 kg is attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
Complete Question:
A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
(a) What is the moment of inertia of the system?
(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?
(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?
(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?
Answer:
a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J
Explanation:
a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.
As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.
The total rotational inertia will be as follows:
I = M*L²/12 + m₁*r₁² + m₂*r₂²
⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²
⇒ I = 8.53 kg*m²
b) The rotational kinetic energy of the rigid body composed by the rod and the point masses m₁ and m₂, can be expressed as follows:
Krot = 1/2*I*ω²
if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:
Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²
⇒ Krot = 31.1 J
c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:
I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²
I = 7.90 kg*m²
d) The new rotational kinetic energy will be as follows:
Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²
Krot= 28.8 J
a. The moment of inertia of this system is equal to [tex]8.53\;kgm^2[/tex].
b. The kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s is 31.13 Joules.
c. The moment of inertia of the rod and masses combined, if the rod is considered to be of negligible mass is [tex]7.9\;kgm^2[/tex].
d. The kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s and the rod is considered to be of negligible mass is 28.80 Joules.
Given the following data:
Mass of rod = 1.90 kilograms.Length of rod = 2.00 meters.Mass, [tex]m_1[/tex] = 5.40 kilograms.Mass, [tex]m_2[/tex] = 2.50 kilograms.a. To determine the moment of inertia of this system:
Treating the two masses as point particles, the total moment of inertia of this system is given by the formula:
[tex]I = \frac{ML^2}{12} + m_1 r_1^2 + m_2 r_2^2\\\\I = \frac{1.9 (2)^2}{12} + (5.4 \times 1^2) + (2.5 \times 1^2)\\\\I=\frac{7.6}{12} + 5.4+2.5\\\\I=0.6333+ 7.9\\\\I=8.53\; kgm^2[/tex]
b. To determine the kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s:
Mathematically, the rotational kinetic energy of a system is given by the formula:
[tex]K.E_R = \frac{1}{2} I\omega^2\\\\K.E_R = \frac{1}{2} \times 8.53 \times 2.7^2\\\\K.E_R = 4.27\times 7.29\\\\K.E_R = 31.13\;Joules[/tex]
c. To determine the moment of inertia of the rod and masses combined, if the rod is considered to be of negligible mass:
[tex]I =m_1 r_1^2 + m_2 r_2^2\\\\I = (5.4 \times 1^2) + (2.5 \times 1^2)\\\\I= 5.4+2.5\\\\I=7.9\; kgm^2[/tex]
d. To determine the kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s and the rod is considered to be of negligible mass:
[tex]K.E_R = \frac{1}{2} I\omega^2\\\\K.E_R = \frac{1}{2} \times 7.9 \times 2.7^2\\\\K.E_R = 3.95\times 7.29\\\\K.E_R = 28.80\;Joules[/tex]
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Complete Question:
A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
(a) What is the moment of inertia of the system?
(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?
(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?
(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?
There is a 9.0 earthquake that just hit San Diego! There have been multiple aftershocks at a 6.0 magnitude. How much weaker are these aftershocks compared to the original earthquake?
a. 3000x weaker
b. 1000x weaker
c. 100x weaker
d. 10x weaker
The magnitude is used to quantify the size of the earthquakes (measures the energy released during the breakdown of a fault) while the intensity is a qualitative description of the effects of the earthquakes (it involves the perception of people as well as damage material and economic suffered due to the event). The relation between intensity and magnitude is
[tex]\frac{I_1}{I_2} =10^{(M1-M2)}[/tex]
Here,
M = Magnitude
I = Intensity of each one
Given
[tex]M_1 = 9[/tex]
[tex]M_2 =6[/tex]
Replacing,
[tex]\frac{I_1}{I_2} =10^{(9-6)}[/tex]
[tex]\frac{I_1}{I_2} = 10^3[/tex]
[tex]\frac{I_1}{I_2} = 1000[/tex]
So the aftershocks are 1000x weaker compared to the original earthquake. The correct answer is B.
. The mass flow rate through the nozzle of a rocket engine is 200 kg/s. The areas of the nozzle inlet and exit planes are 0.7 and 2.4 m2, respectively. On the nozzle inlet plane, the pressure and velocity are 1600 kPa and 150 m/s, respectively, whereas on the nozzle exit plane, the pressure and velocity are 80 kPa and 2300 m/s, respectively. Find the thrust force acting on the nozzle.
