Answer:
[tex]a=2,5m/s^2[/tex]
Explanation:
From the question we are told that:
Coefficient of kinetic friction [tex]\mu= 0.200[/tex]
Vertical Mass [tex]M_v=3kg[/tex]
Horizontal mass [tex]M_h=3.00kg[/tex]
Generally the equation for kinetic force [tex]F_k[/tex] is mathematically given by
[tex]F_k=\mu N\\F_k=0.2*3\\F_k=0.6[/tex]
Generally the equation for T is mathematically given by
[tex]For M_v=3kg3g-T=3a[/tex]
For [tex]M_h=3kg[/tex]
[tex]T=M_v V+F_k\\T=3.0a+0.6[/tex]
Therefore substituting
[tex]3-3a-0.6=3a\\2.4g=6a[/tex]
[tex]a=2,5m/s^2[/tex]
Choose a tool or a device you can investigate. It must not have any source of power other than the person using it. It must use one or more of the six simple machines. Explain how the tool or device is used. What work does a person do while using it? What kind of motion do we see when it’s used?
Answer:
Wheelbarrows, fishing rods, shovels, brooms, arms, legs, boat oars, crow bars, and bottle openers are all examples of levers. Levers may be one of the most used simple machine. As with all simple machines like the lever, they are designed to help make work easier to do.
Explanation:
Answer:
Wheelbarrows- This tool/device is used for a variety of things, such as moving rock, mulch or compost to the garden, moving trees or large shrubs from one spot to another, hauling bricks, disposing of garden debris, for mixing concrete or fertilizers, and even used on farms for mucking horse stalls. The work you do is pushing it whcih also goes along with the montion you will see, forward motion.
Explanation:
Question 1 of 10 The energy diagram shows the changes in energy during a chemical reaction. Which statement best describes the total energy change of the system? Potential energy Reaction progress O A. Energy is absorbed, and the products have higher potential energy. O B. Energy is absorbed, and the reactants have higher potential energy C. Energy is released, and the reactants have higher potential energy. O D. Energy is released, and the products have higher potential energy.
Answer: The Answer is C
Explanation:
I just got this question and took a guess I got it correct lol. Higher potential energy and energy is released.
The statement which best describes the total energy change of the system is energy is absorbed, and the reactants have higher potential energy.
What is potential energy?Potential energy is the energy which a body posses because of its position. The potential energy is the stored energy.
The energy diagram shows the changes in energy during a chemical reaction.
Potential energy and Reaction progress are shown in the problem. The reactants are the substance which reacts and form one or more than one product.
The reactants in a chemical reaction increases the potential energy for the reaction.
Thus, the statement which best describes the total energy change of the system is energy is absorbed, and the reactants have higher potential energy.
Learn more about the potential energy here;
https://brainly.com/question/15896499
The result of the evolutionary process that preserves traits that enhance the adaptation of an organism and suppresses traits that do not is called
Answer:
Natural selection.
Explanation:
Natural selection can be defined as a biological process in which species of living organisms having certain traits that enable them to adapt to environmental factors such as predators, competition for food, climate change, sex mates, etc., tend to survive and reproduce, as well as passing on their genes to subsequent generations.
Simply stated, natural selection entails the survival of the fittest. Therefore, the species that are able to adapt to the environment will increase in number while the ones who can't adapt will die and go into extinction.
On the other hand, artificial selection is also known as selective breeding and it is a process that involves humans (breeders) selecting the animal or plant with desirable traits in order to reproduce favorable offspring having phenotypic traits.
