Answer:
337k
Explanation:
First, let us find the difference between the given two temperatures.
Difference = 85°C - 21°C
= 64°C
And now we have to write the temperature in kelvins.
To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.
Let us find it now.
64°C + 273 = 337k
Therefore,
64°C ⇒ 337k
find the direction of this vector.
The direction of this vector is 170° from the positive x-axis.
Vectors are physical quantities with both magnitude and directionTo identify the direction, we usually use angles and take a reference axisThe vector in the diagram is 80° to the left of the positive y-axis, so the direction of the vector is 80° + 90° = 170° anti-clockwise from the positive x-axis.Scalars are physical quantities which have only magnitudeExamples of vector quantities are displacement, velocity, acceleration, force etcExamples of scalar quantities are distance, speed, volume, density, mass, time etcThe direction of the given vector is 170° from the positive x-axis.Learn more about vectors here:
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what's the difference between coplanar forces and resultant forces?
Answer:
coplanar When all forces are acting in the same
resultunt the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body via vector addition
The value of acceleration due to gravity (g) on Pluto is about 0.61 meters/second2. How much will an object that weighs 250 newtons on Earth weigh on Pluto? Note that the value of acceleration due to gravity on Earth is 9.8 meters/second2.
The weight of the object on pluto will be 15.56 N
What is Newton's law of gravitation?Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.
Given data;
[tex]\rm g_e[/tex] is the acceleration due to gravity on earth.
m is the mass of an object
[tex]\rm g_p[/tex] is the acceleration due to gravity on pluto.
[tex]\rm W_e[/tex] is the weight of the object on earth
[tex]\rm W_ p[/tex] is the weight of an object on pluto
The mass of the object on the earth's surface is found as;
We = m × ge
m = W / ge
m = 250 / 9.81
m = 25.51 kg
The weight of the object on pluto.
Wp = m × gp
Wp = 25.51 kg × 0.61 m/s²
Wp =15.56 N.
Hence the weight of the object on pluto will be 15.56 N
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Answer: B.
15.6 newtons
Explanation: edmentum
(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord can withstand a maximum tension of 64.0 N, a. What is the maximum speed at which the ball can move before the cord breaks? Assume the string remains horizontal during the motion. (5 marks) b. Suppose the ball moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ( i ) . A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long . The ball moves in a horizontal circle . If the cord can withstand a maximum tension of 64.0 N , a . What is the maximum speed at which the ball can move before the cord breaks ? Assume the string remains horizontal during the motion . ( 5 marks ) b . Suppose the ball moves in a circle of larger radius at the same speed v . Is the cord more likely or less likely to break ?
(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]
where [tex]R[/tex] is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]
so that
[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]
Solve for [tex]v[/tex] :
[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]
(b) The net force equation in part (a) leads us to the relation
[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]
so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.
water
0.6m
water wave
tank
How long does it take for the wave to return to the
position XY, but moving to the right?
[3]
b A man is cutting down a tree with an axe. He
hears the echo of the impact of the axe hitting
the tree after 1.6 s.
i What sort of obstacle could have caused the
echo?
ii The speed of sound is 330 m/s. How far is
the tree from the obstacle?
c Distinguish between the nature of the sound
wave in b and the water wave in a.
[2]
ii the amplitude of
b The cone of a louds
diagram shows how
out in front of the c
loudspeaker
P is a compression,
i
Describe how
changes from
ii Describe the
the sound w
iii Copy the di
and mark a
wavelength
5 a The first diagrams
Answer:
Discrimination is the worst thing in the world you can't even do a thing so you will do such a physical thing or do a mathematics problem ok done it's ok
A spring is compressed by 0.0880 m and is used to launch an object horizontally with a speed of 2.76 m/s. If the object were attached
to the spring, at what angular frequency (in rad/s) would it oscillate?
Answer:
Approximately [tex]3.14\; {\rm rad \cdot s^{-1}}[/tex].
Explanation:
Fact: the angular velocity [tex]\omega[/tex] of a simple harmonic oscillator is the ratio between the maximum velocity [tex]v_{\text{max}}[/tex] and the maximum displacement [tex]x_\text{max}[/tex] of this oscillator. In other words:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}[/tex].
Derivation of the previous equation:
Let [tex]A[/tex] denote the amplitude of this oscillation, and let [tex]\omega[/tex] denote the angular velocity.
