Answer:
1. c
2. e
3. d
Explanation:
1.
From Einstein's Photoelectric Equation, we know that:
Energy given up by photon = Work Function + K.E of Electron
hc/λ = φ + K.E
where,
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m
φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J
Therefore,
(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E
K.E = (9.939 - 8.16) x 10⁻¹⁹ J
K.E = 1.778 x 10⁻¹⁹ J
The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:
c. less energetic photo-electrons (on average)
2.
The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,
e. more photo-electrons ejected
3.
X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.
d. more energetic photo-electrons (on average)
When an aluminum bar is connected between a hot reservoir at 860 K and a cold reservoir at 348 K, 2.40 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir
(a) In this irreversible process, calculate the change in entropy of the hot reservoir.
_______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?
Answer:
a) [tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex], b) [tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex], c) [tex]S_{gen} = 4.106\,\frac{J}{K}[/tex], d) Due to irreversibilities due to temperature differences.
Explanation:
a) The change in entropy of the hot reservoir is:
[tex]\Delta S_{in} = \frac{2400\,J}{860\,K}[/tex]
[tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex]
b) The change in entropy of the cold reservoir is:
[tex]\Delta S_{out} = \frac{2400\,J}{348\,K}[/tex]
[tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex]
c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:
[tex]\Delta S_{in} - \Delta S_{out} + S_{gen} = 0[/tex]
[tex]S_{gen} = \Delta S_{out} - \Delta S_{in}[/tex]
[tex]S_{gen} = 6.897\,\frac{J}{K} - 2.791\,\frac{J}{K}[/tex]
[tex]S_{gen} = 4.106\,\frac{J}{K}[/tex]
d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.
A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simultaneously from a height of 1.8m. If the larger ball rebounds elastically from the floor and the small ball rebounds elastically from the larger ball what value of m results in the larger ball stopping when it collides with the small ball?
Help with this answer please
Answer:
Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL
AAAAAAAAAAAA is the answer
A projectile is defined as
Answer:
By definition, a projectile has a single force that acts upon it - the force of gravity.
Explanation:
A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.
// have a great day //
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide
Answer:
The distance covered by puck A before collision is [tex]z = 8.56 \ m[/tex]
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is [tex]v_A = 3.90 \ m/s[/tex]
The speed of puck B is [tex]v_B = 4.30 \ m/s[/tex]
The distance covered by puck A is mathematically represented as
[tex]z = v_A * t[/tex]
=> [tex]t = \frac{z}{v_A}[/tex]
The distance covered by puck B is mathematically represented as
[tex]18 - z = v_B * t[/tex]
=> [tex]t = \frac{18 - z}{v_B}[/tex]
Since the time take before collision is the same
[tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]
substituting values
[tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]
=> [tex]70.2 - 3.90 z = 4.3 z[/tex]
=> [tex]z = 8.56 \ m[/tex]
A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.
Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
55.3
Explanation:
The computation of the number of bright-dark-bright fringe shifts observed is shown below:
[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]
where
d = [tex]3.95 \times 10^{-2}m[/tex]
[tex]\lambda = 400 \times 10^{-9}m[/tex]
n = 1.00028
Now placing these values to the above formula
So, the number of bright-dark-bright fringe shifts observed is
[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]
= 55.3
We simply applied the above formula so that the number of bright dark bright fringe shifts could come
Two parallel, vertical, plane mirrors, 38.8 cm apart, face each other. A light source at point P is 30.1 cm from the mirror on the left and 8.7 cm from the mirror on the right.
(a) How many images of point P are formed by the mirrors?
(b) Find the distance from the mirror on the right to the two nearest images behind the mirror.
first nearest image=
second nearest image=
(c) Find the number of reflections of light rays for each of these images.
first nearest image=
second nearest image=
Answer:
Explanation shown below.
Explanation:
1.The number of images formed by 2 parallel mirrors is an infinite number of images.
2. The characteristics of a plane mirror is such that the object distance equals the image distance.
Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.
Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.
Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .
Hence the shortest distances are 8.7cm and 68.9cm
3. The number of reflections is infinite for both cases.
The amount of friction divided by the weight of an object forms a unit less number called the
Answer:
Coefficient of friction.
Explanation:
The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :
[tex]F=\mu N[/tex]
N is normal force.
[tex]\mu[/tex] = coefficient of friction
[tex]\mu=\dfrac{F}{N}[/tex]
A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.20 cm on a side and 1.70 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 96.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.
Answer:
U = 218 nJ
Explanation:
We are given;
Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m
Voltage across the capacitor; V = 96 V
Dimension of the square plates is 7.2cm x 7.2cm.
So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²
Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²
From relative permeability table;
Dielectric constant of Pyrex; k1 = 5.6
Dielectric constant of polystyrene; k2 = 2.56
Now, formula for capacitance of a capacitor with Dielectric is;
C = kC_o
Where, C_o = ε_o(A/d)
Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)
Since we have 2 capacitor, thus ;
C1 = k1*ε_o*(A/d)
C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C1 = 1.51 × 10^(-10) F
Similarly;
C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C2 = 0.691 × 10^(-10) F
For capacitors in series, formula for total capacitance(Cs) is;
1/Cs = (1/C1) + (1/C2)
Simplifying this, we have;
Cs = (C1*C2)/(C1 + C2)
Plugging in the relevant values ;
Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))
Cs = 0.474 × 10^(-10) F
The formula for energy stored in a capacitor with 2 Dielectrics is given as;
U = ½Cs*V²
So,
U = ½ × 0.474 × 10^(-10) × 96²
U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ
A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce.
What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.
Answer:
J = 0.800 kg m/s
Fmax = 291 N
Explanation:
During the fall, energy is conserved.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × 9.81 m/s² × 1.45 m)
v = 5.33 m/s
Alternatively, you can use kinematics to find the velocity.
Impulse = change in momentum
J = Δp
J = mΔv
J = (0.150 kg) (5.33 m/s)
J = 0.800 kg m/s
Impulse = area under F vs t graph
J = ∫ F dt
J = ½ Fmax Δt
(0.800 kg m/s) = ½ Fmax (0.0055 ms)
Fmax = 291 N
What is the frequency if 140 waves pass in 2 minutes?
Answer:
1.16 Hz
Explanation:
frequency, basically, is the number of wave on 1 second
so, in math we write like this
f = n/t
n = number of waves
t = time to do that (in sec)
f = 140/120 = 7/6 Hz
f = 1.16 Hz
A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the bottom of an inclined plane of height h=3.5m, as shown in the figure. The spherical shell rolls without slipping. Find wif the shell comes to rest at the top of the inclined plane. (Take g-9.81 m/s2, Ispherical shell = 2/3 MR2 ).Express your answer using one decimal place.
Answer:
47.8rad/s
Explanation:
For energy to be conserved.
The potential energy sustain by the object would be equal to K.E
P.E = m× g× h = 2 × 9.81× 3.5= 68.67J
Now K.E = 1/2 × I × (w1^2 - w0^2)
I = 2/3 × M × R2
= 2/3 × 2 × (0.23)^2= 0.0705
Hence
W1 = final angular velocity
Wo = initial angular velocity
From P.E = K.E we have;
68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)
(w1^2 - w0^2) = 1948.09
W1^2 = 1948.09 + (18.3^2)
W1^2=2282.98
W1 = √2282.98
=47.78rad/s
= 47.8rad/s to 1 decimal place.
Countries create quotas and tariffs to increase the volume of trade with their neighbors.
Oooooo, that statement is not true. Countries create quotas and tariffs to LIMIT the volume of trade with other countries, including their neighbors.
Answer:
False
Explanation:
I took the text :)
An aluminum wing on a passenger jet is 30 m long when its temperature is 27 C. At what temperature would the wing be 0.03 shorter?
Answer:2000
Explanation:
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s? Express to 3 sig figs.
Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
[tex]L_{f}[/tex] = I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
Which nucleus completes the following equation?
Se+?
