Answer:
c. More than 2 s
Explanation:
First, we will find the length of the pendulum:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\2\ s = 2\pi \sqrt{\frac{l}{9.81\ m/s^2}}\\\\4\ s^2 = 4\pi^2 (\frac{l}{9.81\ m/s^2})\\\\l = \frac{(4\ s^2)(9.81\ m/s^2)}{4\pi^2} \\\\l = 0.99\ m[/tex]
Now, the value of g becomes 1.625 m/s² on the surface of the moon. So the time period will be:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\T = 2\pi \sqrt{\frac{0.99\ m}{1.625\ m/s^2}}\\\\[/tex]
T = 4.9 s
Therefore, the correct option is:
c. More than 2 s
what is effort arm
don't say the answer of gogle
Answer:
effort arm mean the use of any work by using your hand force motion or by hand power
To a man running east at the rate of 3m/s vain appears to fall vertically with a speed of 4m/s. Find the actual speed and direction of rain...
Answer:
The actual speed of the rain is 5 m/s and its direction is -53.13°
Explanation:
The actual speed of the rain V = speed of man, v + speed of rain relative to man, v'.
V = v + v'
We add these vectorially.
Since the man's speed is 3 m/s east, in the positive x - direction, we have v = 3i and the rain's speed is falling vertically at 4 m/s, in the negative y- direction, we have v' = -4j
So, V = v + v'
V = 3i + (-4j)
V = 3i - 4j
So, the magnitude of V which is its speed is V = √(3² + (-4)²) = √(9 + 16) = √25 = 5 m/s
The direction of V, Ф = tan⁻¹(vertical component/horizontal component) = tan⁻¹(-4/3) = tan⁻¹(-4/3) = tan⁻¹(-1.333) = -53.13°
So, the actual speed of the rain is 5 m/s and its direction is -53.13°
calculate the work function that requires a 455 nm photon to eject an electron of a value 0.73 eV
Answer:
W = 2 eV
Explanation:
Given that,
The wavelength of a photon = 455 nm
The kinetic energy of a photon, K = 0.73 eV
We need to find the work function of the electron. It can be solved using Einstein's equation such that,
[tex]W=E-K[/tex]
E is the energy of the photon
So,
[tex]W=\dfrac{hc}{\lambda}-K\\\\W=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{455\times 10^{-9}\times 1.6\times 10^{-19}}-0.73\\\\W= 2.73\ eV-0.73\ eV\\\\W=2\ eV[/tex]
So, the work function of the metal is 2 eV.
planet smaller than earth but larger than mercury
Answer:
venus......................
who won the battle of Buxar
Answer:
British East India Company
The battle was fought at Buxar, a "small fortified town" within the territory of Bihar, located on the banks of the Ganga river about 130 kilometres (81 mi) west of Patna; it was a decisive victory for the British East India Company.
A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.
Answer:
V = 240V
Explanation:
V = I*R
V = 15A*16ohms
V = 240V
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed into elastic potential energy of the bending board; if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the force. Ultimately the board breaks at an applied force of 870 N and a deflection of 1.4 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a distance of 1.2 cm, what is the average force on the hand?
Answer:
A. The hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand = 1025 N
Explanation:
A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.
Workdone = force × distance
Minimum force required = 870 N;
Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M
Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J
This work done = Kinetic energy of the hand
Kinetic energy = mv²/2 ; where m is mass and v is velocity
Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J
v² = 2 KE/m
v = √2KE/m
v = √(2 × 12.18/0.50)
v = 6.98 m/s
Therefore, the hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand
This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration
From the equation of motion, a = v² - u²/2s
At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m
a = 0² - 6.98²/ 2 × 0.012
a = -2030 m/s²
Force = 2030 m/s² × 0. 50 kg = 1015 N
Therefore, Average force on the hand = 1025 N
HELP ME PLEASE!!!
Which 2 statements are true about this chemical reaction that forms acid rain?
Answer:
B.
Explanation:
HNO2 is less stable thus dissociates easily to HNO3 + NO + H2O while HNO3 is a strong acid. Thus when they react with H2O they form acid rain
Answer:
B
Explanation:
dont have one just trust me
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.6 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
a. How far from the wall is she?
b. What is the period of her up-and-down motion?