The displacement of the oscillator at time [tex]t[/tex] would be:
[tex]x(t) = A\, \sin(\omega\, t)[/tex].
The maximum displacement of this oscillator would be [tex]x_\text{max} = A[/tex].
The velocity of this oscillator at time [tex]t[/tex] is the derivative of displacement with respect to time:
[tex]\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}[/tex].
The maximum velocity of this oscillator would be [tex]v_\text{max} = A\, \omega[/tex].
Notice that dividing [tex]v_\text{max} = A\, \omega[/tex] by [tex]x_\text{max} = A[/tex] would give:
[tex]\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega[/tex].
It is given that [tex]v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}[/tex] while [tex]x_\text{max} = 0.0880\; {\rm m}[/tex]. Therefore:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}[/tex].
(Radians per second.)
What is the relative velocity of two beta particles moving in opposite directions at a speed of 0.8c?
That depends on where YOU are when you measure it.
If you're motionless in the laboratory, then you measure the particles flying apart at 1.6c .
If you're riding on one of the particles, you measure the other one flying away from you at less than c
Can anyone solve? Thank you
Answer:
The rod is stretched because there is a larger gravitational force on the mass closest to mass M.
F = G M m / r^2 - G M m / (r + l)^2 the difference of the forces on m's
F = G M m [1 / r^2 - 1 / (r + l)^2]
[1 / r^2 - 1 / (r + l)^2] = (r^2 + 2 r l + l^2 - r^2 / [r^2 (r + l)^2]
= (2 r l + l^2) / [r^2 (r + l)^2]
if l is small compared to r = (2 r l ) / [r^2 (r + l)^2]
= (2 r l ) / [r^4 + 2 r^3 l ]
The following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s.
Find
A, the average velocity of the particle in the time interval t₁=2sec and t₂=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)
(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.
(b) The velocity and acceleration at any time t is v = (4ti + j) m/s and a = a = 4i m/s²
(c) The average acceleration in the time interval given in part (a) is 3.98 m/s².
Position of the particlex = at²i + btj
x = 2t²i + tj
Average velocity, at t₁=2sec and t₂=3secΔv = Δx/Δt
x(2) = 2(2)²i + 2j
x(2) = 8i + 2j
|x(2)| = √(8² + 2²) = 8.246
x(3) = 2(3)²i + 3j
x(3) = 18i + 3j
|x(3)| = √(18² + 3²) = 18.248
Δv = (18.248 - 8.246)/(3 - 2)
Δv = 10 m/s
Velocity and acceleration at any time, tv = dx/dt
v = (4ti + j) m/s
a = dv/dt
a = 4i m/s²
Average accelerationv(2) = 4(2)i + j
v(2) = 8i + j
|v(2)| = 8.06 m/s
v(3) = 4(3)i + j
v(3) = 12i + j
|v(3)| = 12.04 m/s
a = (12.04 - 8.06)/(3 - 2)
a = 3.98 m/s²
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which has more KE, a 2 g bee flying at 1 m/s, or a 1 g wasp flying at 2 m/s
Answer:
the 1 gram wasp
Explanation:
To start off with this problem, write down every piece of information and do neccessary conversions.
mass of bee = 2 grams = 0.002 kg
speed of bee = 1 m/s
mass of wasp = 1 gram = 0.001 kg
speed of wasp = 2 m/s
now, we will use the kinetic energy formula and compare the answers
KE BEE = 0.5 (0.002 kg)(1 m/s)^2 = 0.001 Joules
KE WASP = 0.5(0.001 kg)(2 m/s)^2 = 0.002 Joules
0.002 J > 0.001 J
A local municipality placed an advert in the local newspaper for a competition of local high schools to participate to build a security system in the community hall to avoid vandalism and break ins
We will use the design process to meet the requirements of the competition by the given explained in the main answer.
What is the competition?The initiative was taken by two or more parties operating independently to win the business of a third party by competing to hire contractors on the best terms possible.
The complete question is
"A local businessperson placed an advert in the local newspaper for a competition for girls only from different schools to participate in the competition. The competition is about drawing ideas for a shopping complex for the local community. Girls from the different schools are excited about the competition. you are a group of boys in the local schools who saw the advert and feel it is unfair for you to be excluded from the competition.
consider yourself as one of the learners in the local schools.
Explain why the project should be given to boys also who have better skills for this project than the girls do.
1.2 think about the design process skills, investigates, design, make, evaluate and communicate."
Explain how you will use the design process to meet the requirements of the competition.
