Answer:
if we are to go by the first law of thermodynamics, the Inventors claims are not reasonable.
The inventor has developed a perpetual-motion machine { any device that violates the first law of thermodynamics}
Explanation:
Given that;
energy supplied to a room Qh= 1.2 kWh
for each for each kWh of electricity it consumes, i.e work input W = 1 kWh
W know that; the first law of thermodynamics says energy can neither be created nor destroyed but can flow from one body to another;
which means ∑Q should or must be equal to ∑W; ∑Q = ∑W
but in the given system; Qh⇒1.2 kWh is NOT equal to W⇒1 kWh
i.e Qh⇒1.2 kWh ≠ W⇒1 kWh
which simply means his device create energy and that violate the first law of thermodynamics.
Therefore, if we are to go by the first law of thermodynamics, the Inventors claims are not reasonable.
The inventor has developed a perpetual-motion machine { any device that violates the first law of thermodynamics}
which of the following are not units used to measure energy?
a. joules
b. newtons
c. BTU
d. calories
Answer:
The BTU, or British thermal unit, is actually a measure of heat.
A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?
Answer:
A Impulse = – 25 Ns
B. Time = 5 s
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
Time (t) =?
A. Determination of the Impulse.
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
I = Ft = M(v – u)
I = M(v – u)
I = 5 (3 – 8)
I = 5 × – 5
I = – 25 Ns
NOTE: the negative sign indicates that the net force is acting in the negative direction.
B. Determination of the time.
Impulse (I) = 25 Ns
Force (F) = 5 N
Time (t) =?
I = Ft
25 = 5 × t
Divide both side by 5
t = 25 / 5
t = 5 s
Thus, it will take 5 s for the box to slide through the 15 m long ramp.
Chemical energy is the stored form of chemical bonds containing combinations of which elements? Select THREE (3) elements. *
Carbon
Nitrogen
Helium
Hydrogen
Oxygen
Gold
A physics grad student has a machine that measures how
many electrons are whizzing through a ring. If she measures 2
Coulombs of charge flowing through the ring every second,
what is the current passing through the ring?
1 = _A
Answer:
2A
Explanation:
Given parameter:
Quantity of charge = 2C
Time taken = 1s
Unknown:
The current passing through the ring = ?
Solution:
Current can also be defined as the quantity of charge that passes through a conductor per unit of time:
I = [tex]\frac{q}{t}[/tex]
So;
I = [tex]\frac{2}{1}[/tex] = 2A
PLEASE ANSWER-Why are loose electrons needed for heat conduction?
Answer:
Why do metals conduct heat so well? The electrons in metal are delocalised electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.
For which medical procedure would Doppler ultrasound be most useful?
A.
Finding a lung tumor
B.
Fixing a pulled muscle
C.
Locating a broken bone in a finger
D.
Detecting a blockage in a heart artery
Doppler ultrasound would be most useful in detecting a blockage in a heart artery.
What are the clinical uses of Doppler ultrasound?By monitoring the rate of change in pitch, a Doppler ultrasound may calculate how quickly blood flows (frequency). A sonographer with training in ultrasound imaging applies pressure to your skin with a tiny, hand-held instrument (transducer) roughly the size of a bar of soap across the area of your body being scanned, moving from one place to another as required.
As an alternative to more invasive treatments like angiography, which involves injecting dye into the blood arteries to make them visible on X-ray images, this test may be performed.
Your doctor may use a Doppler ultrasound to assess for artery damage or to keep track of specific vein and artery therapies.
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A Discuss the possibility of fracture of two leg bones that have a length of about 70cm and an average area of
about 4cm
2 when a 80kg person jump from a height of 300cm.
Noting: The breaking stress of the bone ϬB =1.5×108 N/m2 , and
Young’s modulus for the bone is Y=1.5×1010 N/m2
Answer: The bones won't fracture.
Explanation: Stress, in Physics, is a quantity describing forces that can cause deformation. Strain is the measure of how muc an object can be stretched or deformed. The ratio between stress and strain is called Young's modulus or elastic modulus
Breaking Stress of Bone is the maximum stress a bone can take before a rupture occur.
To determine if a person will break his/her bones by jumping from a height, we determine the energy necessary for that jump and compare it with the energy necessary to break a bone.
