Answer:
The time taken by the wave to travel 8 m is 0.19 s .
Explanation:
frequency, f = 70 Hz
Amplitude, A = 5 mm
Wavelength = 0.6 m
The wave speed is given by
wave speed, v = frequency x wavelength
v = 70 x 0.6
v = 42 m/s
Distance, d = 8 m
Let the time taken is t.
So, [tex]t =\frac{d}{v}\\\\t =\frac{8}{42}\\\\t = 0.19 s[/tex]
What is air
A. A Buchner substance
B. A compound
C. An element
D. A mixture
Air is classified as a mixture. Option D is the correct answer.
Air is a combination of different gases, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases such as carbon dioxide, argon, and trace elements. These gases are not chemically bonded to each other, but rather exist together in the same space. Option D is the correct answer.
In a mixture, the substances involved retain their individual properties and can be separated by physical means. This is true for air as well. The gases in air can be separated through processes like fractional distillation or filtration. It's important to note that air also contains other components such as water vapor, dust particles, and pollutants, which can vary in concentration depending on the location and environmental conditions. These components further contribute to the complex nature of air as a mixture.
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The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light
Answer:
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Explanation:
The energy of a photon is calculated using the following equation;
E = hf
where;
h is Planck's constant = 6.63 x 10⁻³⁴ Js
f is frequency of the photon
[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]
[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
While riding his bike through a neighborhood, Joe notices a red sign out of
the corner of his eye. He quickly comes to a stop. Which active reading
strategy is he using?
A. Asking questions
B. Making mental images
C. Summarizing
D. Using visual cues
SUBMIT
Answer:
B
Explanation:
because it take to the thought of a situation
Answer:
d i think
Explanation:
and are u using IXL???
An object is 70 micrometer long and 47.66 micrometer wide. How long and wide is the object in km
Answer:
The length of the object in kilometer (km) is 70 x 10⁻⁹ km
The width of the object in kilometers (km) is 47.66 x 10⁻⁹ km
Explanation:
Given;
length of the object = 70 micrometer = 70 μm
the width of the object = 47.66 micrometer = 47.66 μm
The length of the object in meter:
70 micrometer = 70 μm = 70 x 10⁻⁶ m
The length of the object in kilometer (km):
70 x 10⁻⁶ m = 70 x 10⁻⁹ km
The width of the object in meters:
47.66 micrometer = 47.66 μm = 47.66 x 10⁻⁶ m
The width of the object in kilometers (km):
47.66 x 10⁻⁶ m = 47.66 x 10⁻⁹ km
The human ear can respond to an extremely large range of intensities - the quietest sound the ear can hear is smaller than 10-20 times the threshold which causes damage after brief exposure. If you could measure distances over the same range with a single instrument, and the smallest distance you could measure was 1 mm, what would the largest be, in kilometers?
Answer:
the largest distance we can measure is 10¹⁴ km
Explanation:
Given the data in the question;
Threshold hearing = 10⁻²⁰
smallest distance measured = 1 mm
Largest distance measured will be;
⇒ ( threshold hearing )⁻¹ × smallest distance
= ( 1 / 10⁻²⁰ ) × 1 mm
= 10²⁰ × 1mm
= 10²⁰ mm
we know that; 1000 mm = 10⁶ km
Largest distance = ( 10²⁰ / 10⁶ ) km
= 10¹⁴ km
Therefore, the largest distance we can measure is 10¹⁴ km
Question 18/55 (2 p.)
A vibrating object produces ripples on the surface of a liquid. The object completes 20 vibrations
every second. The spacing of the ripples, from one crest to the next, is 3.0 cm.
What is the speed of the ripples?
D
C 60 cm/s
120 cm/s
A 0.15cm/s
B 6.7 cm/s
Answer:
the correct answer is C v = 60 cm / s
Explanation:
The speed of a wave is related to the frequency and the wavelength
v = λ f
They indicate that the object performs 20 oscillations every second, this is the frequency
f = 20 Hz
the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength
λ = 3 cm = 0.03 m
let's calculate
v = 20 0.03
v = 0.6 m / s
v = 60 cm / s
the correct answer is C
a system absorb 500 J of heat and the same time 400J of work is done one the system find change in internal enery ?
