Complete Question
Four small masses of 0.2 kg each are connected by light rods 0.4m long to form a square. What is the moment of inertia of this object for an axis through the middle of the square and parallel to two sides.
Answer:
[tex]I=0.032kgm^2[/tex]
Explanation:
From the question we are told that:
Mass[tex]m=0.2kg[/tex]
Length [tex]l=0.4m[/tex]
Generally the equation for Inertia is mathematically given by
[tex]I=md^2[/tex]
[tex]I=0.8*0.20(\frac{0.40}{2})^2[/tex]
[tex]I=0.032kgm^2[/tex]
What has a wind speed of 240 kph or greater?
Answer:
SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.
Light of intensity I0 and polarized horizontally passes through three polarizes. The first and third polarizing axes are horizontal, but the second one is oriented 20.0� to the horizontal. In terms of I0, what is the intensity of the light that passes through the set of polarizers?
A) 0.442 I0
B) 0.180 I0
C) 0.780 I0
D) 0.883 I0
Answer:
Option C.
Explanation:
Suppose that we have light polarized in some given direction with an intensity I0, and it passes through a polarizer that has an angle θ with respect to the polarization of the light, the intensity that comes out of the polarizer will be:
I(θ) = I0*cos^2(θ)
Ok, we know that the light is polarized horizontally and comes with an intensity I0
The first polarizer axis is horizontal, then the intensity after this polarizer is:
then θ = 0°
I(0°) = I0*cos^2(0°) = I0
The intensity does not change. The axis of polarization does not change.
The second polarizer is oriented at 20° from the horizontal, then the intensity that comes out of this polarizer is:
I(20°) = I0*cos^2(20°) = I0*0.88
And the axis of polarization of the light that comes out is now 20° from the horizontal
Now the light passes through the last polarizer, which has an axis oriented horizontally, so the final intensity of the light will be:
note that here the initial polarization is I0*0.88
and the angle between the axis is 20° again.
Then the final intensity is:
I(20°) = I0*0.88*cos^2(20°) = I0*0.78
Then the correct option is C.
Topic: Chapter 19: Some wiggle room
A hummingbird flaps its wings up to 70 times per second, producing a 70 Hz
hum as it flies by. If the speed of sound is 340 m/s, how far does the sound
travel between wing flaps?
= 4.86 m
= 58.9 m
= 0.206 m
= 23,800 m
Answer:
4.86 m
Explanation:
Given that,
The frequency produced by a humming bird, f = 70 Hz
The speed of sound, v = 340 m/s
We need to find how far does the sound travel between wing flaps. Let the distance is equal to its wavelength. So,
[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]
So, the sound travel 4.86 m between wings flaps.
Which is a mixture?
'a' sodium metal
'b' chlorine gas
'c' sodium metal and chlorine gas
'd' sodium chloride (salt) and water
Answer:
d. Sodium chloride (salt) + water
Explanation:
A mixture is made up of two or more substance combined together (combined chemically).NaCl (salt) can completely dissolve in water and sodium chlorine (aqueous) is a homogeneous mixture.sodium metal when extracted is a soft, silvery white solid.chlorine gas is a pure gas.sodium metal and chlorine gas are at pure state hence they are not mixture.learn more: https://brainly.com/question/2331419
Answer:D. Sodium chloride (salt) and water
Explanation:
I got it right on edge 2023
hope this is helpful!
A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800-N firefighter has climbed 4.00 m along the ladder from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9.00 m from the bottom, what is the coefficient of static friction between ladder and ground
Answer:
a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36
Explanation:
a) This is a balancing act.
Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive
-w x - W x₂ + R y = 0 (1)
usemso trigonometry to find distances
cos 60.08 = x / 7.5
x = 7.5 cos 60.08
x = 3.74 m
fireman
cos 60.08 = x₂ / 4
x2 = 4 cos 60
x2 = 2 m
wall support
sin 60.08 = y / 15
y = 15 are 60.08
y = 13 m
we substitute in equation 1
R y = w x + W x2
R = (w x + W x2) / y
R = (500 3.74 +800 2) / 13
R = 266.92 N
now let's write the expressions for the translational equilibrium
X axis
R -fr = 0
R = fr
fr = 266.92 N
Y Axis
Fy - w-W = 0
fy = 500 + 800
fy = 1300 N
b) ask the friction coefficient
the firefighter's distance is
cos 60.08 = x₃ / 9.00
x₃ = 9 cos 60
x₃ = 5.28 m
from equation 1
R = (w x + W x₃) / y
R = 500 3.74 + 800 5.28) / 13
R = 468.769 N
we saw that
fr = R = 468.769
The expression for the friction force is
fr = μ N
in this case the normal is the ratio to pesos
N = Fy
N = 1300 N
μ N = fr
μ = fr / N
μ = 468,769 / 1300
μ = 0.36
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before coming to rest.