The energy for breaking a bone is calculated as
[tex]E=\frac{Al_{0}\sigma_{B}^{2}}{2Y}[/tex]
A is the area in m²
l₀ is length in m
[tex]\sigma_{B}[/tex] is breaking stress in N/m²
Y is Young's modulus in N/m²
Calculating energy to break a bone:
[tex]E=\frac{4.10^{-4}.7.10^{-1}.(1.5.10^{8})^{2}}{2.(1.5.10^{10})}[/tex]
[tex]E=210[/tex] J
This is the energy necessary to break one leg bone, so as there are 2, energy will be 420 Joules.
Potential energy gained by jumping is calculated as
E = m.g.h
m is mass in kg
g is acceleration due to gravity in m/s²
h is height in m
Calculating
E = 80.(9.8)(0.3)
E = 235.2 J
Comparing the two energies, potential energy for jumping is less than maximum energy a bone can absorve without breaking, so the leg bones won't suffer a fracture.
When catching a baseball, a catcher's glove moves by 11 cm along the line of motion of the ball. If the baseball exerts a force of 464 N on the glove, how much work is done by the ball
Answer:
51.04 joules
Explanation:
The movement by the gloves = 11cm
Force by the baseball = 464N
First of all we are going to convert cm to metres
11cm = 11/100 meters = 0.11m
The formula for workdone is given as:
W = f x d
W = workdone
F = force
D = distance
Workdone = 464 x 0.11
= 51.04 joules
The workdone by the ball is 51.04 joules
Thank you
An athlete runs on track at a constant speed of 75 meters/min for 15 s. What is the total distance he covered during that time?
Answer:
45
Explanation:
Which statements correctly describe the formula or name of a compound? Select all that apply.
OA. The formula of nitrogen trifluoride is NF 3
B. The formula of ammonia is NH3.
C. The name of AlF, is trialuminum fluoride.
D. The formula of calcium chloride is CaCl2
E. The name of Li, Se is lithium selenate.
OF. The formula of dinitrogen monoxide is NO
2
Results
G. The formula of sulfur trioxide is 30.
Cho
Аа
H. The formula of magnesium hydroxide is Mg(OH)2
G.
Answer:
A, B, D, and H
Explanation:
Statements A, B, D and H are all correct except the following:
Statement C is incorrect. The name of [tex] AlF_3 [/tex] is aluminium fluoride NOT "trialuminium fluoride".
Statement E is incorrect. The name of [tex] Li_2Se [/tex] is lithium selenide NOT "lithium selenate".
Statement F is incorrect. Dinitrigen monoxide, also known as nitrous oxide has a formula of [tex] N_2O [/tex] NOT [tex] NO_2 [/tex].
Statement G is incorrect. Sulfur trioxide formula is [tex] SO_3 [/tex].
3. Explain why it is, regardless of the location, at a certain distance between the two points, you will perceive the two points as a single point rather than as two distinct points.
Answer:
Hello your question is missing some parts attached below is the missing part of your question
answer: many primary sensory Neurons will converge and become a single Neuron and the single Neuron will send a single harmonized signal to the Brain.
Explanation:
The reason regardless of the location that will make you perceive the two points as a single point rather than as two distinct points is that many primary sensory Neurons will converge and become a single Neuron and the single Neuron will send a single harmonized signal to the Brain.
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0o. (b) What is the minimum coefficient of friction needed for a frightened dri
Answer:
a) The ideal speed = 16.21 m/s
b) Minimum co-efficient of friction = 0.216
Explanation:
From the given information:
The ideal speed can be determined by considering the centrifugal force component and the gravity component.
[tex]\dfrac{mv^2}{r}cos \theta = mg sin \theta[/tex]
[tex]v = \sqrt {gr \ tan \theta}[/tex]
[tex]= \sqrt{(9.8 \ m/s^2) (100) \ tan 15^0}[/tex]
= 16.21 m/s
(b)
Let assume that it requires 25 km/h to take the same curve.
