Answer:
The cross-sectional area of the larger piston is 392 cm²
Explanation:
Given;
output mass of the piston, m₀ = 2000 kg
input force of the piston, F₁ = 500 N
input area of the piston, A₁ = 10 cm² = 0.001 m²
The output force is given by;
F₀ = m₀g
F₀ = 2000 x 9.8
F₀ = 19600 N
The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;
[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]
Therefore, the cross-sectional area of the larger piston is 392 cm²
What is Matter? write its types
According to definition of general chemistry matter is any substance which has atleast a mass and occupies a volume.
Matter is of three types
SolidLiquidGasA massage technique that consists of applying pressure to specific points of the face and body to release muscle tension, stimulate and restore balance (chi) is known as
Answer:Acupressure
.
.
..
But why post this question in Physics
importance of SI system in points
Answer:
SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.
The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
A 180-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,050 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable
Answer:
27yrs
Explanation:
h= difference in height between the initial position and the bottom position
We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical
h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)
=0.674m
Potential Energy = 28× 9.8×0.674
=184.9J
B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:
E= 0.5mv^2
The cylinder is displaced 0.17 m downward from its equilibrium position and is released at time t = 0. Determine the displacement y and the velocity v when t = 3.1 s. The displacement and velocity are positive if downward, negative if upward. What is the magnitude of the maximum acceleration?
Complete Question
The image of this question is shown on the first uploaded image
Answer:
a
[tex]d =0.161 \ m[/tex]
b
[tex]v = - 0.054 \ m/s[/tex]
c
[tex]a = 6.12 \ m/s^2[/tex]
Explanation:
From the question we are told that
The maximum displacement is A = 0.17 m
The time considered is [tex]t = 3.1 \ s[/tex]
The spring constant is [tex]k = 137 \ N \cdot m[/tex]
The mass is [tex]m = 3.8 \ kg[/tex]
Generally given that the motion which the cylinder is undergoing is a simple harmonic motion , then the displacement is mathematically represented as
[tex]d = A cos (w t )[/tex]
Where [tex]w[/tex] is the angular frequency which is mathematically evaluated as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
substituting values
[tex]w = \sqrt{\frac{137}{ 3.8} }[/tex]
[tex]w =6[/tex]
So the displacement is at t
[tex]d = 0.17 cos (6 * 3.1 )[/tex]
[tex]d =0.161 \ m[/tex]
Generally the velocity of a SHM(simple harmonic motion) is mathematically represented as
[tex]v = - Asin (wt)[/tex]
substituting values
[tex]v = - 0.17 sin ( 6 * 3.1 )[/tex]
[tex]v = - 0.054 \ m/s[/tex]
Generally the maximum acceleration is mathematically represented as
[tex]a = w^2 * A[/tex]
substituting values
[tex]a_{max} = 6^2 * (0.17)[/tex]
substituting values
[tex]a = 6^2 * (0.17)[/tex]
[tex]a = 6.12 \ m/s^2[/tex]
A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 from the speaker.
At what rate does this speaker produce energy?
What is the intensity of this sound 9.50 from the speaker?
What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?
Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
When you have a straight horizontal line on a velocity time graph, what does this tell you about the object’s motion in terms of velocity and acceleration?
Answer:
It tell you that the velocity is constant, what means that there's no acceleration
A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contract- ing when it is cooled from 120.0°C to 10.0°C?
Answer:
42000N
Explanation:
First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.
1) linear thermal expansion coef brass 19e-6 /K
∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm
Second part involves linear elasticity.
for brass, young's modulus is 15e6 psi or 100 GPa
cross-sectional area of rod is π(0.008)² = 0.0002 m²
F = EA∆L/L
F = (100e9)(0.0002)(0.00387) / (1.85)
F = 42000 or 42 kN
what are the property of the image formed by plane mirror
of class 10
Answer:
» The image is laterally inverted.
» The image is upright.
» The image geometry is same as object geometry.
» Image distance is same as object distance.
» Image is not real, it's virtual ( not formed on screen ).
[tex].[/tex]
A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.0300 mm. Monochromatic light is directed toward the double slit and forms an interference pattern on the screen. The first dark fringe is 3.70 cm from the center line on the screen.
Required:
a. Determine the wavelength of light.
b. Calculate the distance between the adjacent bright fringes.
