Answer: I don’t know someone tell me
Explanation: ^
A woman of mass 55 kg stands on the rim of a frictionless merry-go-round of radius 2.0m and rotational inertia 1250 kg-m2 that is not moving. She throws a rock of mass 350g horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 2.0m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the woman
Answer:
a) [tex]\omega=9.10*10^{-4}rad/sec[/tex]
b) [tex]V=1.8217*10^{-3}[/tex]
Explanation:
From the question we are told that:
Mass of woman [tex]M_w=55kg[/tex]
Radius of merry go round [tex]r=2.0m[/tex]
Rotational inertia [tex]i= 1250 kg-m2[/tex]
Mass of rock [tex]M_r=350g \approx 0.350kg[/tex]
Speed of rock [tex]V_r=2.0m/s[/tex]
Tangent angle to the outer edge [tex]\theta=1[/tex]
a)
Generally the equation for conservation of momentum is mathematically given by
[tex]M_r(ucos\theta)r=(I+M_wr^2)\omega[/tex]
[tex]0.350(2.0cos1)(2.0)=(1250+(55)(2.0)^2)\omega[/tex]
[tex]1.3998=1470\omega\\\omega=\frac{1.339}{1470}[/tex]
[tex]\omega=9.10*10^{-4}rad/sec[/tex]
b)
Generally the equation for linear speed V is mathematically given by
[tex]V=r\omega\\V=2.0*9.10*10^{-4}[/tex]
[tex]V=1.8217*10^{-3}[/tex]
Question 6 of 10
What is the length x of the side of the triangle below? (Hint: Use the cosine
function.)
X
330
14
A. 9.1
B. 7.6
C. 11.7
D. 10.2
C (11.7)
Explanation:
cos 33 ÷ 14 = 11.7
1. For a person who has a mass 60 kg, calculate their force due to gravity,
otherwise know as "weight," in newtons if the person is in space.
Answer:
Explanation:
Given that gravity on the Moon has approximately 1/6th of the strength of gravity on Earth, a man who weighs 60kg on Earth would weigh approximately 10kg on the Moon. His mass, however, remains constant.
Electromagnetic waves are commonly referred to as _________
O electricity
O magnetism
O light
What makes water move through the water cycle? A the sun the rain C precipitation the ground
Answer: A. The Sun
Explanation: A. The Sun makes water move through the water cycle
Millikan used tiny oil droplets and adjusted the electric field until the electric force balanced the weight of the droplet, and one of the droplets hovered in air. However, the drops are all different sizes and weights, and the amount of charge on each droplet also varies. So, the tuning of the electric field is different for droplets of different sizes, which carry different charges. Let us assume that the density of the oil Millikan used in his experiment is 800 kg m3
A. Calculate the mass (in kg) of the hovering droplet if its radius is measured to be 65 um.
B. Calculate the electric field (in V/m) required to counter the weight of the above droplet if it carries a charge q = 2e.
C. In the second trial, the hovering droplet has a mass of 10-15 kg. It carries a charge of q = 11e. calculate the electric field (in V/m) required to counter the weight of this droplet.
Answer:
A. 9.203 × 10⁻¹⁰ kg
B. 2.82 × 10¹⁰ V/m
C. 5.56 × 10³ V/m
Explanation:
A. Calculate the mass (in kg) of the hovering droplet if its radius is measured to be 65 um.
We know density, ρ = mass of oil drop, m/volume of oil drop, v
m = ρv
ρ = 800 kg/m³ and v = 4πr³/3 (ince the oil drop is a sphere) where r = radius of oil drop = 65 μm = 65 × 10⁻⁶ m.
So, m = ρv = ρ4πr³/3
= 800 kg/m³ × 4π(65 × 10⁻⁶ m)³/3
= 800 kg/m³ × 4π274625 × 10⁻¹⁸ m³/3
= 878800000π × 10⁻¹⁸ kg/3
= 2760831623.97 × 10⁻¹⁸ kg/3
= 920277207.99 × 10⁻¹⁸ kg
= 9.203 × 10⁻¹⁰ kg
B. Calculate the electric field (in V/m) required to counter the weight of the above droplet if it carries a charge q = 2e.
Since the electric force F = qE where q = charge on oil drop = 2e where e = electron charge = 1.602 × 10⁻¹⁹ C and E = electric field equals the weight of the oil drop W = mg where m = mass of oil drop = 9.203 × 10⁻¹⁰ kg and g = acceleration due to gravity = 9.8 m/s².