Then, using the equilibrium conditions;
[tex]mg \ sin \theta = \dfrac{mv^2}{r} cos \theta + \mu ((\dfrac{mv^2}{r}) sin \theta + mg cos \theta)[/tex]
[tex]\mu = \dfrac{mg sin \theta - \dfrac{mv^2}{r} cos \theta }{((\dfrac{mv^2}{r}) sin \theta + mg cos \theta) }[/tex]
[tex]\mu = \dfrac{g sin \theta - \dfrac{ v^2}{r} cos \theta }{((\dfrac{v^2}{r}) sin \theta + g cos \theta) }[/tex]
[tex]\mu = \dfrac{(9.8 \ m/s^2 ) sin (15^0) - \dfrac{ \dfrac{(25 \times 10^3}{3600} \ m/s)^2 }{100 \ m } cos (15^0) }{((\dfrac{(\dfrac{25 \times 10^3}{3600} )^2}{100}) sin 15^0 + (9.8 \ m/s^2) cos 15^0 ) }[/tex]
[tex]\mathbf{\mu = 0.216}[/tex]
Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal acceleration for each given velocity. Include units.
Answer:
[tex]a=4\ m/s^2[/tex]
Explanation:
Given that,
Radius, r = 2 m
Velocity, v = 1 m/s
We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2[/tex]
So, the magnitude of centripetal acceleration is [tex]4\ m/s^2[/tex].
A dog sees a flowerpot sail up and then back past a window H high. If the total time the pot is in sight is t seconds, find the height above the window that the pot rises. (Let H = 2 m and t = 1.0 s, find the height above the window-- after you have found an algebraic solution.)
Answer:
maximum height = 0.1225 m
Explanation:
given data
H = 2m
t = 1 sec
solution
we consider here velocity of pot at lower side is u
and final velocity is v with acceleration a
time take is t/2
so
height h = u × [tex]\frac{t}{2}[/tex] - 0.5 × g × [tex](\frac{t}{2})^2[/tex] .................1
here
u = [tex]\frac{2}{t} \times ( h + \frac{gt^2}{8} )[/tex]
and
v = u +a t/2 .........................2
v = u + g t/2
v = [tex]\frac{2h}{t} + \frac{gt}{4} - \frac{gt}{2}[/tex]
v = [tex]\frac{2h}{t} - \frac{gt}{4}[/tex]
so that
maximum height is = [tex]\frac{v^2}{2g}[/tex]
maximum height = [tex]\frac{(\frac{2h}{t} - \frac{gt}{4})^2}{2g}[/tex]
put here value of h and t
maximum height = [tex]\frac{(\frac{2(2)}{1} - \frac{g(1)}{4})^2}{2g}[/tex]
maximum height = 0.1225 m
Padded dashboards in cars are safer in an accident than nonpadded ones because an occupant hitting the dash has:________.
(a) increased time of impact.
(b) decreased impulse.
(c) decreased impact force.
(d) a and b above.
(e) a and c above.
Answer:
(e) a and c above.
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = Mass * Velocity [/tex]
The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.
Padded dashboards in cars are safer in an accident than non-padded ones because an occupant hitting the dashboard of an automobile car has an increased time of impact and a decreased impact force because the force or shock experienced is high and happens rapidly over a short period of time, thus, the occupant has less time and velocity while absorbing the momentum of the car in the course of the collision.
A projectile is launched with an initial velocity of
200 meters per second at an angle of 30° above the
horizontal. What is the magnitude of the vertical
component of the projectile's initial velocity by?
(1) 200 m/s x cos 30°
(2) 200 m/s X sin 30°
(3) (200 m/s)/(cos 30 °)
(4) (200 m/s)/(sin 30 °)
The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.
The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.
From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.
where;
|OP| = magnitude of projectile velocity|OA| = magnitude of the horizontal component|OB|/|AP| = vertical component of the projectile∴
Using trigonometric approach for ΔOAP;
[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]
[tex]\mathbf{AP =OP\times sin \theta}}[/tex]
AP = 200 × sin 30°
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A jumbo egg (80 grams) is dropped from a height of 15 meters onto a 1 inch of foam. Using kinematics, determine the velocity of the egg the instant before impact.
Answer:
the velocity of the egg the instant before impact is 17.15 m/s.
Explanation:
Given;
mass of the egg, m = 80 g = 0.08 kg
height through which the egg was dropped, h = 15 m
The velocity of the egg before impact will be maximum, and the final velocity is given by the following kinematic equation;
v² = u² + 2gh
where;
u is the initial velocity of the egg = 0
v is the final velocity of the egg before impact
v² = 0 + 2 x 9.8 x 15
v² = 294
v = √294
v = 17.15 m/s
Therefore, the velocity of the egg the instant before impact is 17.15 m/s.