A 3-m-diameter tank is initially filled with water 2 m above the center of a sharp-edged 10-cm-diameter orifice. The tank water surface is open to the atmosphere, and the orifice drains to the atmosphere through a 100-m-long pipe. The friction coefficient of the pipe is taken to be 0.015 and the effect of the kinetic energy correction factor can be neglected. Determine (a) the initial velocity from the tank (5 points); and (b) the time required to empty the tank (5 points). (Q.8-78, E-book)
Answer:
a) 1.542 m/s
b) t = 38.91 mins
Explanation:
Given data:
Diameter of tank ( D ) = 3 m
height of tank ( initial ) ( h = z1 ) = 2m
Diameter of sharp edged orifice = 0.1 m
pipe length ( L ) = 100 m
coefficient of friction of pipe ( f ) = 0.015
∝1 = ∝2 = 1
A) calculate initial velocity from the tank
since the fluid is open to the atmosphere ; p1 = p2 = Patm ( atm pressure )
the initial velocity of the tank can be determined with the equation below
V = [tex]\sqrt{\frac{2gZ1}{1 + f \frac{L}{d} + K _{L} } }[/tex] ---- ( 1 )
where : Z1 = 2m , f = 0.015, L = 100m , Kl = 0.5m , d = 0.1 m , g = 9.81m/s^2
input given values into equation 1 above
V = 1.542 m/s
b) Determine time required to empty tank
Given velocity = [tex]\sqrt{\frac{2gZ1}{1 + f \frac{L}{d} + K _{L} } }[/tex] and Flow rate = [tex]\sqrt{\frac{2gZ1}{1 + f \frac{L}{d} + K _{L} } }* \frac{\pi }{4} *d^2[/tex]
differentiate the flow rate ( dt ) and then integrate the equation to get the required expression
t = [tex]\frac{2D^2}{d^2} \sqrt{\frac{(1+f\frac{L}{d}+K_{L})Z_{1} }{2g} }[/tex] -------- ( 2 )
where : : Z1 = 2m , f = 0.015, L = 100m , Kl = 0.5m , d = 0.1 m , g = 9.81m/s^2, D = 3m
input values into equation 2
t = 38.91 mins
You leave a pastry in the refrigerator on a plate and ask your roommate to take it out before youget home so you can eat it at room temperature, the way you like it. Instead, your roommateplays video games for hours. When you return, you notice that the pastry is still cold, but thegame console has become hot. Annoyed, and knowing that the pastry will not be good if it ismicrowaved, you warm up the pastry by unplugging the console and putting it in a clean trashbag (which acts as a perfect calorimeter) with the pastry on the plate. After a while, you find thatthe equilibrium temperature is a nice, warmTeq.. You know that the game console has a mass ofm1. Approximate it as having a uniform initial temperature ofT1. The pastry has a mass ofm2and a specific heat ofc2, and is at a uniform initial temperature ofT2. The plate is at the sameinitial temperature and has a mass ofm3and a specific heat ofc3. What is the specific heat ofthe console
Answer:
[tex]c_{e1} = \frac{(m_2 c_{e2} \ + m_3 c_{e3} ) \ (T_{Teq} - T_2) }{m_1 (T_1 - T_{eq}) }[/tex]
Explanation:
This is a calorimeter problem where the heat released by the console is equal to the heat absorbed by the cupcake and the plate.
Q_c = Q_{abs}
where the heat is given by the expression
Q = m c_e ΔT
m₁ c_{e1) (T₁-T_{eq}) = m₂ c_{e2} (T_{eq} -T₂) + m₃ c_{e3} (T_{eq}- T₁)
note that the temperature variations have been placed so that they have been positive
They ask us for the specific heat of the console
[tex]c_{e1} = \frac{(m_2 c_{e2} \ + m_3 c_{e3} ) \ (T_{Teq} - T_2) }{m_1 (T_1 - T_{eq}) }[/tex]
You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to calculate the Rotational Kinetic Energy of the bowling ball as it rolls down a sidewalk without slipping. This means it has both linear kinetic energy and rotational kinetic energy. A bowling ball can be modeled as a solid sphere rotating about its center. This bowling ball has a mass of 6.40 kg and a radius of 0.130 m. You'll need to look up the equation for the Moment of Inertia in your textbook. It is rotating with an angular velocity of 16.0 radians / second in the counter-clockwise (or positive) direction. You can use this to determine the linear velocity of the bowling ball (since it is rolling without slipping). What is the Total Kinetic Energy of the bowling ball
Answer:
K_{total} = 19.4 J
Explanation:
The total kinetic energy that is formed by the linear part and the rotational part is requested
[tex]K_{total} = K_{traslation} + K_{rotation}[/tex]
let's look for each energy
linear
[tex]K_{traslation}[/tex] = ½ m v²
rotation
[tex]K_{rotation}[/tex] = ½ I w²
the moment of inertia of a solid sphere is
I = 2/5 m r²
we substitute
[tex]K_{total}[/tex] = ½ mv² + ½ I w²
angular and linear velocity are related
v = w r
we substitute
K_{total} = ½ m w² r² + ½ (2/5 m r²) w²
K_{total} = m w² r² (½ + 1/5)
K_{total} = [tex]\frac{7}{10}[/tex] m w² r²
let's calculate
K_{total} = [tex]\frac{7}{10}[/tex] 6.40 16.0² 0.130²
K_{total} = 19.4 J
what is the planet that scientists are exploring now?
