Answer:
Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2
x1 = [-m / (m + M)] * L / 2 is the original position of the CM
x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right
[-m x - m L / 2 + m x - M x] / (M + m) * L/2
= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm
Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x
i.Name two commonly used thermometric liquids.
ii.State two advantages each of the thermometric liquids mentioned above
Answer:
mercury and alcohol
ii) used to test temperatures
i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.
ii) It does not wet (cling to the sides of) the tube.
Alcohol:
i) Alcohol has greater value of temperature coefficient of expansion than mercury.
ii) it's freezing point is below –100°C.
4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable
and pulley. Does the 1 kg block move or down the slope? What is the net force on it and
its acceleration? (8 pts)
If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• m = 1 kg:
∑ F (parallel) = mg sin(45°) - T = ma … … … [1]
∑ F (perpendicular) = n - mg cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• m = 0.4 kg:
∑ F (vertical) = T - mg = ma … … … [2]
Adding equations [1] and [2] eliminates T, so that
((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a
(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a
==> a ≈ 2.15 m/s²
The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
Convierta 8.5mW a cal/h (1 cal=4.186 j)
Answer:
[tex] = { \bf{2.03 \times {10}^{ - 6} }}[/tex]
The mass of the sun is 2*10^30 kg and its radius is
6.96*10^8 m. what is the weight of 1kg mass on the
surface of the sun.
Explanation:
Distance d=1.5×108 km=1.5×1011 m
Mass of the sun, m=2×1030 kg
Mass of the earth, M=6×1024 kg
Force of gravitation, F=G×d2m×M
F=6.7×10−11×(1.5×1011)22×1030×6×1024=3.57×1022 N
A child is outside his home playing with a metal hoop and stick. He uses the stick to keep the hoop of radius 45.0 cm rotating along the road surface. At one point the hoop coasts downhill and picks up speed. (a) If the hoop starts from rest at the top of the hill and reaches a linear speed of 6.35 m/s in 11.0 s, what is the angular acceleration, in rad/s2, of the hoop? rad/s2 (b) If the radius of the hoop were smaller, how would this affect the angular acceleration of the hoop? i. The angular acceleration would decrease. ii. The angular acceleration would increase. iii. There would be no change to the angular acceleration.
Answer:
a) [tex] \alpha = 1.28 rad/s^{2} [/tex]
b) Option ii. The angular acceleration would increase
Explanation:
a) The angular acceleration is given by:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular speed = v/r
v: is the tangential speed = 6.35 m/s
r: is the radius = 45.0 cm = 0.45 m
[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)
t: is the time = 11.0 s
α: is the angular acceleration
Hence, the angular acceleration is:
[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]
b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).
Therefore, the correct option is ii. The angular acceleration would increase.
I hope it helps you!
Why is the force of attraction between the Earth and ourselves so huge compared to the attraction between two apples?
Answer:
Answer in explanation
Explanation:
The force of attraction between two bodies is governed by Newton's Law of Gravitation:
[tex]F = \frac{Gm_1m_2}{r^2}[/tex]
where,
G = Universal Gravitational Constant
m₁ = mass of the first body
m₂ = mass of the second body
r = distance between the two bodies
F = Force
Hence, it is clear from the formula that the magnitude of the force is directly proportional to the product of the masses of the objects. So in the case of the earth and ourselves, the mass of the earth is very large in order of 10²⁴ kg. Due to this huge mass, the attraction between the earth and ourselves is so huge as compared to the attraction between two apples. Because the masses of the apple are very small in grams.
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?
Answer:
1. 588 N
2. 738 N
3. 588 N
Explanation:
time, t = 4 s
initial velocity, u = 0
final velocity, v = 10 m/s
mass, m= 60 kg
1.
Weight of passenger before starts
W =m g = 60 x 9.8 = 588 N
2.
When the elevator is speeding up
v = u + a t
10 = 0 + a x 4
a = 2.5 m/s2
Now the weight is
W' = m (a + g) = 60 (9.8 + 2.5) = 738 N
3.
When he reaches the cruising speed, the weight is
W = 588 N
Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm.
Answer:
Answer
Correct option is
A
5×10
−6
tesla
I=5A
x=0.2m
Magnetic field at a distance 0.2 m away from the wire.
B=
2πx
μ
0
I
=
2π×0.2
4π×10
−7
×5
=10×5×10
−7
=5×10
−6
tesla
Two guitar strings, of equal length and linear density, are tuned such that the second harmonic of the first string has the same frequency as the third harmonic of the second string. The tension of the first string is 510 N. Calculate the tension of the second string.