To solve this problem we will define the data obtained in each of the sections. We know that the Net Force is equivalent to the Force in section 1, which can be found through mass flow and velocity, plus the force in section two, which can be found as the product between pressure and Area, so therefore we have
State 1:
[tex]\dot{m} = 200kg/s[/tex]
[tex]v_1 = 150m/s[/tex]
[tex]P_1 = 1600kPa[/tex]
State 2
[tex]v_2 = 2300m/s[/tex]
[tex]A_2 = 2.4m^2[/tex]
[tex]P_2 = 80kPa[/tex]
We have that net force is equal to
[tex]F_{net} = F_1 + F_2[/tex]
[tex]F_1 = \dot{m} (v_2-v_1)[/tex]
[tex]F_2 = (P_1-p_2)A_2[/tex]
Replacing,
[tex]F = \dot{m} (v_2-v_1) + (P_1-p_2)A_2[/tex]
[tex]F = 200(2300-150)+(1600-80)*10^3(2.4)[/tex]
[tex]F = 4078kN[/tex]
Therefore the thrust force acting on the nozzle is 4078kN
A glow-worm of mass 5.0 g emits red light (650 nm) with a power of 0.10 W entirely in the backward direction. To what speed will it have accelerated after 10 y if released into free space and assumed to live
Answer:
a = Np/10 yrs×[3.50^7 yrs /sec]
Explanation:
The energy of the single photon of frequency f or wave length λ is given as
E = hc / λ
since the glow warm emits energy 0.1 J/sec
that is the number of photons n emitted by the photon per sec will be
n = 0.1 W / E
Thus, the number of photons emitted in 10 years
N = n×3.15×10^7 sec/yr ×10 yr
Now, momentum associated with each photon
p= h / λ
and, momentum associated with N photon particles
P= N(h/λ)
hence the change in the momentum of the glow is = Np in 10 years
Therefore, acceleration of the glow
a = Np/10 yrs×[3.50^7 yrs /sec]
Answer:
The speed is 21.06 m/s.
Explanation:
Given that,
mass of glow worm = 5.0 g
Wavelength = 650 nm
Power = 0.10 W
Time = 10 years
The total energy emitted in a period [tex]\tau[/tex] is [tex]P\tau[/tex]
The energy of single photon of frequency or wavelength is
[tex]E=\dfrac{hc}{\lambda}[/tex]
The total number of photons emitted in a interval [tex]\tau[/tex] is then the total energy divided by the energy per photon.
[tex]N=\dfrac{P\tau}{E}[/tex]
[tex]N=\dfrac{P\tau\times\lambda}{hc}[/tex]
[tex]N=\dfrac{P\tau\times\lambda}{hc}[/tex]
We need to calculate the speed
Using de Broglie's relation applies to each photon and thus the total momentum imparted to the glow-worm
[tex]p=\dfrac{Nh}{\lambda}[/tex]
[tex]p=\dfrac{\dfrac{P\tau\times\lambda}{hc}\times h}{\lambda}[/tex]
[tex]p=\dfrac{P\tau}{c}[/tex]
[tex]v=\dfrac{P\tau}{mc}[/tex]
Put the value into the formula
[tex]v=\dfrac{0.10\times3.16\times10^{8}}{5.0\times10^{-3}\times3\times10^{8}}[/tex]
[tex]v=21.06\ m/s[/tex]
Hence, The speed is 21.06 m/s.
In each case indicate whether the quantity in question increased, decreased or stayed the same when the string length is increased. Assume that the tension is unchanged. The function generator is kept at the same frequency, and the string is in resonance in all cases. Part A Number of antinodes ___ Number of antinodes ___ increased. decreased. stayed the same. Request Answer Part B Wavelength ___ Wavelength ___ increased. decreased. stayed the same. Request Answer Part C Fundamental frequency ___ Fundamental frequency ___ increased. decreased. stayed the same. Request Answer Part D Fundamental wavelength ___ Fundamental wavelength ___ increased. decreased. stayed the same. Request Answer Part E Wave speed ___ Wave speed ___ increased. decreased. stayed the same.
Answer:
a) the number of antinodes increases , b) wavelength produced is constant.