Explanation:
Here,
v = 45m/s
t = 5s
d = v × t
Therefore,
d = 45 × 5
= 225m
Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb
Answer:
i₀ = V / R_i
Explanation:
For this exercise we use Ohm's law
V = i R
i = V / R
the equivalent resistance for
[tex]\frac{1}{R_{eq}}[/tex] = ∑ [tex]\frac{1}{R_i}[/tex]
if all the bulbs have the same resistance, there are N bulbs
[tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]
R_{eq} = R_i / N
we substitute
i = N V / Ri
where i is the total current that passes through the parallel, the current in a branch is
i₀ = i / N
i₀ = V / R_i
find the distance travelled by a moving body if it attained acceleration of 2m/s2 after starting from rest in 5min
Answer:
300 meters
Explanation:
a= 2m/s^2
t= 5 min
Convert into seconds, 5*60= 300seconds
v0= 0
x0=0
use x-x0= v0t + 1/2at^2
plug in values
x= 1/2*2*(300)
Solve
x= 300 meters
A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should thisresult be recorded in a table of results? a. 0.2000m b. 0.200m c. 0.20m d. 0.2m
Answer:
C
Explanation:
20 cm = 0.2m
since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty
therefore it's 0.200m
A block of mass M is connected by a string and pulley to a hanging mass m.
The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg.
b. Find the acceleration of the system and tensions on the string.
c. How far will block m drop in the first seconds after the system is released?
d. How long will block M move during the above time?
e. At the time, calculate the velocity of block M
f. Find out the deceleration of block M if the connection string is removal by cutting after the first second. Then, calculate the time taken to contact block M and pulley
How far will block m drop in the first seconds after the system is released?
(b) Use Newton's second law. The net forces on block M are
• ∑ F (horizontal) = T - f = Ma … … … [1]
• ∑ F (vertical) = n - Mg = 0 … … … [2]
where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.
Right away, we see n = Mg, and so f = µn = 0.2Mg.
The net force on block m is
• ∑ F = mg - T = ma … … … [3]
You can eliminate T and solve for a by adding [1] to [3] :
(T - 0.2Mg) + (mg - T ) = Ma + ma
(m - 0.2M) g = (M + m) a
a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)
a = 1.96 m/s²
We can get the tension from [3] :
T = m (g - a)
T = (10 kg) (9.8 m/s² - 1.96 m/s²)
T = 78.4 N
(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of
1/2 (1.96 m/s²) t
(e) Assuming block M starts from rest, its velocity at time t is
(1.96 m/s²) t
(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have
∑ F = -f = Ma
The effect of friction is constant, so that f = 0.2Mg as before, and
-0.2Mg = Ma
a = -0.2g
a = -1.96 m/s²
Then block M slides a distance x such that
0² - (1.96 m/s²) = 2 (-1.96 m/s²) x
x = (1.96 m/s²) / (2 (1.96 m/s²))
x = 0.5 m
(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)
Meanwhile, block m would be in free fall, so after 1 s it would fall a distance
x = 1/2 (-9.8 m/s²) (1 s)
x = 4.9 m
state the story of archimedes
Answer:
Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.
Energy from the sun comes to Earth as radiant energy. Which of these is an example of radiant energy being converted to heat energy?
A Turning windmills transform mechanical energy into electrical energy.
B Black shirts feel hotter than light-colored shirts on a sunny day.
C Solar cells convert sunlight into electrical energy.
D Green plants use sunlight in photosynthesis.
Answer:
B
Explanation:
The radiant energy form the sun is absorbed by the black shirt and is converted to heat energy.
Answer:
B Black shirts feel hotter than light-colored shirts on a sunny day.
Explanation:
The energy from the sun also called solar energy is an energy source which reaches the earth as a form of radiant energy, that is it is transmitted without the movement of mass. Solar cells absorbs radiant energy from the sun into electrical energy for powering electrical devices.
During photosynthesis, sunlight absorbed by the chlorophyll of green plants is converted into chemical energy.
In black body, radiant energy abosrde are stored and converted to heat energy, reason dark colored clothes feels hotter than light colored on sunny days.
how did kepler discoveries contribute to astronomy
Answer:
They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.
