Answer:
If the size and direction of the forces acting on an object are exactly balanced, then there is no net force acting on the object and the object is said to be in equilibrium. Because the net force is equal to zero, the forces in Example 1 are acting in equilibrium.
Equilibrium of forces means that the net force is 0. It can either be when there is no force acting on the object or when the force acting on the object are balanced.
Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above
Answer:
Explanation:no change in surface tension
An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.
Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.
The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.
As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
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A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What
is the magnitude of the impulse?
The magnitude of the impulse of the truck is equal to 4000 Kg.m/s.
What is impulse?Impulse can be described as the integral of a force over the time interval for which it acts. Impulse is also a vector quantity since force is a vector quantity. Impulse can be applied to an object that generates an equivalent vector change in its linear momentum.
The S.I. unit of impulse is N⋅s and the dimensionally equivalent unit of momentum is kg⋅m/s. A resultant force gives acceleration and changes the velocity of an object for as long as it acts.
Given the mass of the truck, m= 2000 Kg
The initial speed of the truck, u = 6 m/s
The final speed of the truck, v = 4 m/s
The change in the linear momentum is equal to the impulse.
I = ΔP = mv - mu
I = 2000 ×4 - 2000 × 6
I = 8000 - 12000
I = - 4000 Kg.m/s²
Therefore, the magnitude of the impulse is 4000 Kg.m/s².
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Electromagnetic waves from the sun carry what to the earth
Answer:
Solar radiation
Explanation:
Visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays.
==> Energy
==> Radio noise, heat, visible light, ultraviolet radiation, X-rays, gamma rays
==> They carry all these kinds of energy wherever they go. Not only to the Earth.
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Answer:
A. Power generated by meteor = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Workdone = 981000 J
Power required = 19620 Watts
Note: The question is incomplete. A similar complete question is given below:
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?
Explanation:
A. Power = workdone / time taken
Workdone = Kinetic energy of the meteor
Kinetic energy = mass × velocity² / 2
Mass of meteor = 1.5 g = 0.0015 kg;
Velocity of meteor = 50 km/s = 50000 m/s
Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J
Power generated = 1875000/2.1 = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Work done by elevator against gravity = mass × acceleration due to gravity × height
Work done = 1000 kg × 9.81 m/s² × 100 m
Workdone = 981000 J
Power required = workdone / time
Power = 981000 J / 50 s
Power required = 19620 Watts
Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.
A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?
a. 0.19 m/s^2
b. 0.14 m/s^2
c. 0.16 m/s^2
d. 0.20 m/s^2
e. 0.17 m/s^2
Answer:
The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²
Option c) 0.16 m/s² is the correct answer.
Explanation:
Given the data in the question;
since the train starts from rest,
Initial velocity; u = 0 m/s
final velocity; v = 42 m/s
distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m
acceleration a = ?
From the third equation of motion;
v² = u² + 2as
we substitute in our values
( 42 )² = ( 0 )² + [ 2 × a × 5600 ]
1764 = 0 + [ 11200 × a ]
1764 = 11200 × a
a = 1764 / 11200
a = 0.1575 ≈ 0.16 m/s² { two decimal place }
Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²
Option c) 0.16 m/s² is the correct answer.
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30.08 as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.
Answer:
2.55 m/s
Explanation:
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.
Solution:
The work done by friction is given as:
[tex]W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J[/tex]
The work done by gravity is:
[tex]W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s[/tex]
A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping
Explanation:
The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.
[tex]y:\:\:\:\:N - mg = 0[/tex]
[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]
or
[tex]m \dfrac{v^2}{r} = \mu mg[/tex]
Solving for [tex]\mu[/tex],
[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]
Determine the magnitude of the minimum acceleration at which the thief can descend using the rope. Express your answer to two significant figures and include the appropriate units.
Answer: hello your question is incomplete below is the missing part
A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.
answer:
To 2 significant Figures = 1.6 m/s^2
Explanation:
Calculate the magnitude of minimum acceleration at which the thief can descend
we apply the relation below
Mg - T = Ma --- ( 1 )
M = 69kg
g = 9.81
T = 58 * 9.81
a = ? ( magnitude of minimum acceleration)
From equation 1
a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69
= 1.5639 m/s^2
To 2 significant Figures = 1.6 m/s^2
the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact with the bat, how many impules of importance are used to find the final velocity of the bat
Solution :
Given :
Mass of the baseball, m = 200 g
Velocity of the baseball, u = -30 m/s
Mass of the baseball after struck by the bat, M = 900 g
Velocity of the baseball after struck by the bat, v = 47 m/s
According to the conservation of momentum,
[tex]Mv+mu=Mv_1+mv_2[/tex]
(900 x 47) + (200 x -30) = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
36300 = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)
[tex]9v_1 = 363 - 2v_2[/tex]
[tex]v_1=\frac{363 - 2v_2}{9}[/tex]
The mathematical expression for the conservation of kinetic energy is
[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]
[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex] ................(ii)
[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]
[tex]21681 = 9v_1^2+2v_2^2[/tex]
Substituting (i) in (ii)
[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]
[tex](363-2v_2)^2+18v_2^2=195129[/tex]
[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]
[tex]22v_2^2-145v_2-63360=0[/tex]
Solving the equation, we get
[tex]v_2=96 \ m/s, -30 \ m/s[/tex]
The negative velocity is neglected.
Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get
[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]
= 19
Thus, only impulse of importance is used to find final velocity.
Who stated that man is an animal
13. What type of lens bends light outwards and away from a point?
concave
Answer:
No,it isn't concave. The correct answer is convex lens.
Explanation:
A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.