So, F = W
qE = mg
E = mg/q
E = mg/2e
substituting the values of the variables into the equation, we have
E = 9.203 × 10⁻¹⁰ kg × 9.8 m/s²/(2 × 1.602 × 10⁻¹⁹ C)
E = 90.1894 × 10⁻¹⁰ kg-m/s²/3.204 × 10⁻¹⁹ C
E = 28.15 × 10⁹ V/m
E = 2.815 × 10¹⁰ V/m
E ≅ 2.82 × 10¹⁰ V/m
C. In the second trial, the hovering droplet has a mass of 10-15 kg. It carries a charge of q = 11e. calculate the electric field (in V/m) required to counter the weight of this droplet.
Since F = W
qE = mg
E = mg/q
E = mg/11e
E = 10⁻¹⁵ kg × 9.8 m/s²/(11 × 1.602 × 10⁻¹⁹ C)
E = 9.8 × 10⁻¹⁵ kg-m/s²/17.622 × 10⁻¹⁹ C
E = 0.556 × 10⁴ V/m
E = 5.56 × 10³ V/m
At its peak, a tornado is 65 m in diameter and has 350 km/h winds. What is its frequency in revolutions per second?
Answer:
Explanation:
circumference = diameter * [tex]\pi[/tex]
circumference = 65 * [tex]\pi[/tex]
circumference = 204.2035225m
350 km/h = 97.222222 m/s
0.4761 revolutions per second
A force F making an angle with the horizontal is acting on an object resting on the table. Which statement is true for the motion of sliding the object on the table? A. It is independent of all the forces acting on the object. B. It depends only on the forces acting along the x-axis. C. It depends only on the forces acting along the y-axis. D. It depends only on the normal force acting on the object. E. It depends only on the frictional force acting on the object.
Answer:
C
Explanation:
it depends only on the forces acting along the y- axis
Answer:
The statement that is true for the motion of sliding the object on the table is : its independent of all forces acting on the object
Explanation:
PLS THIS IS DUE IN 2 MINUTES
Which has more momentum, a 0.5kg toy car moving a 5 m/s or a 1000kg real car that is
parked?
Answer:
The toy car
Explanation:
the real car is parked so yeah but maybe in some way technically the real car has more "momentum"
whats it like to be famous?
Good And Bad
Explanation:
People often want to be famous to be recognized in public or even to meet other famous people.
however what people dont realise is that you life is now entirely public, your love life, relationships, the food you eat, the people you talk to,the way you act ,ect Everything is publicly judged.
yes it may be good to be famous for some reasons however, 9 times out of 10 it may be the worst part of your life.
Cual es la Ecuación de velocidad de una onda en función del índice de la amplitud de una cuerda
Answer:
Well velocity is the varaible with direction
Explanation:
In a cloud chamber experiment, a proton enters a uniform 0.260 T magnetic field directed perpendicular to its motion. You measure the proton's path on a photograph and find that it follows a circular arc of radius 6.42 cm.
Required:
How fast was the proton moving?
Answer:
the proton speed of the proton was 1.6 × 10⁶ m/s
Explanation:
Given the data in the question;
Radius r = 6.42 cm = 0.0642 m
magnetic field B = 0.260 T
we know that; charge of proton q = 1.602 × 10⁻¹⁹ C
And mass of proton m = 1.672 × 10⁻²⁷ kg
we know that; Magnetic Force F = qvBsinθ
where q is the charge of proton, v is velocity, B is the magnetic field and θ is angle ( 90° )
Also the Centripetal force experienced by the particle is;
F = mv² / r
where r is radius, m is mass of proton and v is velocity
hence;
qvBsinθ = mv² / r
we solve for v
rqvBsinθ = mv²
divide both sides by mv
rqvBsinθ / mv = mv² / mv
rqBsinθ / m = v
so we substitute
v = [ 0.0642 m × (1.602 × 10⁻¹⁹ C) × 0.260 T × sin(90°) ] / 1.67 × 10⁻²⁷ kg
v = 2.6740584 × 10⁻²¹ / 1.672 × 10⁻²⁷
v = 1.6 × 10⁶ m/s
Therefore, the proton speed of the proton was 1.6 × 10⁶ m/s
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
A body of m = 6.8kg is launched with a speed of 7.5 m / s towards the top of an inclined plane of 15 ° with respect to the horizontal. in the absence of friction, what displacement does it make before reversing the direction of motion?