A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fast!!!)
Answer:let initial velocity u=14m/s
Final velocity v=20m/s
Time taken t=30
Acceleration =a
V=u +at
a= (20-14)/30
a=0.2m/s^2
Explanation:
Acceleration is the change in velocity with respect to time.
A car accelerates uniformly from rest and reaches a speed of 9.9 m/s in 11.4 s. The diameter of a tire is 86.9 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.
Answer:
Number of revolutions = 20.71 rev.
Explanation:
Given the following data;
Initial speed, u = 0m/s
Final speed, v = 9.9m/s
Time, t = 11.4secs
Diameter = 86.9cm to meters = 86.9/100 = 0.869m
To find the acceleration;
Acceleration, a = (v - u)/t
Acceleration, a = (9.9 - 0)/11.4
Acceleration, a = 9.9/11.4
Acceleration, a = 0.87m/s²
Now we would find the distance covered by the tire using the second equation of motion.
S = ut + ½at²
S = 0(11.4) + ½*0.87*11.4²
S = 0 + 0.435*129.96
S = 56.53m
The circumference of the tire is calculated using the formula;
Circumference = 3.142 * diameter
Circumference = 3.142 * 0.869
Circumference = 2.73m
Number of revolutions = distance/circumference
Number of revolutions = 56.53/2.73
Number of revolutions = 20.71 rev.
Therefore, the number of revolutions the tire makes during this motion is 20.71 revolutions.
A car with mass 1141 kg, moving at 16 m/s, strikes a(n) 2916 kg car at rest. If the two cars stick together, with what speed do they move
Answer:
V = 4.49 m/s
Explanation:
Given that,
Mass of the car 1, m₁ = 1141 kg
Initial speed of car 1, u₁ = 16 m/s
Mass of car 2, m₂ = 2916 kg
Initial speed of car 2, u₂ = 0
We need to find the speed of the cars if they stick together. Let the speed be V. The momentum will remain conserved in the process. Using the conservation of momentum to find it.
m₁u₁ + m₂u₂ = (m₁+m₂)V
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\=\dfrac{1141\times 16+2916 \times 0}{(1141 +2916 )}\\\\=4.49\ m/s[/tex]
So, the required speed of the two cars is 4.49 m/s.
A toy car, mass of 0.025 kg, is traveling on a horizontal track with a velocity of 5 m/s. If
the track then starts to climb upwards, how high up the track can the car reach?
Answer:
1.25 m
Explanation:
This is the vertical height not the distance along the slope.
[tex]K=U\\\frac{1}{2}mv^{2} = mgh\\h = \frac{v^{2}}{2g}=\frac{25}{20}=1.25 m[/tex]
The height the car can reach if the the track starts to climb upwards is 1.2742 meters up.
What is kinetic and potential energy?Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or its relation with its surrounding systems.
P.E. = mass × g × height
K.E. = 0.5 × mass × (velocity)²
Given that the toy car has a mass of 0.025 kg and is traveling on a horizontal track with a velocity of 5 m/s. Now, the car starts to climb up vertically, therefore, the kinetic energy will be converted to potential energy.
Kinetic Energy = Potential Energy
0.5 × mass × (velocity)² = mass × g × height
Cancel mass from both the sides,
0.5 × (velocity)² = g × height
0.5 × (5 m/s)² = 9.81 m/sec² × height
height = 1.2742 meters
Hence, the car will travel 1.2742 meters up.
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Two force vectors are oriented such that the angle between their directions is 46 degrees and they have the same magnitude. If their magnitudes are 2.81 newtons, then what is the magnitude of their sum
Answer:
F = 5.17 N
Explanation:
If we know the magnitudes of both vectors, and the angle between them, we can find the magnitude of their sum, applying the cosine theorem, as follows:[tex]F =\sqrt{F_{1} ^{2} +F_{2} ^{2} + 2*F_{1} * F_{2} * cos \theta} (1)[/tex]
Replacing by the givens, F₁ = F₂ = 2.81 N, θ = 46º, we get:[tex]F =\sqrt{2.81 N ^{2} +2.81 N ^{2} + 2*2.81N* 2.81N* cos 46} = 5.17 N (2)[/tex]
A 2.0 kg block of ice with a speed of 8.0 m/s makes an elastic collision with another block of ice that is at rest. The first block of ice proceeds in the same direction as it did initially, but with a speed of 2.0 m/s. What is the mass of the second block? (Hint: Use the conservation of kinetic energy to solve for the second unknown variable.)