Convex lens is the answer.
See the attached diagram.
Page
E QON
1 What is force ? Write its unit and mention
any
three effects of the force.
Force is a push or a pull that changes or trends to change the state of rest or uniform motion of an object or changes the direction or shape of an object. It causes objects to accelerate. SI unit is Newton.
1) Can change the state of an object : For example, pushing a heavy stone in order to move it.
2) May change the speed of an object if it is already moving. For example, catching a ball hit by a batsman.
3) May change the direction of motion of an object.
A 55 kg person is in a head-on collision. The car's speed at impact is 12 m/s. Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
Tres ladrillos idénticos están atados entre sí por medio de cuerdas y penden de una balanza que marca en total 24 N. ¿Cuál es la tensión de la cuerda que soporta al ladrillo inferior? ¿Cuál es la tensión en la cuerda que se encuentra entre el ladrillo de en medio y el superior?
ayuda!!!!!
answer 8N 8N
la tensión del ladrillo inferior es 8N ya que los ladrillos son idénticos, decido 24 por 3 para darme 8
la tensión entre el ladrillo superior y el medio también es 8N
A car accelerates uniformly from rest to a speed of 55.0 mi/h in 13.0 s. (a) Find the distance the car travels during this time. m (b) Find the constant acceleration of the car. m/s2
Answer:
(a) 159.84 m
(b) 1.89 m/s²
Explanation:
Applying,
(a)
s = (v+u)t/2.................. Equation 1
Where s = distance traveled by the car, u = initial velocity, v = final velocity, t = time.
From the question,
Given: u = 0 m/s ( from rest), v = 55 mi/h = (55/2.237) m/s = 24.59 m/s, t = 13 s
Substitute these values into equation 1
s = (24.59+0)13/2
s = 159.84 m
(b)
Also applying,
a = (v+u)/t................. Equation 2
Where a = acceleration of the car.
substituting into equation 2,
a = (24.59+0)/13
a = 1.89 m/s²
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
Determine the tension in the string that connects M2 and M3.
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Water at 200 C has a bulk modulus of 2.2109 Pa, and the speed of sound in water at this temperature is 1480m/s. For 1000Hz sound waves in water at 200 C, what displacement amplitude is produced if the pressure amplitude is 310-2 Pa?
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?
When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.
What is the Doppler formula?The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.
The frequency increase by the Doppler effect is represented by the formula
f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f
Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀ is 0.
Substituting the value into the equation will give us the velocity of the police car.
[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)
When the car is receding, the frequency of the receiving signal f = 4500 Hz.
[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)
Solving both equation, we get the velocity of a police car.
v = 33 m/s
Therefore, the velocity v of the police car is 33 m/s.
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A golf ball is dropped from rest from a height of 8.40 m. It hits the pavement, then bounces back up, rising just 5.60 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
Answer:
t1= 8.40/10 =.84 s
t2 = 5.60/10 = .56s
t3= 1.4/10 = .14s
total time = 1.54 sec
what is the velocity of a 1.3 kg puppy with a forward momentum of 6 kg m/s
Answer:
by using p = mv equation we can find v,
6 = 1.3 v
4.615 = v
A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal to 3 1.15 / kg m . Find the mass of air and the overall (average) specific volume
Complete question:
A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.
Answer:
The mass of the air is 5.32 kg
The specific volume is 5.52 x 10⁻³ m³/kg
Explanation:
Given;
total volume of the container, [tex]V_t[/tex] = 5 m³
mass of granite, [tex]m_g[/tex] = 900 kg
density of granite, [tex]\rho _g[/tex] = 2,400 kg/m³
density of air, [tex]\rho_a[/tex] = 1.15 kg/m³
The volume of the granite is calculated as;
[tex]V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3[/tex]
The volume of air is calculated as;
[tex]V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3[/tex]
The mass of the air is calculated as;
[tex]m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg[/tex]
The specific volume is calculated as;
[tex]V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg[/tex]
each of the following conversions contains an error. In each case, explain what the error is and find the correct answer to make a true statement .
a 1000 kg mg (1kg/1000g) = 1g.
b. 50m (1cm/100m)=0.5 cm
c. "Nano" is 10^-9 , so there are 10^-9 nm in a meter.
d. micro is 10^-6, so 1kg is 10^6 ug
Answer:
a. [tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b. [tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c. "Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d. micro is 10⁻⁶, so 1kg is 10⁹ ug
Explanation:
a.
The conversion factor is written inverted. The correct statement will be:
[tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b.
The values in the conversion factor used are wrong. The correct statement will be:
[tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c.
Change of units is the mistake here. The correct statement will be:
"Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d.
the conversion will be as follows:
[tex]1\ kg(\frac{1000\ g}{1\ kg})(\frac{1\ \mu g}{10^{-6}\ g}) = 10^9 \mu g[/tex]
therefore, the correct statement will be:
micro is 10⁻⁶, so 1kg is 10⁹ ug
g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is
Answer:
a) T = 2π [tex]\sqrt{\frac{m}{k} }[/tex], b) T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
Explanation:
a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity
w² = k / m
angular velocity and period are related
w = 2π /T
we substitute
4π²/ T² = k / m
T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]
b) We change the spring for another with k ’= 2 k, let's find the period
T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]
T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]
T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
What is the law of conservation of energy
An irregular shape object has a mass of 19 oz. A graduated cylinder with and initial volume of 33.9 mL. After the object was dropped in the graduated cylinder, it had a volume of 92.8 mL. What is the density of object( g/mL)
Explanation:
m = 19 oz × (28.3 g/1 oz) = 537.7 g
V = 92.8 mL
[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]