Answer:
d = 11.1 m
Explanation:
Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:
[tex] \frac{1}{2} m {v}^{2} = mgh[/tex]
Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:
[tex] \sin(15) = \frac{h}{d} \\ or \: h = d \sin(15) [/tex]
Plugging this into the energy conservation equation and cancelling m, we get
[tex] {v}^{2} = 2gd \sin(15)[/tex]
Solving for d,
[tex]d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m[/tex]
PLZ HELP NO SNEAKY LINK OR I WILL REPORT U OR NON ANSWERS
define quant......am physics
Answer:
✍ What is quantum physics?➳ It's the physics that explains how everything works; the best description we have of the nature of the particles that make up matter and the forces with which they interact. Quantum physics underlies how atoms work, and so why chemistry and biology work as they do.The blending of the fundamental tone with overtones produces the __?__of an instrument.
Answer:
The blending of the fundamental tone with overtones produces the Sound Quality of Musical Instruments of an instrument.
The blending of the fundamental tone with overtones produces the sound quality of Musical Instruments of an instrument.
The tone of a sound wave determines the timbre, also known as timber, of the auditory impressions it creates. The wave structure of a sound, which varies with the amount of overtones, or harmonics, present, their frequencies, and their relative intensities, determines the timbre of the sound.Resonance is a common technique used by musical instruments to enhance sound waves and increase volume. When an object vibrates in response to sound waves of a particular frequency, this is known as resonance.What are tones and overtones?Aside from a simple sine wave, the fundamental tone and numerous additional tones of various frequencies make up the waveforms of all sounds. Overtones or harmonics are non-fundamental tones that are whole-number multiples of the fundamental tone.Learn more about sound quality of Musical Instruments brainly.com/question/9571788
#SPJ2
Suppose you observe a star orbiting the galactic center at a speed of 1400 km/s in a circular orbit with a radius of 26 light-days. Calculate the mass of the object that the star is orbiting. Express your answer in solar masses to two significant figures.
Answer:
M = 9.9 x 10⁶ Solar masses
Explanation:
Here the centripetal force is given by the gravitational force between star and the object:
[tex]Gravitational\ Force = Centripetal\ Force \\\frac{mv^2}{r} = \frac{GmM}{r^2}\\\\M = \frac{v^2r}{G}[/tex]
where,
M = Mass of Object = ?
v = orbital speed of star = 1400 km/s = 1400000 m/s
G = Universal Gravittaional Constant = 6.67 x 10⁻¹¹ N.m²/kg²
r = distance between star and object = (26 light-days)(2.59 x 10¹³ m/1 light-day) = 6.735 x 10¹⁴ m
Therefore,
[tex]M = \frac{(1400000\ m/s)^2(6.735\ x\ 10^{14}\ m)}{6.67\ x \ 10^{-11}\ N.m^2/kg^2}[/tex]
M = (1.97 x 10³⁷ kg)(1 solar mass/ 1.989 x 10³⁰ kg)
M = 9.9 x 10⁶ Solar masses
Find the Input force, if the Mechanical Advantage of the simple machine used is 5 and Output force is 50 N.
P.S PLEASE DO NOT POST A LINK THAT I HAVE TO DOWNLOAD EVERY TIME I ASK A QUESTION PEOPLE POST THAT JUST POST A WRITEN ANSWER WITH AN EXPLINATION THAT IS ALL
Answer:
6N
Explanation:
How many times does the kinetic energy of a car increase when traveling 60 mph as opposed to traveling 15 mph?
K.E. increases
5
4
16
20
Answer:
20
Explanation:
i'm not so sure with my answer
The bellows of an adjustable camera can be extended so that the largest distance from the lens to the film is 1.45 times the focal length. If the focal length of the lens is 6.14 cm, what is the distance from the closest object that can be sharply focused on the film
Answer: 19.80 cm
Explanation:
Given
focal length [tex]f=6.14\ cm[/tex] (as focal length is positive, it is converging lens)
Image distance [tex]v=1.5f[/tex]
[tex]v=1.5\times 6.14\\v=9.21\ cm[/tex]
using lens formula
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
insert values
[tex]\dfrac{1}{u}=\dfrac{1}{6.14}-\dfrac{1}{8.9}\\\\\dfrac{1}{u}=0.1628-0.1123\\\\\dfrac{1}{u}=0.0505\\\\u=19.80\ cm[/tex]
Thus, the distance of the object is 19.80 cm
Given:
Focal length, f = 6.14 cmNow,
The image distance will be:
→ [tex]v = 1.5 f[/tex]
[tex]= 1.5\times 6.14[/tex]
[tex]= 9.21 \ cm[/tex]
By using the lens formula, we get
→ [tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]
By putting the values, we get
→ [tex]\frac{1}{u} = \frac{1}{6.14} - \frac{1}{8.9}[/tex]
[tex]\frac{1}{u} = 0.1628-0.1123[/tex]
[tex]\frac{1}{u} = 0.0505[/tex]
[tex]u = 19.80 \ cm[/tex]
Thus the answer above is correct.