Answer:
Mars
Explanation:
In the 1960s, humans set out to discover what the red planet has to teach us. Now, NASA is hoping to land the first humans on Mars by the 2030s. Mars has captivated humans since we first set eyes on it as a star-like object in the night sky.
good luck
please mark me as a brainliest
List the two factors that create orbital motion and describe how each factor affects the motion.
Answer:
The forward motion of a body in space, such as a planet or moon, and the pull of gravity on it from another body in space.
Explanation:
Earth and many other bodies—including asteroids, comets, and the other planets—move around the sun in curved paths called orbits. Generally, the orbits are elliptical, or oval, in shape. Because of the sun’s relatively strong gravity, Earth and the other bodies constantly fall toward the sun, but they stay far enough away from the sun because of their forward velocity to fall around the sun instead of into it. As a result, they keep orbiting the sun and never crash to its surface. The motion of Earth and the other bodies around the sun is called orbital motion. Orbital motion occurs whenever an object is moving forward and at the same time is pulled by gravity toward another object.
Which is not a component of an electromagnet?
A. solenoid
B. power source
C. ferromagnetic core
D. permanent magnet
Two ice skaters, with masses of 40.0 kg and 65.0 kg , are at the center of a 50.0 m -diameter circular rink. The skaters push off against each other and glide to opposite edges of the rink. Part A If the heavier skater reaches the edge in 10.0 s , how long does the lighter skater take to reach the edge
Answer:
6.15 s
Explanation:
From the question,
Applying the law of conservation of momentum
Momentum of the Heavier skater = Momentum of the lighter skater.
Mv = mV................. Equation 1
Where M = mass of the heavier skater, m = mass of the lighter skater, v = Velocity of heavier skater, V = velocity of the lighter skater.
But,
v = r/t........................ Equation 2
V = r/t'................ Equation 3
Where r = radius of the circular rink, t = time taken for the heavier skater to reach the edge, t' = time taken for the lighter skater to reach the edge.
Substitute equation 2 and equation 3 into equation 1
M(r/t) = m(r/t')............. Equation 4
Given: M = 65 kg, m = 40 kg, r = 50/2 = 25 m, t = 10 s.
Substitute into equation 4 and solve for t'
65(25/10) = 40(25/t')
162.5 = 1000/t'
t' = 1000/162.5
t' = 6.15 seconds
Samir wants to be a doctor like both of his parents when he gets older he never thought of doing anything else which of the following best explains Samir decision ?
Why is this true? Why is this true? Photons of lower-frequency light don't have enough energy to eject an electron. The electric field of low-frequency light does not vibrate the electrons rapidly enough to eject them. The number of photons in low-frequency light is too small to eject electrons. Low-frequency light does not penetrate far enough into the metal to eject electrons. Request Answer
Answer:
the only correct answer is: Photons of lower-frequency light don't have enough energy to eject an electron
Explanation:
This question is in the model of the photoelectric effect, where some electrons are expelled from the metal by the action of a ray of light.
This effect was explained by Einstein supposes that the light rays are formed by photons and the energy of these photons is given by the Planco relation
K = h f - Ф
where K is the kinetic energy of the ejected electrons and Ф is the work function, it keeps the electrons inside the material.