Answer:
+ 900 J
Explanation:
Since the total energy change ΔE = internal energy change ΔU since there is no change in kinetic and potential energy,
ΔE = ΔU
ΔE = Q - W where Q = heat absorbed by system and W = work done by system
Now since the system absorbs 500 J of heat, Q = + 500 J and work of 400 J is done on the system, W = -400 J
So, the values of the variables into the equation, we have
ΔE = Q - W
ΔE = + 500 J - (-400 J)
ΔE = + 500 J + 400 J
ΔE = + 900 J
So, the internal energy change, ΔE = + 900 J
PAY ATTENTION MY QUESTION ASK FOR RADIATION!!!
You sit with friends around a campfire, roasting marshmallows. Which
transfer of thermal energy involved in this system is an example of radiation
Answer:
The answer is c
Thermal energy moves within the air from the flames to the marshmallow.
Explanation:
Hope it helps
You sit with friends around a campfire, roasting marshmallows. then the transfer of thermal energy involved in this system is an example of radiation Thermal energy moves within the air from the flames to the marshmallow. Hence option C is correct.
What is thermal Energy ?In physics and engineering, the phrase "thermal energy" is thrown around in a lot of different situations. It can relate to a variety of distinct physical notions. Included in this are the internal energy, or enthalpy, of a body of matter and radiation; heat, which is a form of energy transfer (as is thermodynamic work); and the characteristic energy of a degree of freedom in a system described in terms of its microscopic particulate constituents (where T denotes temperature and k denotes the Boltzmann constant.
Hence option C is correct.
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A car traveling at 14 m/s accelerates at 3.5 m/s² for 5 seconds. How much distance does it travel during that time?
Answer: 113.75
Explanation:
You know
acceleration = a = 3.5 m/s²
time = t = 5 seconds
initial velocity = u = 14 m/s
Unknown is distance = s = ?
Use equation: s = ut + [tex]\frac{1}{2}[/tex] at²
Substitute all the known values inside the equation:
s = (14*5) + 0.5 * 3.5 * 5²
s = 70 + 43.75 = 113.75 m
The car travels 113.75 metres.
ADvantage of friction
Answer:
1. Friction enables us to walk freely.
2. It helps to support ladder against wall.
3. It becomes possible to transfer one form of energy to another.
4. Objects can be piled up without slipping.
Choose the CORRECT statements. A standing wave is resulted from the superposition of
two waves in such a way both waves:
I. have the same direction.
II. are opposite in direction.
III. have the same frequency.
IV. have different frequency.
V. have the same amplitude.
VI. have different amplitude
A. I, III and V
B. II. IV and VI
C. I, IV and V
D. II, III and V
E I and III
F. Ill and V
Answer:
The answer is D
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s.
Required:
a. Compute the magnitude and direction of the velocity of the stone after it is struck.
b. Is the collision perfectly elastic?
Answer:
(a)Magnitude=28.81 m/s
Direction=33.3 degree below the horizontal
(b) No, it is not perfectly elastic collision
Explanation:
We are given that
Mass of stone, M=0.150 kg
Mass of bullet, m=9.50 g=[tex]9.50\times 10^{3} kg[/tex]
Initial speed of bullet, u=380 m/s
Initial speed of stone, U=0
Final speed of bullet, v=250m/s
a. We have to find the magnitude and direction of the velocity of the stone after it is struck.