O A. Ga
B. P
C. 31P
D. CI
Answer:First option
Explanation:
hope it helped
The air flowing over the top of the wing travels
in the same amount of time than the air
flowing beneath the wing.
Answer: Short Answer: NO ( In Most Cases)
Explanation:
If that were true then planes couldn't get off the ground to fly. The front of the wing is cutting/pushing the air. On the top of the wing the air moves faster and on the bottom it moves slower making a upward draft giving the object the ability to fly or glide.
g: To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
I =1.8 kgm^2
Explanation:
In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:
[tex]\tau=I\alpha[/tex] (1)
τ: torque applied to the door
I: moment of inertia of the door
α: angular acceleration = 5 rad/s^2
The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).
F = 10 N
d = 0.9 m
You replace the expression for τ, and solve for I:
[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]
The moment of inertia of the door is 1.8 kgm^2
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.
Answer:
d. 1.00 m
Explanation:
In 1905, Einstein proposed special theory of relativity of light.
This theory had a number of consequences or results. One of them is called "Length Contraction".
According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.
But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.
In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:
d. 1.00 m
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact
A) Find the final angular speed of the door.
answer in rad/s
B) Does the mud make a significant contribution to the moment of inertia?
Yes or No
Answer:
0.19rad/s and Yes
Explanation:
From the principle of conservation of momentum it means momentum before and after collision is the same.
Momentum before collision is 0.700 kg×12 = 8.4Ns
Momentum of the door = mass of door × velocity of door
8.4Ns = mass of door × velocity of door
Velocity of door = 8.4Ns/45 =0.19m/s
But velocity V= w×r ;
w-angular velocity
r- raduis = width
w= 0.19/1m = 0.19rad/s
2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.
A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward.
(A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s
Answer:
a) 98 m/s
b) 19.6 s
c) 1449.8 m
d) 29.6 s
e) 168.6 m/s
f) 46.8 s
Explanation:
Given that
Acceleration of the rocket, a = 5 m/s²
Altitude of the rocket, s = 960 m
a)
Using the equation of motion
v² = u² + 2as, considering that the initial velocity, u is 0. Then
v² = 2as
v = √2as
v = √(2 * 5 * 960)
v = √9600
v = 98 m/s
b)
Using the equation of motion
S = ut + ½at², considering that initial velocity, u = 0. So that
S = ½at²
t² = 2s/a
t² = (2 * 960) / 5
t² = 1920 / 5
t² = 384
t = √384 = 19.6 s
c)
Using the equation of motion
v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that
0 = 98² + 2(-9.8) * s
9600 = 19.6s
s = 9600/19.6
s = 489.8 m
The maximum altitude now is
960 m + 489.8 m = 1449.8 m
d)
Using the equation of motion
v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that
0 = 98 +(-9.8 * t)
98 = 9.8t
t = 98/9.8
t = 10 s
Total time then is, 10 + 19.6 = 29.6 s
e) using the equation of motion
v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,
v² = 0 + 2 * 9.8 * 1449.8
v² = 28416.08
v = √28416.08
v = 168.6 m/s
f) using the equation of motion
S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s
1449.8 = 0 + ½ * 9.8 * t²
2899.6 = 9.8t²
t² = 2899.6/9.8
t² = 295.88
t = √295.88
t = 17.2 s
total time in air then is, 17.2 + 29.6 = 46.8 s
Nuclear fusion in our Sun happens when
- hydrogen atoms combine to make helium atoms and release energy
- uranium atoms break apart and release energy
- hydrogen atoms are burned and release energy
- helium atoms break apart and release energy
Answer:
A
Explanation:
Fussion occurs when elements of lower atomic mass combines to form that of a larger atomic mass, releasing energy in the process .
Hydrogen has a lower atomic mass than Helium.