, c) fundamental frequency is constant., d) fundamental wavelength does not change, e) the speed of the wave is constant
Explanation:
This is a resonance problem where we have a frequency generator, attached to a chain with a weight in its final part, at the two ends there is a node, point if movement. The condition for resonance of this system is
λ = 2 L / n
Where n is an integer
L = n λ / 2
Let's review the problem questions.
A) As the length of the chain increases the number of wavelengths (Lam / 2) should increase, so the number of antinodes increases
B) when seeing the first equation the wavelength remains the same since the change in the length of the chain and the change in the number between are compensated, therefore for a configuration of generator frequency and weight applied the wavelength produced is constant.
C) the speed of the wave is
v = λ f
In a string the speed is constant for a fixed applied weight, with the wavelength we did not change, therefore the fundamental frequency must also be constant.
D) The value of the fundamental wavelength does not change if the weight does not change, but there is a minimum chain length for this resonance to be observed and corresponds to n = 1
L = λ / 2
E) the speed of the wave depends on the chain tension and its density if they do not change the speed of the wave does not change either
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 cm ( 0.509 m) and the flow speed of the petroleum is 11.5 m/s. At the refinery, the petroleum flows at 5.25 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery
Answer:
Flow rate 2.34 m3/s
Diameter 0.754 m
Explanation:
Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.
The area at the well head is
[tex]A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2[/tex]
So the volume flow rate along the pipe is
[tex]\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s[/tex]
We can use the similar logic to find the cross-section area at the refinery
[tex]A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2[/tex]
The radius of the pipe at the refinery is:
[tex]A_r = \pi r^2[/tex]
[tex]r^2 =A_r/\pi = 0.446/\pi = 0.141[/tex]
[tex]r = \sqrt{0.141} = 0.377m[/tex]
So the diameter is twice the radius = 0.38*2 = 0.754m
Callisto, one of Jupiter's moons, has an orbital period of 16.69 days and its distance from Jupiter is 1.88*10^6 km. What is Jupiter's mass?
Answer:
The Jupiter´s mass is approximately 1.89*10²⁷ kg.
Explanation:
The only force acting on Calisto while is rotating around Jupiter, is the gravitational force, as defined by the Newton´s Universal Law of Gravitation:
Fg = G*mc*mj / rcj²
where G = 6.67*10⁻¹¹ N*m²/kg², mc= Callisto´s mass, mj= Jupiter´s mass, and rcj = distance from Jupiter for Callisto= 1.88*10⁹ m.
At the same time, there exists a force that keeps Callisto in orbit, which is the centripetal force, that actually is the same gravitational force we have already mentioned.
This centripetal force is related with the period of the orbit, as follows:
Fc = mc*(2*π/T)²*rcj.
In order to be consistent in terms of units, we need to convert the orbital period, from days to seconds, as follows:
T = 16.69 days* 86,400 (sec/day) = 1.44*10⁶ sec.
We have already said that Fg= Fc, so we can write the following equality:
G*mc*mj / rcj² = mc*(2*π/T)²*rcj
Simplifying common terms, and solving for mj, we get:
mj = 4*π²*(1.88*10⁹)³m³ / ((1.44*10⁶)² m²*6.67*10⁻11 N*m²/kg²)
mj = 1.89*10²⁷ kg.
Answer: Mass of Jupiter ~= 1.89 × 10^23 kg
Explanation:
Given:
Period P= 16.69days × 86400s/day= 1442016s
Radius of orbit a = 1.88×10^6km × 1000m/km
r = 1.88 × 10^9 m
Gravitational constant G= 6.67×10^-11 m^3 kg^-1 s^-2
Applying Kepler's third law, which is stated mathematically as;
P^2 = (4π^2a^3)/G(M1+M2) .....1
Where M1 and M2 are the radius of Jupiter and callisto respectively.
Since M1 >> M1
M1+M2 ~= M1
Equation 1 becomes;
P^2 = (4π^2a^3)/G(M1)
M1 = (4π^2a^3)/GP^2 .....3
Substituting the values into equation 3 above
M1 = (4 × π^2 × (1.88 × 10^9)^3)/(6.67×10^-11 × 1442016^2)
M1 = 1.89 × 10^27 kg
In a binary-star system that produces a nova, the white dwarf pulls matter from the companion star. The matter forms an accretion disk that orbits the white dwarf. Then a specific sequence of events must take place for a nova event to occur. Rank the steps leading up to the observed nova event in chronological order from first to last.