Answer:
The tension in the second string is 226.7 N.
Explanation:
Length is L, mass per unit length = m
T = 510 N
Let the tension in the second string is T'.
second harmonic of the first string = third harmonic of the second string
[tex]2 f = 3 f'\\\\2\sqrt{\frac{T}{m}} = 3 \sqrt {\frac{T'}{m}}\\\\4 T = 9 T'\\\\4\times 510 = 9 T'\\\\T' = 226.7 N[/tex]
What is the escape speed on a spherical asteroid whose radius is 517 km and whose gravitational acceleration at the surface is 0.636 m/s2
Answer:
810.94 m/s
Explanation:
Applying,
v = √(2gR)............. Equation 1
Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth
From the question,
Given: g = 0.636 m/s², R = 517 km = 517000 m
Substitute these values into equation 1
v = √(2×0.636×517000)
v = √(657624)
v = 810.94 m/s
Hence, the escape velocity is 810.94 m/s
Think about a thermos bottle. It consists of an inner bottle with a shiny silver surface separated from an outer container by a space with no air. In what ways does it block conduction, convection, and radiation?
Answer:
Radiation
Explanation:
Conduction, convection and radiation are the three modes of heat transfer.
1. Conduction: When the one end is heated of any rod, the heat transfer to the other end by the vibrational motion of the molecules, it is called conduction.
The heat transfer in a solid is due to the conduction.
2. Convection: When the liquid or gas is heated, the molecules which is in contact to the heat, heated first and due to the decrease in density they moves up and the molecules on the upper side are higher in density so they moves down. These are called convection currents. The process continues till the entire liquid becomes heated. It generally takes place in liquids and gases.
3. Radiation: The process of heat transfer in which no molecules takes place is called radiation. The heat coming from sun is due to the radiation. It does not require any medium.
In the thermos bottle, as there is no air between the two layers, so the heat transfer is due to the radiation.
A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction between the crate and the truck bed is 0.645. What is the minimum stopping time for the truck in order to prevent the crate from sliding?
By Newton's second law,
• the net force acting vertically on the crate is 0, and
∑ F = n - mg = 0 ==> n = mg = 1470 N
where n is the magnitude of the normal force; and
• the net force acting in the horizontal direction on the crate is also 0, with
∑ F = f - b = 0 ==> b = f = µn = 0.645 (1470 N) = 948.15 N
where b is the magnitude of the braking force, f is (the maximum) static friction, and µ is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.
Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.
Let a be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for a :
f = ma ==> a = (948.15 N) / (150.0 kg) = 6.321 m/s²
With this acceleration, the truck will come to a stop after time t such that
0 = 50.0 km/h - (6.321 m/s²) t ==> t ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s
and this is the smallest stopping time possible.
how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?
Assuming no heat lost to the surrounding,
-5⁰C ice → 0⁰C ice
Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(2.0 x 10³) x (0-(-5))
Q = 50000J
0⁰C ice → 0⁰C water
Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg
Q = mLf
Q = 5(3.34 x 10⁵)
Q = 1670000J
0⁰C water → 100⁰C water
Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(4.2 x 10³) x (100-0)
Q = 2100000J
100⁰C water → 100⁰C steam
Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Q = mLv
Q = 5(2.26 x 10⁶)
Q = 11300000J
Total amount of heat required
= 50000 + 1670000 + 2100000 + 11300000
= 15120000J
There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their
A temperatures.
B volumes.
C densities.
D masses.
Answer: masses
Explanation:
Trust me
After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off what
Answer:
Scattering is an interaction that can happen when a given particle or wave, like an electron, impacts a target or material. Then the electron changes it's original path and leaves some energy in the process. (This is a really simplified explanation of scattering, this is a really complex phenomenon, but let's not dive into that path)
Particularly, Davisson and Germer used a beam of electrons against a target of nickel, and these scattered electrons were detected by a detector. All of that in a vacuum chamber.
Then the correct answer is a nickel target.
"After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off a nickel target"
An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
Answer:
T = 13.25°C
Explanation:
From the law of conservation of energy:
Heat Lost by Watermelon = Heat Gained by Cans
[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]
where,
[tex]m_w[/tex] = mass of watermelon = 6.48 kg
[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg
[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C
[tex]C_c[/tex] = specific heat capacity of cans = 4200 J/kg.°C
[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T
[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C
T = final temperature = ?