Answer:
a) [tex]d=11m[/tex]
b) [tex]T=4.68s[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=22m[/tex]
Velocity [tex]v=4.6m/s[/tex]
a)
Generally the equation for distance between her and the wall d is mathematically given by
Since
The First Anti node distance is [tex]\frac{\lambda}{2}[/tex]
Therefore
[tex]d= \frac{\22}{2}[/tex]
[tex]d=11m[/tex]
b)
Generally the equation for her up-and-down motion is mathematically given by
[tex]T=\frac{22}{4.7}[/tex]
[tex]T=4.68s[/tex]
A 620 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 870 kg. As the elevator starts moving, the scale reads 450 N.
Required:
a. Find the acceleration of the elevator (magnitude and direction).
b. What is the acceleration if the scale reads 670 N?
c. If the scale reads zero, should the student worry? Explain.
d. What is the tension in the cable in parts (a) and (c)?
Answer:
(a) 9.28 m/s2
(b) 9.03 m/s2
(c) 9.8 m/s2
(d) 450 N, 670 N
Explanation:
mass of elevator + student, m = 870 kg
Reading of scale, R = 450 N
(a) When the elevator goes down, the weight decreases.
Let the acceleration is a.
By the Newton's second law
m g - R = m a
870 x 9.8 - 450 = 870 a
a = 9.28 m/s2
(b) R = 670 N
Let the acceleration is a.
870 x 9.8 - 670 = 870 a
a = 9.03 m/s2
(c) If the scale reads zero, it mean the elevator is falling freely. The acceleration is downwards and its value is 9.8 m/s2.
(d) Tension in cable is 450 N and 670 N.
A 2.7 kg mass is connected to a spring (k=159 N/m) and is sliding on a horizontal frictionless surface. The mass is given an initial displacement of +19 cm and released with an initial velocity of -13 cm/s. Determine the acceleration of the spring at t=3.4 seconds.
Answer:
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
Explanation:
The mass-spring system experiments a Simple Harmonic Motion, whose kinematic expression is the following:
[tex]x(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t + \phi\right)[/tex] (1)
Where:
[tex]x(t)[/tex] - Position, in centimeters.
[tex]A[/tex] - Amplitude, in centimeters.
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]m[/tex] - Mass, in kilograms.
[tex]\phi[/tex] - Phase, in radians.
By Differential Calculus, we derive expression for the velocity and acceleration of the mass-spring system:
[tex]v(t) = -\sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (2)
[tex]a(t) = -\frac{k\cdot A}{m}\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (3)
Where [tex]v(t)[/tex] and [tex]a(t)[/tex] are the velocity and acceleration of the system, in centimeters per second and centimeters per square second, respectively.
If we know [tex]m = 2.7\,kg[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]t = 0\,s[/tex], [tex]x(t) = 19\,cm[/tex] and [tex]v(t) = -13\,\frac{cm}{s}[/tex], then we have the following system of nonlinear equations:
[tex]A \cdot \cos \phi = 19[/tex] (1)
[tex]-7.674\cdot A \cdot \sin \phi = - 13[/tex] (2)
If we divide (2) by (1):
[tex]-7.674\cdot \tan \phi = -0.684[/tex]
[tex]\tan \phi = 0.089[/tex]
[tex]\phi = \tan^{-1} 0.089[/tex]
[tex]\phi \approx 0.028\,rad[/tex]
By (1), we get the value of the amplitude:
[tex]A = \frac{19}{\cos \phi}[/tex]
[tex]A = 19.075\,cm[/tex]
If we know that [tex]A = 19.075\,cm[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]m = 2.7\,kg[/tex], [tex]t = 3.4\,s[/tex] and [tex]\phi \approx 0.028\,rad[/tex], then the acceleration of the spring is:
[tex]a(t) = -29339.947\,\frac{cm}{s^{2}}[/tex]
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
If distance between two charges increased by 2 times then force
Explanation:
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
You take your pulse and observe 80 heartbeats in a minute. What is the period of your heartbeat? What is the frequency of your heartbeat?