Using conservation of momentum
[tex]mu+ MU=mv+ MV[/tex]
Substitute the values
[tex]9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V[/tex]
[tex]3.61i=2.375j+0.150V[/tex]
[tex]3.61 i-2.375j=0.150V[/tex]
[tex]V=\frac{1}{0.150}(3.61 i-2.375j)[/tex]
[tex]V=24.07i-15.83j[/tex]
Magnitude of velocity of stone
=[tex]\sqrt{(24.07)^2+(-15.83)^2}[/tex]
|V|=28.81 m/s
Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s
Direction
[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
=[tex]tan^{-1}(\frac{-15.83}{24.07})[/tex]
[tex]\theta=tan^{-1}(-0.657)[/tex]
=33.3 degree below the horizontal
(b)
Initial kinetic energy
[tex]K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2[/tex]
[tex]K_i=685.9 J[/tex]
Final kinetic energy
[tex]K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2[/tex]
=[tex]\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2[/tex]
[tex]K_f=359.12 J[/tex]
Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.
Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of fall at the earth's surface is given by g = Gm/R2 . What is the acceleration of a satellite moving in a circular orbit round the earth of radius 2R
Explanation:
The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.
[tex]W=F_G[/tex]
[tex]mg = G \dfrac{mM}{R^2}[/tex]
which gives us an expression for the acceleration due to gravity g as
[tex]g = G\dfrac{M}{R^2}[/tex]
At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity [tex]g_h[/tex] at this height is
[tex]mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}[/tex]
Simplifying this, we get
[tex]g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g[/tex]
Consider the heaviest box of 150 lb that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.45, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 60.0° above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.45.
If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.)
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
What is meant by kinetic friction ?Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.
Here,
Mass of the box, m = 150 lb = 68.1 kg
Coefficient of kinetic friction, μ = 0.45
Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,
F(k) = μmg
F(k) = 0.45 x 68.1 x 9.8
F(k) = 300.32 N
Now, the box sits on a ramp inclined at 60°
Coefficient of kinetic friction, μ = 0.45
The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.
So,
Frictional force, F(k)' = μmgcosθ
F(k)' = 0.45 x M x 9.8 x cos 60
F(k)' = 2.2M
Weight of the box acting horizontally,
W = Mgsinθ
W = M x 9.8 x sin60
W = 8.5M
Therefore, net force,
Fn = W - F(k)'
Fn = 8.5M - 2.2M
Fn = 6.3M
The total force acting on the box is
F = F(k) - Fn
ma = 300.32 - 6.3M
Since, the box is moving with constant speed, the acceleration, a = 0
Therefore,
300.32 - 6.3M = 0
6.3M = 300.32
M = 300.32/6.3
M = 47.7 kg = 105.16 pound
Hence,
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
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How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
Explanation:
From the given information:
the car's momentum = momentum of the truck
∴
(a) 816 kg × v = 2650 kg × 16.0 km/h
v = (2650 kg × 16.0 km/h) / 816 kg
v = 51.96 km/hr
(b) 816 kg × v = 9080 kg × 16.0 km/h
v = (9080 kg × 16.0 km/h) / 816 kg
v = 178.04 km/hr
A single loop of wire with an area of 0.0900 m^2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.160 T/s.
Reqiured:
a. What emf is induced in this loop?
b. If the loop has a resistance of 0.600Ω, find the current induced in the loop.
Answer:
a) [tex]E=0.0144[/tex]
b) [tex]I=0.024A[/tex]
Explanation:
From the question we are told that:
Area [tex]A=0.09m^2[/tex]
Magnetic Field [tex]B=3.80T[/tex]
Rate [tex]\frac{dB}{dt}=0.160T/s[/tex]
Generally the equation for EmF E is mathematically given by
[tex]E=-A\frac{dB}{dt}[/tex]
[tex]E=-(0.0900*0.160)[/tex]
[tex]E=0.0144[/tex]
b)
at Resistance R=0.60
Generally the equation for Current I is mathematically given by
[tex]E=IR[/tex]
[tex]I=\frac{0.0144}{0.600}[/tex]
[tex]I=0.024A[/tex]
Match the atmospheric energy transfer process that best fits each of the following scenarios:
Warming of the Earth's surface on a sunny day
[ Choose ] convection conduction radiation advection
On a sunny afternoon, you watch cumulus clouds forming
[ Choose ] convection conduction radiation advection
A very shallow layer of air in contact with the ground is warmed
[ Choose ] convection conduction radiation advection
A south wind carries warm air into the central United States
[ Choose ] convection conduction radiation advection
Answer:
a) RADIATION, b) CONVECTION, c) CONDUCTION, d) CONVECTION
Explanation:
In the heating processes there can be three types: conduction, convention and radiation.