Question
20
what would be the advantages if your body had magnetic properties science subject
Answer:
Some of the advantages if our body had magnetic properties are as follows:
Magnetic properties can have health benefits such as recovering quickly from a stroke, resolving bladder problems, and reducing blood pressure.Brain will be able to control more activities of the nervous system and other organs of the body using magnetic power.Heart will have many benefits of magnetic properties and able to provide more energy to the entire body through the circulation of blood.Magnetic properties in body will be able to maintain the production of melatonin that controls the sleep patterns.Magnetic properties will be able to kill cancer causing cells.Hence, magnetic properties are somehow beneficial for humans.
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
A) 4.55 m.
B) 1.05 m.
C) 3.54 m.
D) 2.25 m.
Answer:
Letter A. [tex]y=4.55 m[/tex]
Explanation:
Let's use the wave equation:
[tex]y=Asin(kx-\omega t)[/tex]
A is the amplitude (A=6.44 m)t is the time (t=0.71 s)k is the wave number (k=2.34 1/m)ω is the angular frequency (ω=2.88 rad/s)x is the propagation of the x direction (x=1.21 m)Therefore the displacement y will be:
[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]
[tex]y=4.55 m[/tex]
The answer is letter A.
I hope it helps you!
Answer:
Explanation:
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
Amplitude (A) of the simple harmonic wave = 6.44 m
wave number (k) of the given wave = 2.34 m-1
Angular frequency (ω) of the given wave = 2.88 rad/s
Displacement x = 1.21 m and time t = 0.71 s
Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by
y = Asin(kx - ωt)
= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]
Y=6.44sin(0.7866 rad)
0.7866rad*(180 degrees/pi rad) =45.1
Y=6.44sin(45.1)
Y=4.55m
Someone please helpp me out thanks !
Answer:
Silver.
Explanation:
To determine the identity of the metal, we need to calculate the density of the metal. This is illustrated below:
Mass of metal (m) = 18.15g
Length (L)= 1.2cm
Volume (V) = L³ = 1.2³ = 1.728cm³
Density =.?
The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, it is expressed as:
Density = Mass /volume
With the above formula, we can obtain the density of the metal as follow:
Mass = 18.15g
Volume = 1.728cm³
Density =.?
Density = Mass /volume
Density = 18.15g/1.728cm³
Density of the metal = 10.50g/cm³
Comparing the density of metal obtained with the densities given in the table above, we can see that the density of the metal is the same with that of silver.
Therefore, the metal is silver.
HELPP MEE
Which image illustrates the desired interaction of a sound wave with
soundproofing material in a recording studio?
Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.
What is the basis of soundproofing?Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.
Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.
Thus, the correct option is C, as the D image is showing the absorption.
For more details regarding soundproofing, visit:
https://brainly.com/question/8980142
#SPJ2
Answer: C.D
Explanation:...
when the same amount of heat is added to equal masses of water and copper at the same temperature the copper is heated to a higher final temperature than water. on a molecular level what explains this difference
a. the average kinetic energy of water molecules is greater than the average kinetic energy of the copper
b.more of the heat is transferred to the potential energy of the water molecules than the potential energy of the copper atoms
c.the intermolecular forces between copper atoms are stronger than those between water molecules
d.more of the heat is transferred to the kinetic energy of the water molecules than to the kinetic energy of the copper atoms
Answer:
C
Explanation:
The intermolecular forces between the water molecule is less binding than that of the copper molecule. Hence the water would take a shorter time to be converted to vapour where the temperature of boiling is constant however the temperature of that of the copper molecule keeps increasing.
You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 85 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
Force = 60.08 N
Explanation:
Given that
Diameter d = 30 mm
Holding pressure = 85 % of Atmospherics pressure
Solution
As we know that here 1 atm = 10⁵ N/m²
and pressure is known as force per unit area
pressure = [tex]\frac{F}{A}[/tex] ................1
put here value and we will get
F = [tex]0.85\times 10^5\times \frac{\pi}{4}\times 0.03^2\ N[/tex]
solve it we get
Force = 60.08 N
In each pair, select a substance that is a better heat conductor.
1. copper wire / wood 3. water / iron
2. water / air 4. iron / glass
Answer:
1)copper wire
Explanation:
it is the best electric conductor