When analyzing this expression there is a minimum frequency (threshold) for which K = 0
hf = Ф
Below this frequency the photons in the light beam do not have the energy to expel the electrons from the material.
Let's examine the answers
a) True. You agree with the above
b) False. The analysis is in terms of individual shock
c) False. The expulsion does not have to do with the number of photons but with the energy of each one
therefore the only correct answer is: Photons of lower-frequency light don't have enough energy to eject an electron
Does time stand still in space
Answer:
[tex]\underline{\textsf{\textbf{Answer -}}}[/tex]
No & Yes - the case depends actually.
You can't make the velocity zero and expect to stay in space. you will need to constantly fire the engine to hold youself in position (this is possible, until you run out of fuel).
Yes, it is possible to fire the engine and lift off earth into space, and then fire in the opposite direction to earths movement until you are still, relative to the sun. But if you now turn the engine off, the suns gravity will pull you down and directly into the sun.
its the earths forward motion that prevents it falling into the sun
Answer:
No & Yes - the case depends actually.
You can't make the velocity zero and expect to stay in space. you will need to constantly fire the engine to hold youself in position (this is possible, until you run out of fuel).
Yes, it is possible to fire the engine and lift off earth into space, and then fire in the opposite direction to earths movement until you are still, relative to the sun. But if you now turn the engine off, the suns gravity will pull you down and directly into the sun.
its the earths forward motion that prevents it falling into the sun
41. 2072 Set E Q.No. 11 A source of sound produces a note of
512 Hz in air at 17°C with wavelength 66.5 cm. Find the ratio
of molar heat capacities at constant pressure to constant
volume at NTP. Densities of air and mercury at NTP are
1.293 kg/m3 and 13600 kg/m3 respectively.
Ans: 1.36
A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20 gg bullets at 970 m/sm/s. The mass of the hunter (including his gun) is 69.5 kgkg, and the hunter holds tight to the gun after firing it. You may want to review (Page) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Collision along a straight line. Part A Find the recoil speed of the hunter if he fires the rifle horizontally.
Answer:
[tex]0.059\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of bullet = 4.2 g
[tex]u_1[/tex] = Initial velocity of bullet = 0
[tex]m_2[/tex] = Mass of hunter with rifle = 69.5 kg
[tex]u_2[/tex] = Initial velocity of hunter with rifle
[tex]v_1[/tex] = Final velocity of the bullet = 970 m/s
[tex]v_2[/tex] = Final velocity of the hunter with the rifle
As the momentum of the system is conserved we have
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{0-0-0.0042\times 970}{69.5}\\\Rightarrow v_2=-0.059\ \text{m/s}[/tex]
The recoil speed of the hunter is [tex]0.059\ \text{m/s}[/tex] in the opposite direction of the bullet.
1. Swordfish are capable of stunning output power for short bursts. A 650 kg swordfish has a cross-sectional area of 0.92 m2 and a drag coefficient of 0.0091- very low due to some evolutionary adaptations. Such a fish can sustain a speed of 30 m/s for a few seconds. Assume seawater has a density of 1026 kg/m3. a) How much power does the fish need to put out for motion at this high speed
Answer:
[tex]P_{sp}=178.4W/kg[/tex]
Explanation:
From the question we are told that:
Mass of fish [tex]m_f=650kg[/tex]
Cross-sectional area [tex]A=0.92 m^2[/tex]
Drag coefficient of [tex]\mu= 0.0091[/tex]
Seawater density [tex]\rho= 1026 kg/m^3.[/tex]
Speed of Fish [tex]v=30 m/s[/tex]
Generally the equation for Drag force F_d is mathematically given by
[tex]F_d = \mu * \rho *A v^2 /2[/tex]
[tex]F_d = 0.0091* 0.92* 1026* 30^2/2 \\F_d= 3865. 35 N[/tex]
Generally the equation for high speed Power [tex]P_{sp}[/tex] is mathematically given by
[tex]P_{sp}=3865*35*\frac{v}{m_f}[/tex]
[tex]P_{sp}=F_d*35*\frac{30}{650}[/tex]
[tex]P_{sp}=178.4W/kg[/tex]
The 180-mm disk is at rest when it is placed in contact with a belt moving at a constant speed. Neglecting the weight of the link AB and knowing that the coefficient of kinetic friction between the disk and the belt is 0.54, determine the angular acceleration of the disk while slipping occurs.