Answer:
120 beats per minute.
Explanation:
If I take your pulse and observe 80 heartbeats in a minute. Then the period of your heartbeat is 0.8 s and frequency is 1.3Hz.
What is Heartbeat ?A pulse is the term used in medicine to describe the tactile arterial palpation of the cardiac cycle (heartbeat) by skilled fingertips. Any location where an artery can be compressed close to the surface of the body, such as the carotid artery in the neck, the radial artery in the wrist, the femoral artery in the groyne, the popliteal artery behind the knee, the posterior tibial artery near the ankle joint, and on the foot, can be used to palpate the pulse (dorsalis pedis artery). Heart rate may be determined by monitoring pulse, or the number of arterial pulses per minute. Auscultation, which is the process of counting the heartbeats while listening to the heart using a stethoscope, is another way to determine the heart rate. Typically, three fingers are used to gauge the radial pulse.
Given,
heart beat = 80 beats/min = 1.3 beats/s
Frequency is nothing but how much beats is heart having in one second and that is 1.3 beats/s. Hence frequency of heart is 1.3Hz.
The Period is reciprocal of frequency,
T = 1/f = 0.8 s
To know more about frequency :
https://brainly.com/question/29739263
#SPJ2.
A crucible (container) of molten metal has an open top with an area of 5.000 m^2. The molten metal acts as a blackbody radiator. The intensity spectrum of its radiation peaks at a wavelength of 320 nm. What is the temperature of that blackbody?
Answer:
T = 9056 K
Explanation:
In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation
λ T = 2,898 10⁻³
where lam is the wavelength of the maximum emission
T = 2,898 10⁻³ /λ
let's calculate
T = 2,898 10⁻³ / 320 10⁻⁹
T = 9.056 10³ K
T = 9056 K
A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1. Which of the following describes the electricfield inside this conductor?
a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.
2. The charge density inside theconductor is:
a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinite
Answer:pp
Explanation:
ii
what affects our utility
Answer:
Energy Bill fluctuations are inevitable and depend on a variety of different factors. Two of the most important are the current weather your home is experiencing and the current price per Kilowatt Hour (which fluctuates more than you might think).
You are packing for a trip to another star, and on your journey you will be traveling at a speed of 0.99c. Can you sleep in a smaller cabin than usual, because you will be shorter when you lie down? Explain your answer.
Explanation:
No. From your own reference frame, nothing will change. Everything will look the same to you even if you're traveling at 99% speed of light. However, an outside observer will see you shrink to about 14% of your original length along the direction of your motion according to the Lorentz contraction predicted by special relativity:
[tex]L=L_{0} \sqrt{1 - \frac{v^2 }{c^2 }}[/tex]
help plzzzzzzzzzzzz ?
Explanation:
1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:
[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]
[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]
Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so
[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]
Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so
[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]
Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is
[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]
2. The total current in the circuit is
[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]
3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is
[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]
4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us
[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]
5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is
[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]
This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is
[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]
The position of an object of
mass 5 kg as a function of
time is giving by r = (20m/s4)t4
i + (12 m/s3)t3 j. Find the
force acting on the object as a
function of time. Express the
force in unit vectors. Hint:
Remember that Newton's
second Law relates the force
to the acceleration
Answer:
[tex]F=5(240t^2i+72tj)\ N[/tex]
Explanation:
Given that,
The mass of the object, m = 5 kg
The position vector is, [tex]r=20t^4i+12t^3j[/tex]
Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]
Newton's second law of motion is given as follows:
F = ma
Put all the values,
[tex]F=5(240t^2i+72tj)\ N[/tex]
Hence, this is the required solution.
A person rolls a 7 kg bowling ball down a lane in a bowling alley. The lane is
18 m long. The ball is traveling at 7 m/s when it leaves the person's hand.
What is the ball's kinetic energy at this point?
Answer:
171.5J
Explanation:
K=1/2 *m *U²
K=1/2 *7 *7²
K=171.5 J
A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.
A positive point charge, Q = 4.00 nC, is located a distance of 5.00 cm above the midpoint of the
rod. What will be the electrical force on the point charge?