Answer:
6kg
Explanation:
According to conservation of kinetic energy
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses of the bodies
u1 and u2 are the initial velocities
v is the final velocity
Given
m1 =2kg
u1 = 8.0m/s
m2 = ?
u2 = 0m/s (second ice at rest)
v = 2.0m/s
Substitute into the formula
2(8)+m2(0) = (2+m2)(2)
16+0 = 4+2m2
16-4= 2m2
12 = 2m2
m2 =12/2
m2 = 6kg
Hence the mass of the second block is 6kg
You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of force you applied to your friend.
Answer:
4
Units:
Newtons
4. Friction is required for :
(a) Fast movement
(0) Both of the above
(b) Stopping the objec
(d) None of the above
Answer:
b) stopping the object
Explanation:
Friction always slows a moving object down. ... Friction can be a useful force because it prevents our shoes slipping on the pavement when we walk and stops car tyres skidding on the road. When you walk, friction is caused between the tread on shoes and the ground. This friction acts to grip the ground and prevent sliding
A plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.
Answer:
-22.1
Explanation:
1 / 4
Step 1. List horizontal (xxx) and vertical (yyy) variables
xxx-direction yyy-direction
t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text
a_x=0a
x
=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a
y
=−9.8
s
2
m
a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction
\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text
v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v
y
=?v, start subscript, y, end subscript, equals, question mark
v_{0x}=120\,\dfrac{\text m}{\text s}v
0x
=120
s
m
v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v
0y
=0v, start subscript, 0, y, end subscript, equals, 0
Note that there is no horizontal acceleration, so v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.
Also, the package has no initial vertical velocity.
Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:
\Delta x=v_xtΔx=v
x
tdelta, x, equals, v, start subscript, x, end subscript, t
Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv
y
v, start subscript, y, end subscript:
\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v
0y
t+
2
1
a
y
t
2
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared
Hint #22 / 4
Step 2. Find ttt from horizontal variables
\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}
Δx
t
t
=v
x
t
=
v
0x
Δx
=
120
s
m
255m
=2.125s
Hint #33 / 4
Step 3. Find \Delta yΔydelta, y using ttt
Using ttt to solve for \Delta yΔydelta, y gives:
\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}
Δy
=v
0y
t+
2
1
a
y
t
2
=
(0)t
+
2
1
(−9.8
s
2
m
)(2.125s)
2
=−22.1m
Hint #44 / 4
The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.
Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both charges are doubled, and the distance between them is halved, the new force between them is
Answer:
The new force between them is increased by a factor of 16.
Explanation:
According to Coulombs law, the force of attraction between two (2) charges is given by the formula;
F = Kq1q2/r²
Given the following data;
q1 = 2q1
q2 = 2q2
r = r/2
Substituting into the equation, we have;
F = 2q1*2q2/(r/2)²
F = 4q1q2/r²/4
F = 4q1q2 * 4/r²
F = 16q1q2/r²
Therefore, the new force between them is increased by a factor of 16.
The volume of a solid object tends to increase as its temperature increases. Which quantities determine how large the change in volume is
Answer:
I. The change in temperature of the object.
II. The initial volume of the object.
Explanation:
Generally, the volume of a solid object tends to increase as its temperature increases and this phenomenon is known as thermal expansion.
Hence, the quantities which determine how large the change in volume of a solid object is includes;
I. The change in temperature of the object.
II. The initial volume of the object.
This ultimately implies that, when a solid object is heated, the atoms of the object vibrate rapidly about their fixed points and thus, causing an increase in the volume of the object.
In conclusion, this scientific phenomenon known as thermal expansion is valid and true for all the three (3) states of matter;
Solid. Liquid. Gas.A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?
Answer:
the answer is b luv .
Explanation:
What is the wavelength of a wave that has a speed of 350 meters second and a frequency of 140 hert??
Wavelength is the distance between identical points (adjacent crests) in the adjacent cycles of a waveform signal propagated in space or along a wire. In wireless systems, this length is usually specified in meters (m), centimeters (cm) or millimeters (mm).
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