As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.
Now the process:
First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.
After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.
As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.
Consider two waves moving past you at the same speed. The wavelength of wave A is half that of wave B. You then know that ____.
The frequency of wave B is half the frequency of wave A
Explanation:
The wavelength, the speed and the frequency of a wave are related by the wave equation:
[tex]v=f \lambda[/tex]
where
v is the speed of the wave
f is its frequency
[tex]\lambda[/tex] is its wavelength
The equation can also be rewritten as
[tex]f=\frac{v}{\lambda}[/tex]
In this problem, we have wave A with wavelength [tex]\lambda_A[/tex] and speed v, so its frequency is
[tex]f_A=\frac{v}{\lambda_A}[/tex]
Then we have wave B, whose wavelength is twice that of wave A:
[tex]\lambda_B = 2 \lambda_A[/tex]
And its speed is the same; Therefore, its frequency is
[tex]f_B = \frac{v}{\lambda_B}=\frac{v}{2\lambda_A}=\frac{1}{2}(\frac{v}{\lambda_A})=\frac{f_A}{2}[/tex]
So, the frequency of wave B is half that of wave A.
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The propeller of a World War II fighter plane is 2.30 m in diameter and spins at 1200 rev/min.
What is the centripetal acceleration of the propeller tip?
Answer: centripetal acceleration
Equals 18350m/s²
Explanation: Centripetal acceleration is given as v²/r
Where v is Velocity measured in rad/s and r is radius
But v= w*r
Therefore,
C.acceleratn = w²*r²/r
So we have
C.accelertn = w²*r
Our w is given as 1200 rev/min so we have to convert to rad/sec
1rad/sec equals 9.5rev/min
Therefore,
1200rev/min = 1200/9.5 rad/sec
w=126.32rad/sec
r = diameter/2
r = 2.3/2 =1.15m
Let's now calculate our centripetal acceleration = w²*r
= {126.32rad/sec}² *1.15
=18350m/s²
The centripetal acceleration of the World War II fighter plane propeller tip is approximately 18167.6 m/s² or about 1853 times the gravitational acceleration. This is determined using the radius, angular velocity, and linear speed of the propeller tip.
The propeller of a World War II fighter plane is 2.30 meters in diameter and spins at 1200 revolutions per minute. We can calculate the centripetal acceleration of the propeller tip as follows:
Angular Velocity: First, we need to convert the angular velocity from revolutions per minute (rev/min) to radians per second (rad/s).Linear Speed: The radius of the propeller is half of its diameter, so r = 2.30 m / 2 = 1.15 meters. The linear speed (v) at the tip is given by v = ωr, where ω is the angular velocity.Centripetal Acceleration: The centripetal acceleration [tex](a_c)[/tex] is given by [tex]a_c = v^2/r[/tex].Therefore, the centripetal acceleration of the propeller tip is approximately 18167.6 meters per second squared or about 1853 times the gravitational acceleration.
Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4 m×5 m× 6 m at 100 kPa and 25 C
Answer : The density, specific gravity, and mass of the air in a room is, 1.16825 g/L, 0.916 and 140.19 kg respectively.
Explanation :
First we have to calculate the volume of air.
[tex]Volume=Length\times Breadth\times Height[/tex]
[tex]Volume=4m\times 5m\times 6m[/tex]
[tex]Volume=120m^3=120000L[/tex] [tex](1m^3=1000L)[/tex]
Now we have to calculate the mole of air.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of air = 100 kPa = 0.987 atm (1 atm = 101.3 kPa)
V = Volume of air = 120000 L
n = number of moles of air = ?
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of air = [tex]25^oC=273+25=298K[/tex]
Putting values in above equation, we get:
[tex]0.987atm\times 120000L=n\times (0.0821L.atm/mol.K)\times 298K[/tex]
[tex]n=4841.04mol[/tex]
Now we have to calculate the mass of air.
[tex]\text{Mass of air}=\text{Moles of air}\times \text{Molar mass of air}[/tex]
As we know that the molar mass of air is, 28.96 g/mol
[tex]\text{Mass of air}=4841.04mol\times 28.96g/mol=140196.5184g=140.19kg[/tex]
Now we have to calculate the density of air.