Explanation:
A dandelion seed floats to the ground in a mild wind with a resultant velocity of 26.0 cm/s. If the horizontal component velocity due to the wind is 10.0 cm/s, what is the vertical component velocity? Show all work.
Answer:
24 cm/s
Explanation:
Applying
Pythagoras theorem,
a² = b²+c²............. Equation 1
Where a = resultant, b = vertical component, c = horizontal component
From the question,
Given: a = 26 cm/s, c = 10 cm/s
Substitute these values into equation 1
26² = b²+10²
676 = b²+100
b² = 676-100
b² = 576
b = √576
b = 24 cm/s
An aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft dears the building by 12 m, how far from the building does the aircraft touch down on the runway
The aircraft is 12 meters higher than the building so it is at 45 + 12 = 57 meters high.
For every 12 meters it travels it drops 1 m.
Divide the height by 12 to find the distance it travels:
57 / 12 = 4.75
It touches down 4.75 meters from the building.
The building is 684 meters away from the aircraft touching down on the runway.
What are trigonometric functions?A right-angled triangle's side ratios are the easiest way to express a function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant. These functions are known as trigonometric functions.
As given in the problem an aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft clears the building by 12 m,
the total height of the aircraft when it clears the building = 45 +12
the total height of the aircraft when it clears the building is 57 meters
It is given that the Glide ratio is 12:1,
The distance of the building from touch down on the runway = 12 ×57
The distance of the building from the touch-down on the runway is 684 meters.
Thus, the building is 684 meters away from the aircraft touching down on the runway.
Learn more about the trigonometric functions here,
brainly.com/question/14746686
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Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities: The collision is instantaneous. The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact. The collision takes a finite amount of time, during which the ball and bat are bending or being compressed. How can two of these be ruled out based on energy or momentum considerations?
The collision is instantaneous.
The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.
How can two of these be ruled out based on energy or momentum considerations?
Answer:
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
Explanation:
These two conditions can be ruled out on the fact that :The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
The rule of energy is been broken here because during the collision of objects energy and momentum is conserved. i.e. the change in shape of the ball when hit by the bat should not be noticed because the compression and returning to normal shape happens instantaneously
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
If a bale of hay behind the target exerts a constant friction force, how much farther will your arrow burry itself into the hay than the arrow from the younger shooter
Answer:
The arrow will bury itself farther by 3S₁
Explanation:
lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter
Given that ; acceleration is constant , Frictional force is constant
A₂ = A₁
Vf²₂ - Vi²₂ / 2s₂ = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )
final velocities = 0
Initial velocities : Vi₂ = 2(Vi₁ )
Back to equation 1
0 - (2Vi₁ )² / 2s₂ = 0 - Vi₁² / 2s₁
hence :
s₂ = 4s₁
hence the Arrow shot by me will burry itself farther by :
s₂ - s₁ = 3s₁
Note : S1 = distance travelled by the arrow shot by the younger shooter
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
What is the value of the charge that experiences a force of 2.4×10^-3N in an electric field of 6.8×10^-5N/C
Hi there!
[tex]\large\boxed{\approx 35.29 C}[/tex]
Use the following formula:
E = F / C, where:
E = electric field (N/C)
F = force (N)
C = Charge (C)
Thus:
6.8 × 10⁻⁵ = 2.4 × 10⁻³ / C
Isolate for C:
C = 2.4 × 10⁻³ / 6.8 × 10⁻⁵
Solve:
≈ 35.29 C
Stationary waves are
A) transverse waves
B) longitudinal waves
C) mechanical waves
Answer:
stationary waves are transverse waves
How much work is required to stretch an ideal spring of spring constant (force constant) 40 N/m from x
Answer:
The work done will be "0.45 J".
Explanation:
Given:
K = 40 N/m
x₁ = 0.20 m
x₂ = 0.25 m
Now,
The required work done will be:
= [tex]\frac{1}{2}k[x_2^2-x_1][/tex]
By putting the values, we get
= [tex]\frac{40}{2}[(0.25)^2-(0.20)^2][/tex]
= [tex]20\times 0.0225[/tex]
= [tex]0.45 \ J[/tex]
Question 7 of 10
Which statement best describes diffraction?
A. Waves bend as they pass through an opening.
B. Waves and vibrations are oriented in a single direction.
.