[ 1.1 ]Boys should also be given the project because some of them have more artistic talent than girls do when it comes to sketching designs for local shopping centers.
Boys can also present a clear picture of how local communities can expand their shopping options.
For instance, they can encourage local merchants to offer seasonal items because doing so will benefit their businesses.
They can also suggest that retailers encourage customers to bring a friend to participate in competitions, which will be highly regarded.
[1.2] In order to fulfill the needs of the competition, I would make sure to explore how a shopping center for the neighborhood functions, for instance, by making sure I am aware of the strategies one may use to enhance his or her firm in the neighborhood.
In order to win the competition, I would make sure that I was confident in what I was saying and that I was audible enough to be heard during the competition.
Finally, I would make sure I met the competition's conditions by staying on-topic and solely addressing the issues raised in the advertisement
Hence will use the design process to meet the requirements of the competition.
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A force of 5.25 newtons acts on an object of a mass 25.5 kilograms. How far is the object from the center of Earth? (The value of G is 6.673 × 10-11 newton meter2/kilogram2. The mass of Earth is 5.98 × 1024 kilograms.)
Answer:
The force between two objects is calculated through the equation,
F = Gm₁m₂/d²
where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is the universal gravitation constant.
In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,
2.5F = Gm₁m₂ / (D²)
D = 0.623d
Subsituting the known value of d,
D = 0.623(6.9 x 10^8) = 4.298 x 10^8 m
In a hydraulic lift whose input line has a cross-sectional area of 1.00 cm² and whose output line has
a cross-sectional area of 22.0 cm², what is the largest mass (kg) that can be lifted by an input force
of 200. N?
By using Newton's Second law of motion, the largest mass that can be lifted by this input force is equal to 2,244 kg.
How to calculate the largest mass?Based on the information provided, we can logically deduce that the output to input force ratio is equal to 22:1. This ultimately implies that, the input force is equal to 1000 N, which would result in an output force of 22,000 N.
Next, we would calculate the largest mass by using Newton's Second law of motion:
Mass = Force/Acceleration
Mass = 22,000/9.8
Mass = 2,244 kg.
Note: Acceleration due to gravity is equal to 9.8 m/s².
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suppose a uniform electric field 4 N/C is in the positive x-direction .
The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
Electric field at position 4 mElectric field at a given distance is calculated as follows;
E = kq/r²
E₂ = (9 x 10⁹ x q)/(2²)
E₂ = 2.25 x 10⁹q
E₂ + E₀ = 0
2.25 x 10⁹q + 4 = 0
2.25 x 10⁹q = - 4
q = -4 / (2.25 x 10⁹)
q = -1.78 x 10⁻⁹
E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)
E₄ = - 1 N/C
|E₄| = 1 N/C
Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
The complete question is below:
Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?
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A stationary 0.750kg ball is thrown up by doing 2.50J of work on it. What is the velocity of the ball?
When a ball is thrown up by doing work, the velocity of the ball will be 2.6 m/s.
What is Work energy theorem?It states that the Work done in moving a body is equal to the change in kinetic energy of the body
Kinetic energy = 1/2 mv²
Given is a ball of mass m = 0.750 kg and the work done on ball W = 2.50 J
The ball is initially at rest. So, initial velocity is zero. Then, change in kinetic energy will be
W= ΔK.E = K.Ef - K.Ei
According to work energy theorem, work done is
W = 2.5J = 1/2 x 0.750 x (v)² -0
v =2.6 m/s
Thus, the velocity of the ball is 2.6 m/s
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How many seconds are in 28 hours?
Answer:
1680 seconds
Explanation:
[tex]28 hrs * \frac{60 s}{1 hr} =1680s[/tex]
A 0.6 kg ball moving to the right at 12 m/s makes a head on elastic collision with a 0.3 kg ball traveling to the left at 3 m/s.
The velocity of the 0.6 kg ball initially after the collision is
6. A metal block increases in temperature from 15 °C to 60 °C when supplied with 13 500 J of heat energy.
a) Calculate the heat capacity of the metal.
b) Calculate the specific heat capacity of the metal, if this sample has a mass of 0.75 kg.