Learn more about focal length here:
https://brainly.com/question/25020197
A nurse asks a doctor about medication for a
patient in a hospital. The doctor accidentally tells
the nurse to give the patient the wrong
medication, and the nurse follows the doctor's
orders. Which type of medical error has occurred?
Answer:
medication error thats easy
A box is at rest on a table. What can you say about the forces acting on the box?
The upward
and the downward
are balanced.
Answer:
normal force and gravitational force
Explanation:
I hope this helps :D
Grandma Sue (mass 80 kg) and her grandson James (mass 40 kg) are on a smooth icy surface. As Grandma Sue whizzes around the icy surface at 3 m/s in a straight line, she is suddenly confronted with scared James standing at rest directly in her path. Rather than knock him over, she picks him up and continues her uniform motion in a straight line without braking. Find the speed of Grandma Sue and James after the collision.
Answer:
v = 2 m/s
Explanation:
Here, we will use the law of conservation of momentum to solve this problem:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of grandma = 80 kg
m₂ = mass of James = 40 kg
u₁ = initial speed of grandma = 3 m/s
u₂ = initial speed of James = 0 m/s
v₁ = v₂ = v = final speed of grandm and James = ?
Therefore,
[tex](80\ kg)(3\ m/s)+(40\ kg)(0\ m/s)=(80\ kg)(v)+(40\ kg)(v)\\\\(120\ kg)v = 240\ Ns\\\\v = \frac{240\ N.s}{120\ kg}\\[/tex]
v = 2 m/s
A horizontal line labeled B has an arrow labeled A strike it from right and above and then another arrow D emerges from the strike point at the same angle as A. A dotted line C evenly divides the angle between A and D.
In the mirror diagram shown, which is the normal?
A
B
C
D
Answer:
c is the actual answer.
Explanation:
Answer:
c is correct
Explanation:
When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the alors will emit electromagnetic radiation and visible light can be observed. If this light passes through a diffraction grating, the resulting spectrum appears as a pattern of four isolated, sharp parallel lines, called spectral lines. Each spectral line corresponds lo one specific wavelength that is present in the light emitted by the source. Such a discrete spectrum is referred to as a line spectrum
By the early 19th century, it was found that discrete spectra were produced by every chemical element in its gaseous slale. Even though these spectra were found to share the common feature of appearing as a set of isolated lines, it was observed that each element produces its own unique pattern of lines. This indicated that the light emitted by each element contains a specific set of wavelengths that is characteristic of that element.
part A What is the wavelength of the line corresponding to n =4 in the Balmer series? Express your answer in nanometers to three significant figures.
Part B What is the wavelength of the line corresponding to t=5 in the Balmer series? Express your answer in nanometers to three significant figures.
Answer:
A) λ = 4.88 10² nm, B) λ = 4.08 10² nm
Explanation:
The spectrum of hydrogen is correctly explained by the Bohr model, where the energy of each level is
Eₙ = -13.606 /n² [eV]
the transition generally occurs from a given level to a lower state nf <no, so a transition is
ΔE = E_f -Eₙ = -13,606 ( [tex]\frac{1}{n_f^2} - \frac{1}{n_o^2}[/tex] )
to find the wavelength let's use the planck relation
ΔE = h f
the speed of light is
c = λ f
we substitute
ΔE = h c /λ
λ = [tex]\frac{h \ c}{ \Delta \lambda}[/tex]
let's apply this equation to our case
the Balmer series has as final state the level n_f = 2
A) initial state n₀ = 4, final state n_f = 2
ΔE = -13.606 ( [tex]\frac{1}{2^2} - \frac{1}{4^2}[/tex] )
ΔE = 2.55 eV
let's reduce to SI units
ΔE = 2.55 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.08 10⁻¹⁹ J
we calculate
λ = 6.63 10⁻³⁴ 3 10⁸ / 4.08 10⁻¹⁹
λ = 4.875 10⁻⁻⁷ m
we reduce to nm
λ = 4.875 10⁻⁷ m (10⁹ nm / 1m)
λ = 487.5 nm
we reduce to three significant figures
λ = 4.88 10² nm
B) initial state n₀ = 5
ΔE = -13,606 ( [tex]\frac{1}{2^2} - \frac{1}{5^2}[/tex] )
ΔE = 2,857 eV
we repeat the process of the previous point
ΔE= 2,857 1.6 10⁺¹⁹ = 4.286 10⁻¹⁹J
we look for the wavelength
λ = 6.63 10⁻³⁴ 3 10⁸ / 4.88 10⁻¹⁹
λ = 4.0758 10⁻⁷ m
we reduce to nm
λ = 4.0758 10² nm
ignificant numbers
λ = 4.08 10² nm
1. Which statement about data attenuation is true? (1 point)
Data carried on a fiber-optic cable experience more attenuation than data broadcast from a radio tower.