C. Waves bounce off a surface.
D. Waves change direction as they enter a new material.
Answer:
A. Waves bend as they pass through an opening.
Explanation:
important word here is "opening"
diffraction example is a CD reflecting rainbow colors
A. Waves bend as they pass through an opening best describes diffraction.
Diffraction is the spreading out or bending of waves as they pass through an aperture or around an object. If we talk about light waves, diffraction of light occurs when a light wave passes by a corner or through a slit or opening. The slit or opening can be physically approximately the size of, or even smaller than that light's physical wavelength. An example of diffraction is the diffraction of sunlight by the clouds.
To know more, refer to,
brainly.com/question/10709914
The attached picture shows the diffraction of light through a single slit.
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You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.
Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?
Answer:
Explanation:
a)
v² = u² + 2 a s
v = 9 m/s
u = 4 m/s
s = 50 m
9² = 4² + 2 x a x 50
a = 0.65 m /s²
Acceleration is 0.65 m /s²
b )
time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²
(v - u ) / t = a
(v - u ) / a = t
(9 - 4 ) / .65 = t
t = 7.7
time when passing the first sign will be 7.7 s earlier .
Reading of time indicator = 45 - 7.7
= 37.3 seconds.
Answer:
(a) 0.45 m/s^2
(b) 33.9 s
Explanation:
initial velocity, u = 4 m/s
final velocity, v = 9 m/s
distance, s = 50 m
(a) Let the acceleration is a.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2[/tex]
(b) Let the time is t.
Use first equation of motion
v = u + at
9 = 4 + 0.45 x t
t = 11.1 s
So, the initial time, t' = 45 - 11.1 = 33.9 s
del tema de fuerza centripeta
1.- Un chico va en bicicleta a 10m/s por una curva plana de 200m de radio.
a) ¿Cuál es la aceleración?
b) si el chico y la bicicleta tienen una masa total de 70kg, ¿Qué fuerza se necesita para producir esta aceleración?
Answer:
a. C = 0.5 m/s²
b. F = 35 Newton
Explanation:
Given the following data;
Radius, r = 200 m
Velocity, v = 10 m/s
Mass, m = 70 kg
a. To find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
C = v²/r
Where:
C is the centripetal acceleration
v is the velocity
r is the radius
Substituting into the formula, we have;
C = 10²/200
C = 100/200
C = 0.5 m/s²
b. To find the force;
F = mv²/r
F = (70*10²)/200
F = (70 * 100)/200
F = 7000/200
F = 35 Newton
3
Select the correct answer.
What is a substance?
Answer:
physical material from which something is made or which has discrete existence
Explanation:
An object accelerates from rest, and after traveling 145 m it has a speed of 420 m/s. What was the acceleration of the object?
I am not sure how to calculate acceleration without being given the time directly.
Explanation:
Here,we've been given that,
Initial velocity (u) = 0 m/s (as it starts from rest)Distance (s) = 145 mFinal velocity (v) = 420 m/sWe've to find the acceleration of the object. By using the third equation of motion,
→ v² - u² = 2as
→ (420)² - (0)² = 2 × a × 145
→ 176400 - 0 = 290a
→ 176400 = 290a
→ 176400 ÷ 290 = a
→ 608.275862 m/s² = a
If you know initial speed and final speed, you can find the average speed. Then, knowing distance, you can find the time.
KimYurii posted the first answer to this question.
That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.
My problem is that I can never remember all the different formulas. I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo. Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.
When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:
Starting speed . . . zero
Ending speed . . . 420 m/s
Formula: Average speed . . . (1/2)·(0 + 420) = 210 m/s
Distance covered . . . 145 m
Formula: Time taken = (distance) / (average speed) = (145/210) second
(Now you have the time.)
Formula: Distance = (1/2)·(acceleration)·(time²)
145 m = (1/2)·(acceleration)·(145/210 sec)²
Acceleration = 290 m / (145/210 s)²
Acceleration = 608.28 m/s²
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
If 5kg Stone and 1kg stone throw the from the building which will land more fa ster and why?
Answer:
Both stones will land at the same time because both stones will fall with the same acceleration through the same height.
Explanation:
We are given that
Mass of stone ,m1=5 Kg
Mass of stone, m2=1 kg
We have to find which stone more faster will land and why.
[tex]h=u+\frac{1}{2}gt^2[/tex]
Initial velocity of both stones=0
[tex]h=\frac{1}{2}gt^2[/tex]
[tex]t^2=\frac{h}{g}[/tex]
[tex]t=\sqrt{\frac{h}{g}}[/tex]
[tex]t_1=t_2=\sqrt{\frac{h}{g}}[/tex]
Because both stones are thrown from the same height.
Both stones will land at the same time because both stones will fall with the same acceleration through the same height and the acceleration does not depend of its mass.