[tex]\text{Density of air}=\frac{\text{Mass of air}}{\text{Volume of air}}[/tex]
[tex]\text{Density of air}=\frac{140.19kg}{120000L}=1.16825\times 10^{-3}kg/L=1.16825g/L[/tex]
Now we have to calculate the specific gravity of air.
[tex]\text{Specific gravity of air}=\frac{\text{Air density at given condition}}{\text{Air density at STP}}[/tex]
As we know that air density at STP is, 1.2754 g/L
[tex]\text{Specific gravity of air}=\frac{1.16825g/L}{1.2754g/L}=0.916[/tex]
The two spheres pictured above have equal densities and are subject only to their mutual gravitational attraction. Which of the following quantities must have the same magnitude for both spheres? (A) Acceleration (B) Velocity (C) Kinetic energy (D) Displacement from the center of mass (E) Gravitational force........ I understand the answer is E but dont know how to explain it
Answer:
Gravitational force
Explanation:
If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.
It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.
Hence, the correct option is (E) "Gravitational force".
The gravitational force has the same magnitude for both spheres.
By the Newton's law of gravitation and the Newton's third law we understand that gravitational force is directly proportional to the masses of the spheres and inversely proportional to the square of distance between centers, also that gravitational force experimented by one sphere has the same magnitude but opposite direction than the gravitational force experimented by the other sphere.
Therefore, we conclude that gravitational force has the same magnitude for both spheres. (Correct choice E)
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A spring operated dart gun fires 10 g darts. Arming the gun requires 185 N of force and results in the shortening of the spring by 10 cm. Find the spring constant. Find the energy stored in the spring. Find the muzzle velocity of the dart.
Answer:
(a) Spring constant = 1850 N/m.
(b) Energy stored in the spring = 9.25 J.
(c) Muzzle velocity of the dart = 43.01 m/s.
Explanation:
(a) Spring constant
From hook's law,
F = ke ...................... Equation 1
Where F = force on the gun, k = spring constant of the gun, e = extension of the gun's spring.
Making k subject of the equation,
k = F/e ............... Equation 2
Given: F = 185 N, e = 10 cm = 0.1 m.
Substitute into equation 2
k = 185/0.1
k = 185/0.1
k = 1850 N/m.
Hence the spring constant = 1850 N/m.
(b) Energy stored in the spring,
E = 1/2ke²............... Equation 3
Where E = Energy stored in the spring, k = spring constant, e = extension.
Given: k = 1850 N/m, e = 0.1 m.
Substitute into equation 3
E = 1/2(1850)(0.1)²
E = 925(0.01)
E = 9.25 J.
(c) Muzzle velocity of the dart.
Kinetic energy of the dart = 1/2mv²
Note: The kinetic energy of the dart is equal to the energy stored in the sprig.
E = 1/2mv²
Where m = mass of the dart, v = velocity of the dart.
Making v the subject of the equation,
v = √(2E/m)............... Equation 4
Given: E = 9.25 J, m = 10 g = 0.01 kg
Substituting these values into equation 4
v = √(2×9.25/0.01)
v = √(18.5/0.01)
v = √1850
v = 43.01 m/s.
A vertical spring stores 0.962 J in spring potential energy whena 3.5-kg mass is suspended from it.(a)by what multiplicative factordoes the spring potential energy change if the mass atttached tothe spring is doubled?(b)verify your answer to part (a)bycalculating the spring potential energy when a 7.0-kg mass isattached to the spring.
Answer:
a) If the mass get double then the potential of the spring gets four times.
b) P'=3.848 J
Explanation:
Given that
P= 0.962 J
m = 3.5 kg
m'= 7 kg
Lets take extension in the spring is x when the mass 3.5 kg is attached to the spring.
m g = K x
K=Spring constant
[tex]x=\dfrac{mg}{K}[/tex]
We know that potential energy given as
[tex]P=\dfrac{1}{2}Kx^2[/tex]
[tex]P=\dfrac{1}{2}K\times \dfrac{m^2g^2}{K^2}[/tex]
[tex]P=\dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]
If the mass get double then the potential of the spring gets four times.