Answer:
Since 1 cal = 4.19 J the heat Q received by the metal is
Q = 13,500 J * 1 C / 4.19 J = 3,222 calories
S = ΔQ / (ΔT * ΔM) = 3222 cal / (45 deg C / 750 gm)
S = .095 cal / gm deg C
Note that specific heat capacity for Cu is .093 cal / gm deg C
the question in photo
Answer:
a
Explanation:
A 17.0-m-high and 11.0-m-long wall and its bracing under construction are shown in the figure. Calculate the force, in newtons, exerted by each of the 10 braces if a strong wind exerts a horizontal force of 655 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall.
The force, in newtons, exerted by each of the 10 braces is 2.135 x 10⁴ N.
What is force?The force is defined as the shear stress or pressure applied per unit area.
F = P/A
Given is a 17.0-m-high and 11.0-m-long wall and its bracing under construction are shown in the figure(attached). The force is exerted by each of the 10 braces, if a strong wind exerts a horizontal force of 655 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths.
Considering the pivot at the base of wall.
From the equilibrium of forces, we have
r₁ x Fwind = r₁ x Fbsinθ
Put the values, we get
Fb = Fwind /10sin35°.............(1)
The wind force is also given by
Fwind = Horizontal force or pressure x Area
Fwind = F/A x hw
=655 x 17 x 11
F wind = 122,485 N
From equation (1), we have
Fb = Fwind /10sin35°
Fb = 122,485 /10sin35°
Fb = 21,354.6080 N
0r Fb = 2.135 x 10⁴ N
Thus, the force exerted on each of the 10 brace is 2.135 x 10⁴ N.
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During a NASCAR race a car goes 58 m/s around a curved section of track that has a radius of 260 m. What is the car's acceleration?
Answer:
12.9m/s^2
Explanation:
As, a=(v^2)/r
=(58^3)/260
=12.9
Which graph shows constant acceleration with correctly placed independent and dependent variables?
Answer:
a = (V2 - V1) / t equation for constant acceleration
if t the independent variable is placed on the abcissa or x-axis and V the dependent variable is placed on the ordinate or y-axis then the acceleration would be the slope at the point considered.
Lester plays middle linebacker for dices football team. During one play, he made the following movements. First, he back-pedaled in the southern direction for 2.6 meters
Answer:
Explanation:
missing info .....
A spaceprobe in outer space is flying with a constant speed of 1.795 km/s. The probe has a payload of 1635.0 kg and it carries 4092.0 kg of rocket fuel. The rocket engines of the probe are capable of expelling propellant at a speed of 4.161 km/s. Then the rocket engines are fired up. How fast will the spaceprobe travel when all the rocket fuel is used up?
The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
What is the law of conservation of linear momentum?
According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.
Unit conversion;
1 km/sec = 1000 m/sec
Given data;
Spaceprobe speed = 1.795 km/s = 1795 m /sec
Probe mass = 635.0 kg
Fuel mass = 4092.0 kg
Expelled propellent velocity = 4.161 km/s = 41461 m/sec
From the momentum conservation principle;
[tex]\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec[/tex]
Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
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8. A charged ball of mass m = 0.265 kg and unknown charge q is hanging by a light thread from a ceiling. A fixed charge Q = +5.00 μC on an insulated stand is brought close to the unknown charge. As a result, the unknown charge hangs at an angle 0 = 38.0° to the vertical as shown in the diagram below. The distance between the two charges is r = 22.0 cm. (a) What is the sign of the unknown charge? Explain how you know this. (b) What is the magnitude of the unknown charge? 8. A charged ball of mass m = 0.265 kg and unknown charge q is hanging by a light thread from a ceiling . A fixed charge Q = +5.00 μC on an insulated stand is brought close to the unknown charge . As a result , the unknown charge hangs at an angle 0 = 38.0 ° to the vertical as shown in the diagram below . The distance between the two charges is r = 22.0 cm . ( a ) What is the sign of the unknown charge ? Explain how you know this . ( b ) What is the magnitude of the unknown charge ?
(a) The sign of the unknown charge is negative.
(b) The magnitude of the unknown charge is 2.18 μC.
What is charge?The charge is the physical quantity which defines the object's electric field. The charged objects creates the electric field around it and attracts or repels another objects coming into that field.
Given is a charged ball of mass m = 0.265 kg and unknown charge q is hanging by a light thread from a ceiling. A fixed charge Q = +5.00 μC on an insulated stand is brought close to the unknown charge. As a result, the unknown charge hangs at an angle 0 = 38.0° to the vertical as shown in the diagram below. The distance between the two charges is r = 22.0 cm.