Fiber-optic cables experience very little attenuation over large distances.
Data carried on a fiber-optic cable experience more attenuation than data carried on a copper wire.
Data carried on a copper wire and data broadcast from a radio tower experience very little attenuation over large distances.
2. Which statement is an example of information being carried by an electromagnetic wave that has been interpreted by humans?(1 point)
A sound wave in an ultrasound machine transmits information about the parts of the body it travels through.
A seismic wave changes speed, conveying information about the materials it has passed through.
A wave in the ocean changes, indicating a change in the depth of the water.
A light wave from a distant galaxy experiences redshift, indicating that the galaxy is moving away from Earth.
3. Which statement is true about a parabolic dish? (1 point)
It reflects waves and focuses them to a point.
It refracts waves in the dish.
It absorbs the wave energy.
It filters the type of electromagnetic radiation.
4. Which natural phenomenon describes guitar strings vibrating at certain natural frequencies?(1 point)
interference
beats
oscillation
resonance
5. What will increase the power output when doing work?(1 point)
doing less work in the same amount of time
doing the same work in less time
doing the same work in more time
doing less work and increasing the amount of time
Answer:
1. c
2. d
3. a
4. d
5. c
Explanation:
im pretty sure im right
A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a force of 34 N over 0.6 seconds to the pedal, which has a length of 16.5 cm. If the back wheel has a radius of 33.0 cm and moment of inertia of 1200 kg cm^2, what is the tangential velocity of the rim of the back wheel in m/s
Answer:
[tex]V=9.2565m/s[/tex]
Explanation:
From the question we are told that:
Force [tex]F = 34 N[/tex]
Time [tex]t = 0.6 s[/tex]
Length of pedal [tex]l_p=16.5cm \approx0.165m[/tex]
Radius of wheel [tex]r = 33 cm = 0.33 m[/tex]
Moment of inertia, [tex]I = 1200 kgcm2 = 0.12 kg.m2[/tex]
Generally the equation for Torque on pedal [tex]\mu[/tex] is mathematically given by
[tex]\mu=F*L\\\mu=34*0.165[/tex]
[tex]\mu=5.61N.m[/tex]
Generally the equation for angular acceleration [tex]\alpha[/tex] is mathematically given by
[tex]\alpha=\frac{\mu}{l}[/tex]
[tex]\alpha=\frac{5.61}{0.12}[/tex]
[tex]\alpha=46.75[/tex]
Therefore Angular speed is \omega
[tex]\omega=\alpha*t[/tex]
[tex]\omega=(46.75)*(0.6)[/tex]
[tex]\omega=28.05rad/s[/tex]
Generally the equation for Tangential velocity V is mathematically given by
[tex]V=r\omega[/tex]
[tex]V=(0.33)(28.05)[/tex]
[tex]V=9.2565m/s[/tex]
If the speed of a wave increases...
A. The frequency decrease
B. The frequency increases
Answer:
A. The frequency decrease
In transverse waves, the movement of the particles is _________
O in a circle
O left to right
O diagonal
O up and down
Answer:
the correct answer is up and down
Part A : A cylindrical water tank 10cm high and 520cm in diameter is filled with solution with a density of 0.75 g/cm^3
a) What is the water pressure on the bottom of the tank? (pressure in psi)
Show calculations
b) What is the average force on the bottom?
Answer: [tex]0.1066\ psi, 15.611\ kN[/tex]
Explanation:
Given
the height of the tank is [tex]h=10\ cm[/tex]
The diameter of the tank is [tex]d=520\ cm[/tex]
Density of solution [tex]\rho=0.75\ g/cm^3\ or\ 750\ kg/m^3[/tex]
[tex](a)[/tex] Water pressure at the bottom of the tank is
[tex]P=\rho gh\\P=750\times 9.8\times 0.1\\P=735\ Pa[/tex]
[tex]1\ Pa=0.000145038\ psi[/tex]
[tex]\Rightarrow P=735\ Pa\ or\ 0.1066\ psi[/tex]
[tex](b) \text{Average force on the bottom is the product of pressure and area of the base}[/tex]
[tex]F_{avg}=735\times \pi \cdot (\frac{520}{200})^2\\\\F_{avg}=735\times 3.142\times 6.76\\F_{avg}=15,611.34\ N\ or\ 15.611\ kN[/tex]