Answer: the length of the pendulum should be 1.4% longer
Explanation:
Given that;
when its noon, the clock reads 11:50 am,
i.e we have 10 minutes delay ⇒ 10min × 60 = 600secs
we know that in simple pendulum
T = 2π√(l/g)
d means delay and c means correct;
24hrs = 86400 secs
Now
Td/Tc = (86400-600) / 86400 = 0.993 = [2π√(ld/gc)] / [2π√(ld/gc)] = √(ld/lc)
so ld/lc = 0.98616
lc = 1.014 ld
lc/l.d = -1 + 1.014 = 0.014 × 100 = 1.4% longer
therefore the length of the pendulum should be 1.4% longer
Which type of circuit would be best to use for lights used for decorations? Question 1 options: Series circuit. One bulb could go out and the strand will stay on. Series circuit. One bulb could go out and the rest go out. Parallel circuit. One bulb goes out and the rest go out. Parallel circuit. One bulb could go out and the strand will stay on.
Answer:
One bulb could go out and the strand will stay on.
Explanation:
In series circuit, there is only one path provided for the current to flow. So, all the lights are required to be in working condition, for the others to work. And if anyone light bulb goes out, the circuit will become incomplete and the rest of the strand will also go out. Because there is only one path for current flow which is now broken.
On the other hand, in parallel circuits, each light bulb has a separate connection with the source. Current path to each bulb is independent of the others. Therefore, if one bulb goes out, the rest of the strand will stay on.
So, the correct option is:
One bulb could go out and the strand will stay on.
What is the significance of Nucleotides in Chromosomes?
Answer:
it comprises of the DNA/RNA bipolymer molecules
When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The player then lands on the floor with their feet and is quickly brought to a stop. If the the floor deforms by 0.025 m while the player lands on it, what was the average force under the player's feet during the landing
Answer:
2361.6N
Explanation:
Mass of player = 82kg
Velocity = 1.2m/s
Kinetic energy of player:
= 1/2mv²
= 1/2*82*1.2²
= 41x1.44
= 59.04J
Final kinetic energy = 0
Change in kinetic energy
|∆k| = |0-59.04|
= 59.04
Workdone by the feet = fd
d = 0.025
Fd = 59.04
F = 59.04/0.025
= 2361.6N
This is his average force.
HELP PLZ!!!!!!!!!!!!!!
Answer:
Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)
Explanation:
In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)
For example, let's say you're trying to send the message "11001"
on on off off on
0 1 2 3 4 5 Numbers represent seconds
As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.
a guitar string is 0.620 m long, and oscillates at 234 Hz. if a player uses his finger to shorten the string to 0.480 m, what is the new frequency?
Answer:
the new frequency is : 302.25Hz
using the formula F2 = [tex]\frac{F1 L1}{L2}[/tex]
Explanation:
meaning of frequencyfrequency of a string is the number of vibrations of a plucked string per second. it is measured in Hertz.
the frequency of a string is inversely proportional to twice the length of the string. which means the longer the string, the smaller the freqency and the higher the string the higher the frequency.
f ∝ 1/2L.
f = k/2L
where f = frequency
L = length of string
k = constant
k = 2fL is a constant
given data
L1 = 0.62m
f1 = 234Hz
L2 = 0.48m
2f1L1 = 2f2L2
f1L1 = f2L2
f2 = f1L1/L2
f2 = [tex]\frac{234 x 0.62}{0.48}[/tex] = 302.25Hz
in conclusion, the new frequency is 302.25Hz
learn more about frequency of a vibrating string: brainly.in/question/1149252
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draw simple vector diagram and resolve 60N at an angle of 30° from the horizontal.. plz help guys
Explanation:
The attatched figure shows the vector diagram for a force that has magnitude of 60 N and it is acting at an angle of 30° from the horizontal.
When it is resolved, the horizontal and vertical components are given by :
[tex]F_x=F\cos\theta\\\\=60\times \cos30\\\\=51.96\ N[/tex]
And
[tex]F_y=F\sin\theta\\\\=60\times \sin30\\\\=30\ N[/tex]
Hence, this is the required solution.
Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls have a head-on, inelastic collision and the 2.0-kg ball recoils with a speed of 8.0 m/s, how much kinetic energy is lost in the collision
Answer:
The kinetic energy lost in the collision is 48 J
Explanation:
Given;
mass of the first ball, m₁ = 2.0 kg
mass of the second ball, m₂ = 6.0 kg
initial speed of the first ball, u₁ = 12 m/s
initial speed of the second ball, u₂ = 4 m/s
let v be the final velocity of the two balls after the inelastic collision
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 12 + 6 x 4 = v(2 + 6)
48 = 8v
48 / 8 = v
v = 6 m/s
The initial kinetic energy of the balls is calculated as;
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²
K.E₁ = 144 + 48
K.E₁ = 192 J
The final kinetic of the balls is calculated as;
K.E₂ = ¹/₂(m₁ + m₂)(v²)
K.E₂ = ¹/₂(2 + 6)(6²)
K.E₂ = ¹/₂(8)(6²)
K.E₂ = 144 J
The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J
Therefore, the kinetic energy lost in the collision is 48 J
Which of the following is a definition of motion?
Answer:
sorry I don't get it so pls can u repeat the question .
Movement of any object from one position to another position with respect to the observer is called as Motion. Motion Along a Straight Line: When an object moves along a straight line, the motion of the object is called rectilinear motion. For example; motion of a car on highway.
QUESTION:
Which of the following is a definition of motion?
ANSWER:
In physics, motion is the phenomenon in which an object changes its position over time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time.
The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
[tex]dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2[/tex]
[tex]For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2[/tex]
The second distance, r₂, can be determined from sound intensity formula given as;
[tex]I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m[/tex]
Therefore, the second distance of the sound from the source is 431.78 m.
True or False: Solids always have a higher density than liquids and gases.
Answer:
TRUE THE ANSWER IS TRUE
If vector A= aj^ and vector B = bj^, then vector A×B is equal to
Answer:
0 because j×j =0
Explanation:
so a×b is 0
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.9 rad/s in 3.10 s. (a) Find the magnitude of the angular acceleration of the wheel.
Answer:
The value is [tex]\alpha = 3.84 \ rad/s^2[/tex]
Explanation:
From the question we are told that
The constant angular speed is [tex]w = 11.9 \ rad/s[/tex]
The time taken is [tex]t = 3.10 \ s[/tex]
Generally the magnitude of the angular acceleration is mathematically represented as
[tex]\alpha = \frac{w}{t}[/tex]
=> [tex]\alpha = \frac{11.9}{ 3.10 }[/tex]
=> [tex]\alpha = 3.84 \ rad/s^2[/tex]
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
Answer:
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
An observer stands 150 meters from a fireworks display rocket, which is fired directly upward. When the rocket reaches a height of 200 meters, it is traveling at a speed of 12 meters/second. At what rate is the angle of elevation formed with the observer increasing at that instant
Answer:
[tex]-0.288\ \text{rad/s}[/tex]
Explanation:
x = Distance of observer from the initial location of the rocket = 150 m
y = Vertical displacement of the rocket from the ground = 200 m
r = Distance between observer and rocket
[tex]\dfrac{dy}{dt}[/tex] = Rate of change in height of rocket = 12 m/s
[tex]\dfrac{dx}{dt}[/tex] = Rate of change in x = 0
Distance between observer and rocket at y = 200 m
[tex]r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{150^2+200^2}\\\Rightarrow r=250\ \text{m}[/tex]
[tex]\tan\theta=\dfrac{y}{x}[/tex]
Differentiating with respect to time
[tex]\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}}{\sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{0-150\times 12}{150^2}}{(\dfrac{250}{150})^2}\\\Rightarrow \dfrac{d\theta}{dt}=-0.288\ \text{rad/s}[/tex]
The rate of change of the angle of elevation is [tex]-0.288\ \text{rad/s}[/tex].
As a freely falling object speeds up, what is happening to its acceleration - does it increase, decrease, or stay the same? (a) Ignore air resistance. (b) Consider air resistance.
Essay ANSWER
Answer:
(a)When ignoring air resistance its accelerating increases steadily .
(b)When considering air resistance then its acceleration decreases this could either be uniformly or unevenly.
Hope this helped.
An object that floats in water weighs 20 N in air.
a. What is the weight of the object in water?
b.What is the Upthrust acting on the object in water?
c. What is the weight of the water displaced by the object?
Answer:
a. Weight of Object in Water = 20 N
b. Up thrust = 20 N
c. Weight of Water Displaced = 20 N
Explanation:
a.
The weight of the object remains same in the water as well. Because, the same force of gravity is acting there as well. Hence,
Weight of Object in Water = 20 N
b.
Since, the object floats on the water. Therefore, according to Archimedes' principle the up thrust force acting on the object must be equal to the weight of object:
Up thrust = Weight of object
Up thrust = 20 N
c.