Answer:
The wavelength of this light is approximately [tex]427\; \rm nm[/tex] ([tex]4.27\times 10^{-7}\; \rm m[/tex].)The distance between the first and central maxima is approximately [tex]7.40\; \rm cm[/tex] (about twice the distance between the first dark fringe and the central maximum.)Explanation:
WavelengthConvert all lengths to meters:
Separation of the two slits: [tex]0.0300\; \rm mm = 3.00\times 10^{-5}\; \rm m[/tex].Distance between the first dark fringe and the center of the screen: [tex]3.70\; \rm cm = 3.70\times 10^{-2}\; \rm m[/tex].Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:
The angle between the filter and the beam of light from the lower slit, andThe angle between the screen and that same beam of light.These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].
The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)
Therefore:
[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].
On the other hand:
[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].
Because the cotangent of [tex]\theta[/tex] is very close to zero,
[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].
[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].
[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].
Distance between two adjacent maximaIf the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.
The path difference required for the central maximum is [tex]0[/tex].The path difference required for the first maximum is [tex]\lambda[/tex].The path difference required for the second maximum is [tex]2\,\lambda[/tex].On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:
[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].
Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion.
(a) Should the sail be absorbing or reflective? Why?
(b) How large a sail is necessary to propel a 10000kg
spacecraft against the gravitational force of the sun? Express your result in square kilometers.
(c) Explain why your answer to part (b) is independent of the distance from the sun.
The gravitational constant is G=6.67×10−11m3⋅s−2⋅kg−1.
The mass of the sun is Ms=1.99×1030kg.
Answer:
The complete question is
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.
For totally reflective, F = (2I/c)A ....1
for totally reflective, F = (I/c)A ....2
where I is the intensity of the light
c is the speed of light = 3 x 10^8 m/s
A is the area the sail
b) The intensity of the light from the sun = power/area
==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]
where r is the distance from the sun and the sail
The Force from the sail from equation 1 is therefore
[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]
gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]
where
G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.
m is the mass of the sail = 10000 kg
M is the mass of the sun = 1.99 x 10^30 kg.
==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
Equating the forces, we have
[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
the distance cancels out
A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2
==> 6428.2 km^2
c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation is inversely proportional to the square of the distance, so they both cancel out.
PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is 9.0 cm. What is the height of the image?
Answer:
The image height is 3.0 cm
Explanation:
Given;
object distance, [tex]d_o[/tex] = 15.0 cm
image distance, [tex]d_i[/tex] = 5.0 cm
height of the object, [tex]h_o[/tex] = 9.0 cm
height of the image, [tex]h_i[/tex] = ?
Apply lens equation;
[tex]\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm[/tex]
Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.
the atomic number of a nucleus increases during which nuclear reactions
Answer:
Answer A : Fusion followed by beta decay (electron emission)
Explanation:
Notice that you want the Atomic number to increase, that is the number of protons in a nucleus. So if all four cases given experience the same fusion of nuclei, the only one that net increases the number of protons in the last stage, is the reaction that undergoes a beta decay (with emission of an electron) thus leaving a positive imbalance of positive charge (proton generated in the beta decay of a neutron).
Therefore, answer A is the correct one.
Answer:
A : Fusion followed by beta decay (electron emission)
Explanation:
Ap3x
A block weighing 400 kg rest on a horizontal surface and supports on top of it another block of weight 100 kg placed on the top of it as shown. The block W2 is attached to a vertical wall by a string 6 m long. Ifthe coefficient of friction between all surfaces is 0.25 and the system is in equilibrium find the magnitude of the horizontal force P applied to the lower block.
The horizontal force applied to the lower block is approximately 1,420.85 Newtons
The known parameters are;
The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N
The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N
The length of the string attached to the block, W₂, l = 6 m
The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m
The coefficient of friction between the surfaces, μ = 0.25
Let T represent the tension in the string
The upward force on W₂ due to the string = T × sin(θ)
The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)
The tension in the string, T = N₂ × μ × cos(θ)
∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)
sin(θ) = √(6² - 5²)/6
cos(θ) = 5/6
∴ T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
T ≈ 183.27 N
The normal reaction on W₂, N₂ = T/(μ × cos(θ))
∴ N₂ = 183.27/(0.25 × 5/6) = 879.7
N₂ ≈ 879.7 N
The friction force, [tex]F_{f2}[/tex] = N₂ × μ
∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N
The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂
[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N
The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ
∴ [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N
The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = 1,200.925 N + 219.925 N = 1,420.85 N
The horizontal force applied to the lower block, P ≈ 1,420.85 N
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer?