Answer:
[tex]\mathbf{\alpha = 25.88 \ rad/s^2 }[/tex]
Explanation:
Consider the force due to friction:
[tex]F = \mu_kN -------- (1)[/tex]
where;
N = normal reaction
[tex]\mu_k[/tex] = coefficient of kinetic friction.
Via the horizontal direction, the forces of equilibrium are:
[tex]\sum F_x = \sum(F_x) _{eff} \\ \\ N- F_{AB} \ cos \theta = 0 \\ \\ F_{AB} \ cos \theta = N ------ (2)[/tex]
where;
[tex]\theta[/tex] = angle of AB is associated with the horizontal;
[tex]F_{AB}[/tex] = force exerted by AB
Let's take a look at the equilibrium of forces along the vertical direction.
[tex]\sum F_y = \sum (F_y)_{eff} \\ \\ F+F_{AB} sin \theta - W = 0\\ \\ \mu_k N + F_{AB} sin \theta -mg =0 \\ \\ F_{AB} sin \theta = mg - \mu_kN-------(3)[/tex]
By dividing (3) by (2), we get:
[tex]\dfrac{F_{AB} sin \theta }{F_{AB} cos \theta} = \dfrac{mg - \mu_k N}{N} \\ \\ tan \theta = \dfrac{mg-\mu_kN}{N} \\ \\ Ntan \theta = mg - \mu_kN \\ \\ N(tan \theta + \mu_k ) = mg \\ \\ N = \dfrac{mg}{tan \theta + \mu_k}[/tex]
By replacing the obtained value of N into:
[tex]F = \mu_k N[/tex]
we have:
[tex]F = \mu_k (\dfrac{mg}{tan \theta + \mu_k}) \\ \\ \\ F = \dfrac{mg \mu_k}{tan \theta + \mu_k}[/tex]
Moment about center A can be expressed as:
[tex]\sum M_A = \sum (M_A) _{eff} \\ \\ Fr = I \alpha\\ \\ \alpha = \dfrac{Fr}{I}[/tex]
where;
[tex]\alpha[/tex] = angular acceleration of the disc
I = mass moment of inertia
r = radius of disc
By replacing F and I; ∝ becomes:
[tex]\alpha = \dfrac{\bigg( \dfrac{mg \mu_k}{tan \theta + \mu_k} \bigg) r}{\dfrac{1}{2}mr^2}[/tex]
[tex]\alpha = \dfrac{2g}{r} \dfrac{\mu_k}{tan \ \theta + \mu_k}------- (4)[/tex]
where;
g = 9.8 m/s²
r = 0.18 m
[tex]\mu_k = 0.54[/tex]
θ = 60°
[tex]\alpha = \dfrac{2(9.8)}{0.18} (\dfrac{0.54}{tan \ 60+ 0.54})[/tex]
[tex]\mathbf{\alpha = 25.88 \ rad/s^2 }[/tex]
what is a black hole what does it do
Answer:
What's a black hole?=> Black holes are points in space that are so dense they create deep gravity sinks. Beyond a certain region, not even light can escape the powerful tug of a black hole's gravity.
What does it do?=> Anything that ventures too close—be it star, planet, or spacecraft—will be stretched and compressed like putty in a theoretical process aptly known as spaghettification.
[tex] \infty \infty [/tex]
3. An airplane is flying at 10 km altitude in the standard atmosphere. The internal pressure of the aircraft interior is 100 kPa. Estimate the outward force on the window. The window is flat and has an elliptical shape with lengths of 300 mm along the major axis and 200 mm along the minor axis.
Answer:
The correct response will be "13.755 kN".