The conduction process occurs when the movement of atoms or thermal agitation of molecules creates the transfer of thermal energy.
The convention process occurs when there is a movement of matter creating the transfer of energy
The process of Radiation an electromagnetic wave indexes on a material and is absorbed, creating the process of energy transfer.
Now let's examine each situation>
a)Warming of the Earth's surface on a sunny day
in this case the sunlight heats the earth as it is absorbed, which is why it is a RADIATION process
b) On a sunny afternoon, you watch cumulus clouds forming
in this case the nines rise from the surface, since it is a moving mass, the process is CONVECTION
c) A very shallow layer of air in contact with the ground is warmed
in this case the thermal movement of the layer molecules heat the earth, for which the process of CONDUCTION
d) A south wind carries warm air into the central United States
As we have a movement of a mass of matter, the process is CONVECTION.
Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of 33 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.
1. What is the ratio of the electric force on the bee to the bee's weight?
2. What electric field strength would allow the bee to hang suspended in the air?
3. What electric field direction would allow the bee to hang suspended in the air?
Answer:
A) 3.367 × 10^(-6)
B) 2.97 × 10^(7) N/C
C) Upwards
Explanation:
We are given;
Mass of bee; m = 100 mg = 100 × 10^(-6) kg
Charge on bee;q=33 pC = 33 × 10^(-12)C
Electric field strength; E = 100 N/C
A) Formula for weight of bee; W = mg = 100 × 10^(-6) × 9.8 = 9.8 × 10^(-4) N
Electric force on Bee; F = qE = 33 × 10^(-12) × 100 = 33 × 10^(-10) N
ratio of the electric force on the bee to the bee's weight; F/W = (33 × 10^(-10))/(9.8 × 10^(-4)) = 3.367 × 10^(-6)
B) For the bee to be suspended in the air, it means the weight of the bee must be equal to the electric force. Thus;
mg = qE
100 × 10^(-6) × 9.8 = 33 × 10^(-12) × E
E = (100 × 10^(-6) × 9.8)/(33 × 10^(-12))
E = 2.97 × 10^(7) N/C
C) From Newton's law, sum of forces = 0.
Thus;
F_n + F + W = 0
Where F is the normal force.
Thus;
F_n = -(F + W)
F_n = - ((33 × 10^(-10)) + (9.8 × 10^(-4)))
F_n = -9.8 × 10^(-4) N
Thus, applied electric field is;
E_a = F_n/q = (-9.8 × 10^(-4))/(33 × 10^(-12)) = -2.97 × 10^(7) N/C
This is negative and so it means the direction will be opposite the Earth's electric filed which is upwards.
Planets closer to a star will have what type of average temperature
Answer:
Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018
If the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change
Answer:
A neutral atom has no net charge and should not be affected by a point charge, and the distance of separation is not a factor.
The force will remain the same and is equal to zero.
We have a point charge and a neutral atom.
We have to investigate if the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change.
State Coulomb's Law of Electrostatic force.The Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically -
[tex]$F=\frac{q _{1}\cdot q _{2}}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
According to question, we have -
A point charge and a neutral atom.
If initially the distance between the point charge and neutral atom is r meters, then -
q(1) = Q (say)
q(2) = 0 ( Neutral atom has zero charge)
Using Coulomb's law -
[tex]$F=\frac{Q\times 0}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
F = 0 Newtons.
Now, if you will increase the distance by the factor of six, still the force will remain zero as there is only one point charge and other is neutral atom. There is no electrostatic force between a charged and uncharged particle.
Hence, the force will remain the same and is equal to zero.