[tex]P'=\dfrac{1}{2}\times \dfrac{(2m)^2g^2}{K}[/tex]
[tex]P'=4\times \dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]
P'= 4 P
When mass ,m' = 7 kg
Then potential will be
[tex]0.962=\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} [/tex] -----1
[tex]P'=\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K}[/tex] -------2
From equation 1 and 2
[tex]\dfrac{0.962}{P'}=\dfrac{\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} }{\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K} }[/tex]
P'= 4 x 0.962 J
P'=3.848 J
Answer:
[tex]k=611.517\ N.m^{-1}[/tex]
[tex]U_7=3.8477\ J[/tex]
Explanation:
Given:
spring potential energy stored due to hanging mass, [tex]U=0.962\ J[/tex]mass attached to the spring, [tex]m=3.5\ kg[/tex]Now the force on the mass due to gravity:
[tex]F=m.g[/tex]
[tex]F=3.5 \times 9.8[/tex]
[tex]F=34.3\ N[/tex]
This force pulls the spring down, so:
[tex]F=k.\delta x[/tex]
[tex]34.3=k\times \delta x[/tex] ....................(1)
For the spring potential:
[tex]U=\frac{1}{2} k.\delta x^2[/tex]
[tex]0.962=0.5\times k\times \delta x^2[/tex]
[tex]1.924=k\times \delta x^2[/tex] .........................(2)
Using eq. (1) & (2)
[tex]\frac{1.924}{x^2} =\frac{34.3}{x}[/tex]
[tex]x=0.05609\ m[/tex]
a.
Now the spring factor:
using eq. (1)
[tex]34.3=k\times \delta 0.05609[/tex]
[tex]k=611.517\ N.m^{-1}[/tex]
b.
when mass attached is 7 kg.
The spring potential energy:
[tex]U_7=\frac{1}{2} \times k.\delta x'^2[/tex] ............(3)
Now the force on the mass due to gravity:
[tex]F=m.g[/tex]
[tex]F=7\times 9.8[/tex]
[tex]F=68.6\ N[/tex]
This force pulls the spring down, so:
[tex]F=k.\delta x[/tex]
[tex]68.6=611.517\times \delta x[/tex]
[tex]x=0.11218\ m[/tex]
Using eq. (3)
[tex]U_7=\frac{1}{2}\times 611.517\times 0.11218^2[/tex]
[tex]U_7=3.8478\ J[/tex]
A bag containing originally 60 kg of flour is lifted through a vertical distance of 9 m. While it is being lifted, flour is leaking from the bag at such rate that the number of pounds lost is proportional to the square root of the distance traversed. If the total loss of flour is 12 kg find the amount of work done in lifting the bag.
Answer:
The amount of work done in lifting the bag is -20109.6 N-m
Explanation:
Given that,
Mass of bag = 60 kg
Distance = 9 m
Loss of mass = 12 kg
The number of pounds lost is proportional to the square root of the distance traversed
Mass of the bag containing flour at height is
[tex]m(y)=60-k\sqrt{y}[/tex]
Put the value into the formula
[tex]60-k\sqrt{y}=12[/tex]
[tex]k=144[/tex]
We need to calculate the work done
Using formula of work done
[tex]W=\int_{0}^{9}{m(y)gdy}[/tex]
Put the value into the formula
[tex]W=\int_{0}^{9}{(60-k\sqrt{y})gdy}[/tex]
[tex]W=((60y-\dfrac{2k}{3}\times y^{\frac{3}{2}}})_{0}^{9})\times9.8[/tex]
Put the value of y
[tex]W=(60\times9-\dfrac{2\times144}{3}\times 9^{\frac{3}{2}})\times9.8[/tex]
[tex]W=-20109.6\ N-m[/tex]
Hence, The amount of work done in lifting the bag is -20109.6 N-m
When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If that charge is 4.0 ?C , by what percentage does the mass of a 33 g comb change during charging?
The concept required to solve this problem is quantization of charge.
First the number of electrons will be calculated and then the total mass of the charge.
With these data it will be possible to calculate the percentage of load in the mass.
[tex]Q= ne[/tex]
Here Q is the charge, n is the number of electrons and e is the charge on the electron
[tex]n = \frac{Q}{e}[/tex]
Replacing,
[tex]n = \frac{4*10^{-6}C}{1.6*10^{-19}}[/tex]
[tex]n = 2.5 * 10^{13}77[/tex]
According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron
The total mass of the charge is
[tex]m= nm_e[/tex]
Here,
m = Mass of the charge
n = Number of electrons
[tex]m_e[/tex] = Mass of the electron
[tex]\text{Percentage change} = \frac{nm_e}{M}*100[/tex]
Replacing we have
[tex]\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100[/tex]
[tex]\text{Percentage change} = 6.9*10^{-14} \%[/tex]
The percentage change in the mass of the 33 g comb during charging is extremely small, approximately [tex]\(6.91 \times 10^{-16}\%\),[/tex] which is negligible.