Explanation:
According to the question,
The given values are:
a = 300 mm
i.e.,
= 0.3 cm
b = 200 mm
i.e.,
= 0.2 dm
Internal pressure,
[tex]P_{in}=100[/tex]
Now,
The area of the elliptical shape window will be:
⇒ [tex]A = \pi ab[/tex]
On substituting the values, we get
⇒ [tex]=3.14\times 0.300\times 0.200[/tex]
⇒ [tex]=0.1885 \ m^2[/tex]
By using the table,
At 10 km, the atmospheric pressure will be
⇒ [tex]p_o=27.03 \ kPa[/tex]
Now,
The outward force will be:
⇒ [tex]F_{net}=p_{in}A-p_{0}A[/tex]
⇒ [tex]=(p_{in}-p_{0})A[/tex]
⇒ [tex]=(100-27.03)\times (0.1885)[/tex]
⇒ [tex]=13.755 \ kN[/tex]
What do microwaves have In common with light waves?
Answer:
Both are electromagnetic waves
Explanation:
Hope this helped!!
Which has the least gravitational force?
Answer: Mount Nevado Huascarán
Explanation: Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2, while the highest is at the surface of the Arctic Ocean, at 9.8337 m/s2
HELP PLEASEEEEEEEEEEE
Answer:
dolphin and wolf
Explanation:
the sharks jaw is way to small for it and is shaped differently
A molecule with a charge of 9 x 10-8 C is moving with a velocity of 7 x 107 m/s perpendicular to a magnetic field with a value of 0.57 Tesla. What is the magnitude force (in N) on the molecule?
Answer:
3.591 N
Explanation:
Applying,
F = Bvq................. Equation 1
Where F = magnitude of force on the molecule, B = Magnetic field, v = Velocity of the molecule, q = charge of the molecule.
From the question,
Given: B = 0.57 Tesla, q = 9×10⁻⁸ C, v = 7×10⁷ m/s
Substitute these values into equation 1
F = 0.57×9×10⁻⁸×7×10⁷
F = 35.91×10⁻¹
F = 3.591 N
Hence the force on the molecule is 3.591 N
A container in the shape of a cube 11.6 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pressure and temperature 291 K. (a) Find the mass of the gas. kg (b) Find the gravitational force exerted on it. mN (c) Find the force it exerts on each face of the cube. kN (d) Why does such a small sample exert such a great force
Answer:
a. 0.00189 kg
b. 18.552 mN
c. 1.363 kN
d. Since the molecular density is high, the force exerted by the sample is thus high.
Explanation:
(a) Find the mass of the gas. kg
Using PV = mRT/M where P = pressure on gas = atmospheric pressure = 1.013 × 10⁵ Pa, V = volume of gas = L³ where L = length of cube = 11.6 cm = 0.116 cm,m = mas of gas, R = molar gas constant = 8.314 J/mol-K, T = temperature of gas = 291 K and M = molar mass of gas = 28.9 g/mol
So, m = PVM/RT = PL³M/RT
Substituting the values of the variables into the equation, we have
m = PL³M/RT
= 1.013 × 10⁵ Pa × (0.116)³ × 28.9 g/mol/ 8.314 J/mol-K × 291 K
= 0.0457 × 10⁵ Pa g/mol/2419.374J/mol
= 1.89 × 10⁻⁵ × 10⁵ g
= 1.89 g
= 1.89 × 10⁻³kg
= 0.00189 kg
(b) Find the gravitational force exerted on it. mN
The gravitational force, F exerted on it is its weight W
So, F = W = mg where m = mass of gas = 1.89 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²
F = mg
= 1.89 × 10⁻³ kg × 9.8 m/s²
= 18.522 × 10⁻³ kgm/s²
= 18.552 × 10⁻³ N
= 18.552 mN
(c) Find the force it exerts on each face of the cube. kN
Since pressure, P = F/A where F = force exerted on each face and A = area of each face = L² where L = length of side of cube = 11.6 cm = 0.116 m
So, F = PA since P = atmospheric pressure = 1.013 × 10⁵ Pa,
F = PL²
= 1.013 × 10⁵ Pa (0.116 m)²
= 0.01363 × 10⁵ N
= 1.363 × 10³ N
= 1.363 kN
(d) Why does such a small sample exert such a great force
To answer this question, we need to find the density of the gas in the cube.