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A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgsinθ−ma, where g=9.80meter/second2, a=3.60meter/second2, θ=27.0∘, and m is not given. Which of the following represents a simplified expression for μk?A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: , where , , , and is not given. Which of the following represents a simplified expression for ?tanθ− agTo avoid making mistakes, the expression should not be simplified until the numerical values are substituted.gsinθ−agcosθThe single equation has two unknowns and cannot be solved with the information given.
Solution :
Given expression :
[tex]$\mu_k$[/tex]mgcosθ = mgsinθ − ma
Here, g = 9.8 [tex]m/s^2[/tex] , a = 3.60 [tex]m/s^2[/tex] , θ = 27°
Therefore,
[tex]$\mu_k mg \cos \theta = mg \sin \theta - ma$[/tex]
[tex]$\mu_k mg \cos \theta = m(g \sin \theta - a)$[/tex]
[tex]$\mu_k g \cos \theta = (g \sin \theta - a)$[/tex]
[tex]$\mu_k =\frac{(g \sin \theta-a)}{g \cos \theta}$[/tex]
Mow calculating the coefficient of kinetic friction as follows :
[tex]$\mu_k=\frac{g \sin \theta-a}{g \cos \theta}$[/tex]
[tex]$\mu_k=\frac{9.8 \times \sin 27^\circ-3.60}{9.8 \times \cos 27^\circ}$[/tex]
[tex]$\mu_k=0.097$[/tex]
A student wishes to construct a mass-spring system that will oscillate with the same frequency as a swinging pendulum with a period of 3.45 S. The student has a spring with a spring constant of 72.0 N/m. What mass should the student use to construct the mass-spring system?
Answer:
21.73 kg
Explanation:
Applying,
T = 2π√(m/k)............... Equation 1
Where T = period, m = mass on the spring, k = spring constant, π = pie.
make m the subject of the equation
m = T²k/4π²................. Equation 2
From the question,
Given: T = 3.45 s, k = 72.0 N/m, π = 3.14
Substitute these values into equation 2
m = (3.45²×72)/(4×3.14²)
m = 21.73 kg.
Hence the mass should be 21.73 kg
Lonnie pitches a baseball of mass 0.500 kg. The ball arrives at home plate with a speed of 35.0 m/s and is batted straight back to Lonnie with a return speed of 50.0 m/s. If the bat is in contact with the ball for 0.055 s, what is the impulse experienced by the ball
Answer:
Explanation:
The impulse equation is
Δp = FΔt, where Δp = final momentum - initial momentum, F is the Force exerted on an object, and Δt is the change in time. In this equation,the entire right side defines the impulse. In other words, FΔt is the impulse; thus the change in momentum an object experiences is due to its change in impulse and is directly proportional to it.
Therefore, once we find the change in momentum, that is the impulse the object experiences. Δp = final momentum - initial momentum, where
p = mv and p is momentum.
[tex]p_f=(.500)(50.0)[/tex] so
[tex]p_f=25.0[/tex] and
[tex]p_i=(.500)(35.0)[/tex] so
[tex]p_i=17.5[/tex]; therefore,
Δp = 25.0 - 17.5 = 7.5[tex]\frac{kg*m}{s}[/tex] which is the unit for momentum
The magnitude of the impulse experienced by the ball is equal to 7.5 Kg.m/s.
What is impulse?Impulse experienced by an object can be described as the integral of a force over a time interval. Impulse can be defined as a vector quantity as the force is a vector. Impulse generates an equivalent vector change in the linear momentum of the object.
The S.I. unit of impulse can be defined as N⋅s and the dimensionally similar to the unit of momentum is kg⋅m/s. A resultant force offers acceleration and changes the velocity as long as it acts.
Given the mass of the baseball, m= 0.500 Kg
The initial speed of the baseball, u = 35 m/s
The final speed of the baseball, v = 50 m/s
The change in the linear momentum gives the impulse experienced by the ball.
I = ΔP = mv - mu
I = 0.500 ×50 - 0.500 × 35
I = 25 - 17.5
I = 7.5 Kg.m/s
Therefore, the magnitude of the impulse is 7.5 Kg.m/s.