To find the percentage change in mass of the comb due to charging, we need to understand the relationship between the amount of charge transferred and the mass of the electrons involved. Here's the step-by-step process:
1. Calculate the number of electrons corresponding to the given charge:
The charge of one electron[tex](\(e\))[/tex] is approximately[tex]\(1.6 \times 10^{-19}\)[/tex] coulombs.
Given:
- Net charge [tex](\(Q\)) = 4.0 μC = \(4.0 \times 10^{-6}\) C[/tex]
The number of electrons n transferred can be calculated using the formula:
[tex]\[ n = \frac{Q}{e} \][/tex]
[tex]\[ n = \frac{4.0 \times 10^{-6} \text{ C}}{1.6 \times 10^{-19} \text{ C/electron}} \][/tex]
[tex]\[ n = 2.5 \times 10^{13} \text{ electrons} \][/tex]
2. Calculate the mass of the transferred electrons:
The mass of one electron [tex](\(m_e\))[/tex] is approximately [tex]\(9.11 \times 10^{-31}\)[/tex] kg.
The total mass[tex](\(m\))[/tex] of the electrons transferred is:
[tex]\[ m = n \times m_e \][/tex]
[tex]\[ m = 2.5 \times 10^{13} \times 9.11 \times 10^{-31} \text{ kg} \][/tex]
[tex]\[ m = 2.28 \times 10^{-17} \text{ kg} \][/tex]
3. Convert the mass of the comb to kilograms:
Given:
- Mass of the comb = 33 g = 0.033 kg
4. Calculate the percentage change in mass:
The percentage change in mass is given by:
[tex]\[ \text{Percentage change} = \left( \frac{\text{Change in mass}}{\text{Original mass}} \right) \times 100\% \][/tex]
[tex]\[ \text{Percentage change} = \left( \frac{2.28 \times 10^{-17} \text{ kg}}{0.033 \text{ kg}} \right) \times 100\% \][/tex]
[tex]\[ \text{Percentage change} = 6.91 \times 10^{-16} \% \][/tex]
Can you tell from your experiment so far whether the tapes carry a positive charge or a negative charge? Briefly explain your answer.
Answer:
Explanation:
Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge: positive and negative (commonly carried by protons and electrons respectively). Like charges repel each other and unlike charges attract each other. An object with an absence of net charge is referred to as neutral. Early knowledge of how charged substances interact is now called classical electrodynamics, and is still accurate for problems that do not require consideration of quantum effects.
Electric charge is a conserved property; the net charge of an isolated system, the amount of positive charge minus the amount of negative charge, cannot change. Electric charge is carried by subatomic particles. In ordinary matter, negative charge is carried by electrons, and positive charge is carried by the protons in the nuclei of atoms. If there are more electrons than protons in a piece of matter, it will have a negative charge, if there are fewer it will have a positive charge, and if there are equal numbers it will be neutral. Charge is quantized; it comes in integer multiples of individual small units called the elementary charge, e, about 1.602×10−19 coulombs,[1] which is the smallest charge which can exist freely (particles called quarks have smaller charges, multiples of
e, but they are only found in combination, and always combine to form particles with integer charge). The proton has a charge of +e, and the electron has a charge of −e.
An electric charge has an electric field, and if the charge is moving it also generates a magnetic field. The combination of the electric and magnetic field is called the electromagnetic field, and its interaction with charges is the source of the electromagnetic force, which is one of the four fundamental forces in physics. The study of photon-mediated interactions among charged particles is called quantum electrodynamics.
A 2.4 kg toy oscillates on a spring completes a cycle every 0.56 s. What is the frequency of this oscillation?
Answer:
[tex]f=1.79Hz[/tex]
Explanation:
The period is defined as the time taken by an object to complete a cycle in a simple harmonic motion. As the frequency of the motion increases, the period decreases. Therefore, they are inversely proportional. The frequency does not depend on the mass of the object.