O The nail exerts a much smaller force on the hammer in the opposite direction
O The nail exerts a much smaller force on the hammer in the same direction.
The nail exerts an equal force on the hammer in the same direction.
O The nail exerts an equal force on the hammer in the opposite direction.
Answer:
reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
This is an action-reaction problem or Newton's third law, which states that forces in naturals occur in pairs.
This is the foregoing, the hammer exerts a force on the nail of magnitude F and it will direct downwards, if we call this action and the nail exerts a force on the hammer of equal magnitude but opposite direction bone directed upwards, each force is applied in one of the bodies.
The difference in result that each force is that the force between the nail exerts a very high pressure (relation between the force between the nail area), instead the area of the hammer is much greater, therefore the pressure is small.
When reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that:_______
a. no magnetic field exists in that region of space.
b. the particle must be moving parallel to the magnetic field.
c. the particle is moving at right angles to the magnetic field.
d. either no magnetic field exists or the particle is moving parallel to the magnetic field.
e. either no magnetic field exists or the particle is moving perpendicular to the magnetic field.
Answer:
b. the particle must be moving parallel to the magnetic field.
Explanation:
The magnetic force on a moving charged particle is given by;
F = qvBsinθ
where;
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field
θ is the angle between the magnetic field and velocity of the moving particle.
When is the charge is stationary the magnetic force on the charge is zero.
Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.
Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.
b. the particle must be moving parallel to the magnetic field.
Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact
Answer:
Explanation:
Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.
While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.
Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''
The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m], RD= 6 [m], RE=12 [m].
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
g suppose he used an alpha particle with an energy of 8.3 MeV, what would be the speed of this alpha particle
Answer:
speed of the alpha particle is 2 x 10^7 m/s.
Explanation:
energy of alpha particle = 8.3 Mev
1 Mev = 1.602 x 10^-13 J
8.3 Mev = [tex]x[/tex]
solving, [tex]x[/tex] = 8.3 x 1.602 x 10^-13 = 1.329 x 10^-12 J
mass of a alpha particle = 6.645 x 10^−27 kg
The energy of the alpha particle is the kinetic energy KE of the alpha particle
KE = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the alpha particle
v is the velocity of the alpha particle
substituting values, we have
1.329 x 10^-12 = [tex]\frac{1}{2}*6.645*10^{-27}*v^{2}[/tex]
[tex]v^{2}[/tex] = 4 x 10^14
[tex]v = \sqrt{4*10^{14} }[/tex] = 2 x 10^7 m/s
5. A body falls freely from rest. It covers as much distance in the last second of its
motion as covered in the first three seconds. The body has fallen for a time of:
B) 5s
C) 7s
D) 9s
A) 35
Answer:
B 5s
Explanation:
Because of the Displacement in the nth second of the free fall is
Snth=21g(t12−t22)
Given that (tn−tn−1)=1
Displacement in 3 seconds of the free fall
S=21gt2
S=21×10×32
S=45m
Given that: Snth=45
On solving that we get:
t1=5sec
In _____ research, a group of people of one age is compared to a group of people who are another age.
Answer:
cross-sectional
Explanation:
The full definition of this is ''a research design in which several different age-groups of participants are studied at one particular point in time.''
A 500 kg rocket sled is coasting in reverse at 10 m/s (to the left). It then turns on its rocket engines for 10.0 s, with a thrust of 1500 N (to the right). What is its final velocity? (Remember velocity has magnitude and direction)
Explanation:
F = ma
[tex]a = \frac{f}{m} [/tex]
[tex]a = \frac{1500}{500} = 3[/tex]
[tex]a = \frac{v2 - v1}{t} [/tex]
[tex]3 = \frac{v2 - 10}{10} [/tex]
v2 (final) = 40 m/s to the right direction
Find the current through a person and identify the likely effect on her if she touches a 120 V AC source in the following circumstances. (Note that currents above 10 mA lead to involuntarily muscle contraction.)
(a) if she is standing on a rubber mat and offers a total resistance of 300kΩ
(b) if she is standing barefoot on wet grass and has a resistance of only 4000kΩ
Answer:
A) 0.4 mA
B) 0.03 mA
Explanation:
Given that
voltage source, V = 120 V
to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.
mathematically, Ohms Law, V = IR
V = Voltage
I = Current
R = Resistance
from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have
120 = I * 300*10^3 Ω
making I the subject of the formula,
I = 120 / 300000
I = 0.0004 A
I = 0.4 mA
Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.