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a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
Answer:
The surface tension is 190.2 N/m.
Explanation:
Initial radius, r = 4 cm
final radius, r' = 6 cm
Work doen, W = 15 J
Let the surface tension is T.
The work done is given by
W = Surface Tension x change in surface area
[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]
Describe an experiment to find the density of copper turning using a density bottle and kerosene
The density is the ratio of mass to volume of a substance.
What is the density bottle?The density bottle is used to obtain the density of substance by measuring the volume of the fluid displaced.
If the mass of copper turnings are previously weighed and known, the volume of the fluid displaced in the density bottle is the volume of the copper turning.
Hence;
Density = mass/ volume
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Hooke's law describes a certain light spring of unstretched length 38.0 cm. When one end is attached to the top of a door frame and a 6.00-kg object is hung from the other end, the length of the spring is 42.0 cm.
a. Find its spring constant.
b. The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 150 N. Find the length of the spring in this situation.
Answer:
(a) 1470 N/m
(b) 48.2 m
Explanation:
Applying,
(a) F = ke.................... Equation 1
Where F = force applied to the spring, k = spring constant, e = extension
make k the subject of the equation
k = F/e............... Equation 2
But,
F = mg............. Equation 3
Where m = mass, g = acceleration due to gravity
Substitute equation 3 into equation 2
k = mg/e.............. Equation 4
From the question,
Given: m = 6 kg, e = 42-38 = 4 cm = 0.04 m
Constant: g = 9.8 m/s²
Substitute these values into equation 4
k = (6×9.8)/0.04
k = 1470 N/m
(b) Consider the end of the spring to the left which exert a force to the right
Then,
e = F/k............. Equation 5
Given: F = 150 N, k = 1470 N/m
Substitute these values into equation 5
e = 150/1470
e = 0.102 m
Hence the length of the spring is
L = 0.38+0.102 = 0.482 cm = 48.2 m
a motor car reaches a velocity of 15m/s in 6s from rest on a perfect test track . what is the average acceleration
Answer:
[tex]{ \tt{initial \: velocity, \: u = 0}} \: (at \: rest) \\ { \tt{final \: velocity, \: v = 15 { {ms}^{ - 1} }}} \\ { \tt{time, \: t = 6s}} \\ { \bf{from \: first \: newtons \: equation \: of \: motion : }} \\ { \bf{v = u + at}} \\ { \tt{15 = 0 + (a \times 6)}} \\ { \tt{6a = 15}} \\ { \tt{acceleration, \: a = 2.5 \: {ms}^{ - 2} }}[/tex]
Identify the reactants in the combustion of methane: CH4 + O2 CO2 + O°H
The process of finding the age of fossil is called which answer carbondating,urinumdating ,both of them,none
Answer:
the process of finding the age of fossils is called carbon dating because they use samples of carbon and compare it's characteristics to the samples of carbon from each set time
if a cyclist is travelling a road due east at 12km/h and a wind is blowing from south-west at 5km/h. find the velocity of the wind relative to the cyclist.
Answer:
The velocity of wind with respect to cyclist is [tex]-15.5 \widehat{i} - 3.5 \widehat{j}[/tex].
Explanation:
speed of cyclist = 12 km/h east
speed of wind = 5 km/h south west
Write the speeds in the vector form
[tex]\overrightarrow{vc} = 12 \widehat{i}\\\\overrightarrow{vw} = - 5 (cos 45 \widehat{i} + sin 45 \widehat{j})\\\\\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}[/tex]
The velocity of wind with respect to cyclist is
[tex]\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}\overrightarrow{v_(w/c)} = \overrightarrow{vw}-\overrightarrow{vc}\\\\\overrightarrow{v_(w/c)} = - 3.5 \widehat{i} - 3.5 \widehat{j} - 12 \widehat{i}\\\\\overrightarrow{v_(w/c)} =-15.5 \widehat{i} - 3.5 \widehat{j}[/tex]