[tex]f=\frac{1}{T}\\f=\frac{1}{0.56 s}\\f=1.79Hz[/tex]
Jane is pulling a chain of 2 boxes with a massless rope at an angle of 20 degrees above the horizontal with a force of 30N. Jane and the boxes are all on horizontal ground. The boxes are connected horizontally by a length of massless rope. The box closest to Jane is 45 kg and the box farthest from Jane is 30kg.
a. Draw a picture of the situation.
b. Draw a free-body diagram for each box.
c. What is the normal force acting on the 45kg box?
d. What is the acceleration of the boxes?
e. What is the tension in the rope connecting the boxes?
Find the Cartesian coordinates for the point whose distance from the origin is 2 m and the angle measured from the positive horizontal axis is 15°
x =
y =
The Cartesian coordinates for the point are approximately: x ≈ 1.93185 m and, y ≈ 0.523599 m.
Here, we have to find the Cartesian coordinates (x, y) for a point given its distance from the origin (r) and the angle (θ) measured from the positive horizontal axis, we can use trigonometric functions.
In this case, r = 2 m and θ = 15°.
The Cartesian coordinates can be calculated as follows:
x = r * cos(θ)
y = r * sin(θ)
Given:
r = 2 m
θ = 15°
First, convert the angle from degrees to radians, since trigonometric functions in most programming languages use radians:
θ_rad = 15° * (π / 180) ≈ 0.261799 radians
Now, calculate the coordinates:
x = r * cos(θ_rad) = 2 m * cos(0.261799) ≈ 1.93185 m
y = r * sin(θ_rad) = 2 m * sin(0.261799) ≈ 0.523599 m
So, the Cartesian coordinates for the point are approximately:
x ≈ 1.93185 m
y ≈ 0.523599 m
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The Cartesian coordinates for the point with a distance of 2 m from the origin and an angle of 15° from the positive horizontal axis are approximately (1.93 m, 0.52 m), calculated using the cosine and sine of the angle.
Explanation:To find the Cartesian coordinates for a point given its distance from the origin and the angle from the positive horizontal axis, you can use trigonometric functions. Since the distance from the origin is 2 m and the angle is 15°, you can calculate the x and y coordinates using the cosine and sine functions, respectively.
The x-coordinate is found by multiplying the distance by the cosine of the angle: x = 2 m × cos(15°). The y-coordinate is found by multiplying the distance by the sine of the angle: y = 2 m × sin(15°).
Using standard values for cos(15°) and sin(15°), the calculations would yield:
x = 2 m × 0.9659 (approximately) = 1.9318 my = 2 m × 0.2588 (approximately) = 0.5176 mTherefore, the Cartesian coordinates are approximately (1.93 m, 0.52 m).
What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from each other?
Answer:
[tex]F=1.38*10^{-9}N[/tex]
Explanation:
According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
Here k is the Coulomb constant. In this case, we have [tex]q_1=-e[/tex], [tex]q_2=e[/tex] and [tex]d=4.09*10^-10m[/tex]. Replacing the values:
[tex]F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N[/tex]
The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:
[tex]F=1.38*10^{-9}N[/tex]
In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point d = 2.8 m from his foot at the edge of the pool a) Where does the spot of light hit the bottom of the h = 2.1-m-deep pool? Measure from the bottom of the wall beneath his foot.
According to the question,
Length, [tex]l_1 = 2.1 \ m[/tex]Height, [tex]h_1 = 1.3 \ m[/tex] Distance, [tex]d = 2.8 \ m[/tex]The angle of incidence will be:
→ [tex]tan \Theta_1= \frac{l_1}{h_1}[/tex]
[tex]= \frac{2.1}{1.3}[/tex]
[tex]= 2.076[/tex]
then,
[tex]\Theta_1 = 64.3^{\circ}[/tex]
From air into water, we get
→ [tex]n_{air} Sin \Theta_1 = n_{water} Sin \Theta_2[/tex]
[tex](1.00) Sin 64.3^{\circ} = (1.3) Sin \Theta_2[/tex]
[tex]\Theta_2 = 42.6^{\circ}[/tex]
hence,
The horizontal distance,
→ [tex]l_2 = l_1+h_2 tan \Theta[/tex]
By substituting the values, we get
[tex]= 2.1+(2.1)tan 42.6[/tex]
[tex]= 4.6 \ m[/tex]
Thus the above approach is right.
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