B, we have Resistance, R = 4000kΩ
Substituting like in part A, we have
120 = I * 4000*10^3 Ω
I = 120 / 4000000
I = 0.00003 A
I = 0.03 mA
This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA
The current through a person will be:
a) 0.4 mA
b) 0.03 mA
Given:
Voltage, V = 120 V
Ohm's Law:It states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.
Ohms Law, V = I*R
where,
V = Voltage
I = Current
R = Resistance
a)
Given: Resistance= 300kΩ
[tex]120 = I * 300*10^3 ohm\\\\I = 120 / 300000\\\\I = 0.0004 A[/tex]
Thus, current will be, I = 0.4 mA
b)
Given: R = 4000kΩ
[tex]120 = I * 4000*10^3 ohm\\\\I = 120 / 4000000\\\\I = 0.00003 A[/tex]
Thus, current will be, I = 0.03 mA
From calculations, we observe that nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA.
Find more information about Current here:
brainly.com/question/24858512
If 1 kg of each substance in the table absorbs 250 kJ of thermal energy,
which substance will have the smallest change in temperature?
Answer:
Water.
Explanation:
To know which of the substance that will absorbed the greatest amount of thermal energy, we'll simply determine the amount of energy absorbed by each substance.
This is illustrated below:
For Air:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 1.01 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 1.01 x 15
Q = 15.15 KJ
Therefore, the heat absorbed by air is 15.15 KJ.
For Plastic:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 2.60 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 2.60 x 15
Q = 39 KJ
Therefore, the heat absorbed by plastic is 39 KJ.
For Water:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 4.18 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 4.18 x 15
Q = 62.7 KJ
Therefore, the heat absorbed by water is 62.7 KJ.
For Wood:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 1.68 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 1.68 x 15
Q = 25.2 KJ
Therefore, the heat absorbed by Wood is 25.2 KJ.
Summary:
Substance >>>>> Heat Absorbed
Air >>>>>>>>>>>> 15.15 KJ.
Plastic >>>>>>>>> 39 KJ
Water >>>>>>>>>> 62.7 KJ
Wood >>>>>>>>>> 25.2 KJ.
From the above calculations, we can see that water will absorb the greatest amount of thermal energy.
The substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.
Specific heat capacity
The specific heat capacity of each substance can be used to determine the substance with the smallest change in temperature.
Q = mcΔθ
where;
m is massc is specific heat capacityΔθ is change in temperatureΔθ = Q/mc
For airΔθ = (250)/(1.01)
Δθ = 247.5 ⁰C
For plasticΔθ = (250)/(2.6)
Δθ = 96.15 ⁰C
For waterΔθ = (250)/(4.18)
Δθ = 59.81 ⁰C
For woodΔθ = (250)/(1.68)
Δθ = 148.81 ⁰C
Thus, the substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.
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You are outdoors when you hear the constant chirp of a still cricket. You start walking toward the cricket and at some point you are able to detect that the intensity of the chirp of the cricket has increased by a factor of 4. What of the following statements is true at your new position with respect to the cricket?
a. The power delivered by the sound wave you hear has doubled.
b. The speed of the sound wave emitted by the cricket has decreased by a factor of 4.
c. The distance between you and the cricket has decreased by a factor of 2
Answer:
C
Explanation:
intensity = Power delivered by the sound (Watt)/ Surounding Area (m²)
I = P/A
A = πr²
r = is the distance between you and the cricket.
so in other form we can get
I = P/πr²
let take I(1) as first intensitilynyou heard and I(2) as the increased intensity.
I(1) / I(2) = r(2)² / r(1)²
1/4 = r(2)²/r(1)²
1/2 = r(2) / r(1)
r(2) = ½ r(1)
or r(2) is decreaases by a factor of 2.
A student is hammering a nail into a board. Where should he hold the hammer and why?
Answer:
At the end of the handle farthest from the head of the hammer.
Explanation:
The force of the hammer is greatest the longer the radius is on a which would be the length of the handle. Simple mechanical advantage.
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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An oscillating particle has the equation x = 10cos(8πt +π ) (x in cm, t in s). The number of times the object passes through the equilibrium position in one second is
Answer:
x = A sin (wt + theta) where w = angular frequency - basic SHM equation
w = 8 pi = 2 pi f
f = 4 basic frequency
N = 8 number of times thru origin
Each cycle the particle will pass thru the origin +x and -x twice