Answer the following using equations, number substitution and keep units. 1. What is the speed of an object that travels 5m in 10s. 2. What force is on a 10kg mass that accelerates at 3m/s/s. 3. What is the potential energy of a 7kg object 4m off the ground *
show all your work please
Explanation:
1. Distance, d = 5 m
Time, t = 10 s
Speed = distance/time
[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]
2. Mass, m = 10 kg
Acceleration, a = 3 m/s³
Force, F = mass (m) × acceleration (a)
F = 10 × 3
= 20 N
3. Mass, m = 7 kg
Height, h = 4 m
Potential energy, E = mgh
E = 7 × 9.8 × 4
E = 274.4 J
Hence, this is the required solution.
If ATM is 102 kPa, what force does the atmosphere exert on the palm of your hand which has an area of 0.016 meters?
Answer:
Force = 1.632 Newton
Explanation:
Given the following data;
Pressure = 102 kPa
Area = 0.016 m²
To find what force the atmosphere exert on the palm of your hand;
Mathematically, pressure is given by the formula;
[tex] Pressure = \frac {Force}{area} [/tex]
Force = 102 * 0.016
Force = 1.632 Newton
a brick of mass 0.8 kg is accidentally dropped from a high scaffolding. it reaches the ground with a kinetic energy of 240 J. How high is scaffolding ?(Take acceleration due to gravity g be 10 m s-¹)
Answer:
30 m
General Formulas and Concepts:
Energy
Gravitational Potential Energy: [tex]\displaystyle U_g = mgh[/tex]
m is mass (in kg)g is gravityh is height (in m)Kinetic Energy: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]
m is mass (in kg)v is velocity (in m/s²)Law of Conservation of Energy
Explanation:
Step 1: Define
Identify variables
[Given] m = 0.8 kg
[Given] g = 10 m/s²
[Given] U = 240 J
[Solve] h
Step 2: Solve for h
[LCE] Substitute in variables [Gravitational Potential Energy]: (0.8 kg)(10 m/s²)h = 240 JMultiply: (8 kg · m/s²)h = 240 JIsolate h [Cancel out units]: h = 30 mWhat innovation did jethro wood add to plows in the 1800s?
Built by 1800, it was the home of inventor Jethro Wood (1774-1834), whose 1819 invention of an iron moldboard plow revolutionized American agriculture.
In addition to acceleration, what else will be a maximum at the amplitude for SHM?
A. Potential energy
B. Kinetic energy
C. Nuclear energy
D. Chemical energy
It is Potential energy's
A skateboarder is inside of a half pipe, shown here. Explain her energy transformations as she jumps off at point A, slides to point B, and finally reaches point C.
Question 1 of 25
Which equation is an example of a synthesis reaction?
A. HNO3 + KOH → KCI + H20
B. 2Li+ CaCl2 - 2LiCl + Ca
O C. S+ 02 - S02
7
O D. CH4 + 202 - 2H2O + CO2
Answer:
C. S + 02 → S02
Explanation:
A synthesis or combination reaction is that reaction involving two elements as reactants to form a single compound as product.
In the reaction given below;
S + 02 → S02
Sulphur and oxygen are elemental substances that combine to synthesize sulfur IV oxide (SO2), and hence it is an example of synthesis reaction.
Hey, can a physics major help me?
I have been wondering about the exact difference between theories laws facts and hypothosis.
I know the general layout but I am still a bit confused.
100 points for answering and brainly if it is a good one.
Answer:
A hypothesis is a limited explanation of a phenomenon; a scientific theory is an in-depth explanation of the observed phenomenon. A law is a statement about an observed phenomenon or a unifying concept
Answer:
Explanation:
will try 2 explain fact, hypothesis, theory n law
fact is the starting pt: e.g. apple falls from tree
hypothesis tries 2 explain a fact: e.g. there is a force pulling down apple
theory is a complete explanation w/ equations n stuff: e.g. Newton came up w/ theory of gravitational attraction force
law is a theory dat has been proven right through tests n experiments: Newton's gravity theory had been proven right in many many tests.