From Archimedes' Principle, we know that the up thrust or the Buoyant force is equal to the weight of the water displaced by the object. therefore:
Weight of Water Displaced = Up thrust
Weight of Water Displaced = 20 N
1. What is the total distance traveled?
A 3.0m
B 4.0m
C 5.0m
D 6.0m
Answer:
c
Explanation:
What is the magnitude of the velocity after it hits the
ground?
9.3 m/s
12 m/s
41 m/s
73 m/s
Answer: 9.3m/s
Explanation:
Your question isn't complete but let me help out:
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
We would use Newton's law of motion to solve this which goes thus:
F = ma
f = m(v-u)/t
Cross multiply
ft = m(v-u)
where,
f = 55
t = 45/1000 = 0.045
m = 0.0060
u = -32
v = Unknown
Therefore,
55 × 45/1000 = 0.060(v - -32)
55 × 0.045 = 0.060(v + 32)
2.475 = 0.06(v + 32)
2.475 = 0.06v + 1.92
0.06v = 2.475 - 1.92
0.06v = 0.555
v = 0.555/0.06
v = 9.25m/s
v = 9.3m/s Approximately
Answer:
A.9.3 m/s
Explanation:
A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a diameter of 0.15 m, approximately how much mass can be lifted by the jack
Answer:
1400.38N
Explanation:
Step one
Given data
P1= 250N
D1= 0.02m
A1= πD1^2/4
substitute
[tex]A1= 3.142*0.02^2/4\\\\A1=3.142*10^-4[/tex]
D2= 0.15m
A1= πD2^2/4
[tex]A2= 3.142*0.15^2/4\\\\A2=1.76*10^-3[/tex]
Required
The load P2
Step two:
Applying the hydraulic expression for a non-compressible fluid
we know that
Pressure= force/are
P1/A1=P2/A2
[tex]250/3.142*10^-4= P2/1.76*10^-3[/tex]
cross multiply we have
P2= 1.76*10^-3*250/3.142*10^-4
P2=0.44/3.142*10^-4
P2=1400.38N
When an object falls, it is reacting to the force of gravity. true or false
Answer:
The answer is true, as gravity affects everything.
An object that is initially traveling at 17.6 m/s has
1,743 J of work performed on it. If the mass of the
object is 283 kg, its final kinetic energy will be
J.
Find the resultant of the following displacement:
A = 20 Km 30° south of east;
B = 50 Km due west;
C = 40 Km north east;
D = 30 Km 60° south of west.
Answer:
Explanation:
Basically you just have to find the left vectors. To do so divide A, C and D into horizontal and vertical vector. A: 10km to south and 10root3 to east. Just sine and cosine of 30 at 20km. D: 15 km to west and 15root3 to south. Again sine and cosine of 60 at 30 km. C: 45 degrees so 20root2 to north and east. Add all these up with B. Then you have 7.696 km due south and 19.395 km due west. Resultant displacement magnitude = root(7.696^2+19.395^2)=20.866 south of west with angle=arctan(7.696/19.395)=21.644 degrees
A harmonic oscillator starts with an amplitude of 20.0 cm. After 10.0 s, the amplitude decreases to 15.0 cm. If the linear damping coefficient is 2.00 Ns/m, how much mass is oscillating
Answer: the amount of mass is oscillating is 34.8 kg
Explanation:
Given that;
amplitude A = 20.0 cm
time t = 10 s
amplitude decreases x = 15.0 cm
damping coefficient b = 2.00 N.s/m
amount of mass is oscillating = ?
we know that; amplitude can be expressed as;
x = Ae^-(∝t)
we substitute
15 = 20e^-∝(10)
∝ = 0.02877 s⁻¹
Hence mass m will be;
m = b/2∝
we substitute
m = (2 N.s/m) / ( 2 × 0.02877 s⁻¹)
m = 34.8 kg
Therefore the amount of mass is oscillating is 34.8 kg
Which electromagnetic waves have the shortest wavelength and the highest frequency?
i tried gamma and it said it was inncorect plz help
A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic friction coefficient is 0.1.
Draw a FBD of all the forces acting on the sled.
What is the weight of the sled?
What force is needed to start the sled moving?
What force is needed to keep the sled moving at a constant velocity?
Answer:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.
b) The weight of the sled is 490.35 newtons.
c) A force of 147.105 newtons is needed to start the sled moving.
d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
Explanation:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:
[tex]F[/tex] - External force exerted on the sled, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force from the ground on the mass, measured in newtons.