So density of gas,ρ = m/V where m = mass of gas = 1.89 g and V = volume of gas = L³ and L = length of side of cube = 11.6 cm
ρ = m/V = m/L³ = 1.89 g/(11.6 cm)³ = 1.89 g/1560.896 cm³ = 0.00121 g/cm³
We now find the number of moles of gas in a cm³ by dividing its density by its molar mass.
So n = ρ/M = 0.00121 g/cm³ ÷ 28.9 g/mol = 23687.67 mol/cm³
Since there are 6.022 × 10²³/mol, we find the number of molecules in a cm³ which is n × 6.022 × 10²³/mol = 23687.67 mol/cm³ × 6.022 × 10²³/mol
= 143731.1 × 10²³ molecules/cm³
= 1.437311 × 10²⁸ molecules/cm³
≅ 1.44 × 10²⁸ molecules/cm³
Since the molecular density is high, the force exerted by the sample is thus high.
Why is no image formed when an object is at the focal point of a converging lens?
Answer:
the refracted rays neither converge nor diverge. After refracting, the light rays are traveling parallel to each other and cannot produce an image.
Explanation:
Which type of force is needed to lift the weight?
A friction
B gravity
C pull
D push
b.gravity have a great day:)
Why does an energy transfer not always result in phase change?
Answer:
Explanation:
The energy transfer are not changing in kinetic energy. They are changes in bonding energy between the molecules.
Lithium was one of the metals studied by the American physicist Robert Millikan in his research on the photoelectric effect. When illuminated with blue light of frequency 6.64 x 10" Hz, the photoelectrons ejected from a lithium surface have a maximum kinetic energy of 0.332 eV. What is the threshold frequency for lithium? For this problem, let the value of Planck's constant, h, be 6.63 x 10J's
Answer:
f = 7.9487 10¹³ Hz
Explanation:
The photoelectric effect was correctly explained by Einstein assuming that the radiation is composed of photons, which behave like particles.
hf = K + Ф
It indicates the frequency and the kinetic energy, let's look for the work function
Ф = hf - K
let's reduce the magnitudes to the SI system
K = 0.332 eV (1.6 10⁻¹⁹ J / 1 eV) = 0.5312 10⁻⁻¹⁹ J
let's calculate
Ф = 6.63 10⁻⁻³⁴ 6.64 10¹¹ - 0.5312 10⁻¹⁹
Ф = 4.40 10⁻²² - 0.5312 10⁻¹⁹
Ф = 5.27 10⁻²⁰ J
for the minimum frequency that produces photoelectrons, the kinetic energy is zero
hf = Ф
f = Ф / h
f = 5.27 10⁻²⁰ / 6.63 10⁻³⁴
f = 7.9487 10¹³ Hz
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per minute. At point 2 in the pipe, the gauge pressure is 152kPa and the cross-sectional area is 8.00cm2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00cm2. Part A Find the mass flow rate. M
Answer:
[tex]1.302\ \text{kg/s}[/tex]
Explanation:
[tex]\rho[/tex] = Density of water = [tex]1\ \text{kg/L}[/tex]
[tex]dV[/tex] = Change in volume = [tex]220\times 0.355\ \text{L}[/tex]
[tex]dt[/tex] = Time elapsed = 1 minute = 60 seconds
Mass flow rate is given by
[tex]\dot{m}=\rho\dfrac{dV}{dt}\\ =1\times \dfrac{220\times 0.355}{60}\\ =1.302\ \text{kg/s}[/tex]
The mass flow rate is [tex]1.302\ \text{kg/s}[/tex].
How do scientists know the continents were once closer than they are today? Explain your answer in 2-3 complete sentences.
Answer:
Continental drift describes one of the earliest ways geologists thought continents moved over time. Today, the theory of continental drift has been replaced by the science of plate tectonics.
The theory of continental drift is most associated with the scientist Alfred Wegener. In the early 20th century, Wegener published a paper explaining his theory that the continental landmasses were “drifting” across the Earth, sometimes plowing through oceans and into each other. He called this movement continental drift.