Tension in the string has two components, Tsinθ and Tcosθ
(a) The fixed charge Q will attract the movable charge q attached to string. The charge Q is positive. So, the charges must of opposite sign.
Thus, the sign of unknown charge is negative.
(b) From the equilibrium of forces, we get
Electrostatic force Fe = kQq/r²
Fe = Tsinθ .............(1)
and
weight force mg = Tcosθ.............(2)
Dividing both the equations, we get
mg/Fe = tanθ
mg / (kQq/r²) = tanθ
q = mgr²tanθ / kQ
Substitute the values from the question, we have
q = 0.265 x 9.81 x (0.22)² tan38 / (9x10⁹ x 5 x 10⁻⁶)
q = 2.18 x 10⁻⁶ C
or q = 2.18 μC
Thus, the magnitude of charge is 2.18 μC.
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An object is located 26.0 cm from a concave lens with f = -54.0 cm. What is its magnification?
Answer:
28
Explanation:
Or from this equation, when we can write the magnification of talents is F times F plus the object distance knowledge, substitute the value to find out the magnification and that equals minus 54 centimeters Over -54 minus 26 centimeters. And we got the magnification of the image produced by The lens is 0.675.
two rods one aluminum and one brass are is clamped at one end. At zero degrees celsius, the roads are each 50 cm long and separated by 0.024 CM at their unfastened ends. At what temperature will the rod just come into contact.
At 11.3°C the rod will just come into contact
Coefficient of linear expansion of aluminium [tex]\alpha_{Al}[/tex] = [tex]23*10^-^6[/tex] °[tex]C[/tex]
Coefficient of linear expansion of Brass [tex]\alpha_{B} =19*10^-^6[/tex] °[tex]C[/tex]
[tex]\alpha = \Delta{L}/Lo(T2-T1)[/tex]
For aluminium
[tex]\alpha_{Al} = \Delta{L}/Lo(T-0)[/tex]
[tex]\Delta{L_{1}} = (23*10^-^6*50*T)[/tex]
For Brass
[tex]\alphaB = \Delta{L_{2}/Lo(T-0)[/tex]
[tex]\Delta{L_{1} +\Delta{L_{2} =0.024cm(23*10^-^6*50*T)+(19*10-6*50*T) =0.024[/tex]
T =11.43 °C
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An appliance draws 3 amperes at 120 V (plugged into the wall). If the same appliance is run with a 12 V car battery, how many amperes will it need?
The appliance is run with a 12 V car battery and the value of the electric current will be 0.3A.
What is electric current?The pace at which electrons travel through a conductor is known as electric current. The ampere is the SI unit for electric current.
From ohm's law;
Case 1;
V=IR
120 V = 3 A × R
R = 40 ohm
For the same appliances, the value of the resistance is the same;
Case 2;
V=IR
12 V = I × 40 ohm
I = 0.3 A
Hence the value of the current for case 2 will be 0.3 A.
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What is the magnitude of the force exerted on an electron in an electric field with a strength of 6.5 *10^ 8 N/C? Remember, the charge on an electron is 1.6 * 10 ^ - 19 * C
All work must be shown to earn credit
The field exerts 6.5 x 10⁸ Newton of force on EACH COULOMB of charge in the field.
We're putting 1.6 x 10⁻¹⁹ Coulomb of charge into the field.
The force on it will be
(1.6 x 10⁻¹⁹ Coulomb) x (6.5 x 10⁻⁸ Newton/Coulomb).
That's 1.04 x 10⁻¹⁰ Newton.
Exam Which is a true statement about the energy level of gas? A. Gases have very little energy. B. Gases have more energy than liquids and solids. C. Gases have no energy. D. Gases have less energy than solids. Copyright © 2003-2022 International Academy of Science. All Rights Reserved. Exam Which is a true statement about the energy level of gas ? A. Gases have very little energy . B. Gases have more energy than liquids and solids . C. Gases have no energy . D. Gases have less energy than solids . Copyright © 2003-2022 International Academy of Science . All Rights Reserved .
Statement B is true about the energy level of gas, Gases have more energy than liquids and solids.
What is gas?A gas is a sample of matter that adopts the shape of the container in which it is housed and develops a uniform density inside the container.
Liquid molecules are more energetic than solid molecules. Further heating will cause the molecules to move so quickly that they won't stick together at all. The gas molecules have the highest energy content.
Gases have more energy than liquids and solids.
Hence statement B is true about the energy level of gas.
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