[tex]W[/tex] - Weight, measured in newtons.
b) The weight of the sled is determined by the following formula:
[tex]W = m\cdot g[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]m = 50\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the weight of the sled is:
[tex]W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 490.35\,N[/tex]
The weight of the sled is 490.35 newtons.
c) The minimum force needed to start the sled moving on the horizontal ground is:
[tex]F_{min,s} = \mu_{s}\cdot W[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]W[/tex] - Weight of the sled, measured in newtons.
If we know that [tex]\mu_{s} = 0.3[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to start the sled moving is:
[tex]F_{min,s} = 0.3\cdot (490.35\,N)[/tex]
[tex]F_{min,s} = 147.105\,N[/tex]
A force of 147.105 newtons is needed to start the sled moving.
d) The minimum force needed to keep the sled moving at constant velocity is:
[tex]F_{min,k} = \mu_{k}\cdot W[/tex] (3)
Where [tex]\mu_{k}[/tex] is the kinetic coefficient of friction, dimensionless.
If we know that [tex]\mu_{k} = 0.1[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to keep the sled moving at a constant velocity is:
[tex]F_{min,k} = 0.1\cdot (490.35\,N)[/tex]
[tex]F_{min,k} = 49.035\,N[/tex]
A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
Is anyone good at science I need help with 2 tests
Answer:
i am!
Explanation:
A spanner is an example of:
i. screw ii. wheel and axle
iii. pulley
iv. wedge
formula
Answer:
The answer is "wedge"
Explanation:
A wedge is an item that tightens to a meager edge. Pushing the wedge one direction makes a force in a sideways direction. It is normally made of metal or wood and is used for parting, tightening, lifting, or fixing, as in making sure about a hammer head onto its handle.
The wedge was used in ancient occasions to part logs and shakes; an ax is also a wedge, similar to the teeth on a saw. As far as its mechanical capacity, the screw might be considered as a wedge folded over a cylinder.
In order to model the motion of an extinct ape, scientists measure its hand and arm bones. From shoulder to wrist, the arm bones are 0.60 m long and their mass is 4.0 kg. From wrist to the tip of the fingers, the hand bones are 0.10 m long and their mass is 1.0 kg. In the model above, each bone is assumed to have a uniform density. When the arm and hand hang straight down, the distance from the shoulder to the center of mass of the arm-hand system is most nearly
Answer:
0.37 m
Explanation:
Let the shoulder be the origin.
The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.
The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m
Substituting these into the equation for the center of mass, we have
y = (m₁y₁ + m₂y₂)/(m₁ + m₂)
y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)
y = (1.20 kgm + 0.65 kgm)/5.0 kg
y = 1.85 kgm/5.0 kg
y = 0.37 m
The distance of the center of mass from the shoulder is thus y = 0.37 m
How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?
s = 40m - 4m = 36m
W = F × s
= 10N × 36m = 360J
A bit of explanation :W = Work (J)
F = Force / weight (N)
s = distance (m)
Work done in physics is the product of force and displacement. The displacement for the object is 36 m and force acts on it is 10 N. Then the work done is 360 J.
What is work done?Work done is the dot product of force acting on a body and the resultant displacement. When a force applied on an object results in a displacement from its position, the force is said to be work done.
Work done is a vector quantity thus, characterised by a magnitude and direction. The common unit of work done is joule.
Given that force applied on the weight = 10 N
displacement occurred = 40 m - 4 m = 36 m
Work done = F . ds
ds = 36 m and f = 10 N
Then W = 10 N × 36m
= 360 J.
Therefore, the work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth is 360 J.
To find more on work done, refer here:
https://brainly.com/question/13662169
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A pinball machine launches a .045 kg ball at a speed of 9.2 m/s. Determine the potential energy of the spring just before it launched the pinball
Given :
A pinball machine launches a .045 kg ball at a speed of 9.2 m/s.
To Find :
The potential energy of the spring just before it launched the pinball.
Solution :
We know, their is no external force applied on system.
It means that kinetic energy will remains constant.
Initial Energy = Final Energy
[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\0 + P.E_i= \dfrac{mv^2}{2}+ 0\\\\P.E_i = \dfrac{0.045 \times 9.2^2}{2}\ J\\\\P.E_i = 1.9044\ J[/tex]
Therefore, the potential energy of the spring just before it launched the pinball os 1.9044 J.