Therefore,
[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]
T = 13.25°C
For waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths.
a) true
b) false
Answer:
false.
Explanation:
We know that for a wave that moves with velocity V, with a wavelength λ, and a frequency f, we have the relation:
V = λ*f
So, if the velocity is constant and we increase the frequency to:
f' > f
we will have a new wavelength λ'
Such that:
V = f'*λ'
And V = f*λ
Then we have:
f'*λ' = f*λ
Solvinf for λ', we get:
λ' =(f/f')*λ
And because:
f' > f
then:
(f/f') < 1
Then:
λ' =(f/f')*λ < λ
So, if we increase the frequency, we need to decrease the wavelength.
So, for higher frequency waves, we must have proportionally shorter wavelengths.
Then we can conclude that the given statement:
"or waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths"
is false.
A heat engine exhausts 3 000 J of heat while performing 1 500 J of useful work. What is the efficiency of the engine
efficiency=work output/work input×100
since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input
efficiency=1500/3000×100
=50%
boat carrying people more than its capacity is attributes of sinking why
Answer:
Upthrust on boat becomes lesser than Weight of boat
Explanation:
When there are more people than the capacity, The weight of the boat acting downwards increases. However, the upthrust acting on the submerged part of the boat is constant. Since Weight > Upthrust, there is a net force downwards, leading to sinking.
You simultaneously release two balls: one you throw horizontally, and the other you drop straight down. Which one will reach the ground first? Why?
(a) The ball dropped straight down lands first, since it travels a shorter distance.
(b) Neither. Their vertical motion is the same, so they will reach the ground at the same time.
(c) It depends on the mass of the balls—the heavier ball falls faster so lands first
Answer:
Option B.
Explanation:
Remember that we can think on any movement as a sum of a movement in the y-axis, the movement in the x-axis, and the movement in the z-axis. And these are not related, this means that, for example, the movement in x does not affect the movement in y.
So, when we analyze the problem of "how long takes an object to hit the ground"
We do not care for the horizontal motion of the object, we only care for the vertical motion of the object.
So, if an object is dropped, and another has a given initial velocity in the x-axis, in both cases the initial velocity in the y-axis will zero.
And in both cases, the only vertical force acting on the balls will be the gravitational force (so both objects will have the same vertical acceleration and the same vertical initial velocity) with this, we already know that the vertical motion of both objects will be exactly the same.
So, both objects will hit the ground at the same time.
(notice that here we are ignoring things like air resistance and other complex forces)
So here the correct option is b: Neither. Their vertical motion is the same, so they will reach the ground at the same time.
A 5 kg object is moving in a straight-line with an initial speed of v m/s. It takes 13 s for the speed of the object to increase to 13 m/s and it kinetic energy increases at a rate of 15 J/s. What is the initial speed v (in m/s)?
The object's kinetic energy changes according to
dK/dt = 15 J/s
If v is the object's initial speed, then its initial kinetic energy is
K (0) = 1/2 (5 kg) v ²
Use the fundamental theorem of calculus to solve for K as a function of time t :
[tex]K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t[/tex]
After t = 13 s, the object's kinetic energy is
K (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J
Put this as the left side in the equation above for K(t) and solve for v :
[tex]422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)[/tex]
==> v ≈ 9.5 m/s
Diffuse reflection occurs when parallel light waves strike which surface? a mirror a rippling fountain a polished silver plate a still pond
Answer: a rippling fountain
Explanation: diffuse reflection happens on rough surfaces, so using the process of elimination, that leaves us with b, a rippling fountain (I also just took this test I'm pretty sure I'm right)
A car is moving with a velocity of45m/sis brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before it comes to rest.
Explanation:
Given,
v = 45m/s
t = 5s
Therefore,
d = v × t
= 45 × 5
= 225m
The table below describes some features of methods used to generate electricity. Name method 4.
Answer:
Hydroelectricity
Explanation:
Because of flooding of water, we can assume that the electricity was generated by Water which is known as Hydroelectricity.
We can presume that the energy was produced by water because of the flooding of the water, which is a process known as hydroelectricity.
What is hydroelectricity?Hydroelectric power, often known as hydropower, is the name given to electricity generated by turbines that turn the potential energy of falling or swiftly running water into mechanical energy. As of 2019, hydropower accounted for more than 18% of the world's total power generation capacity, giving it the most frequently used renewable power source in the early 21st century.
When water is used to produce energy, it is first gathered or stored at a higher altitude and then transported through extensive pipelines or tunnels (called pen stocks) to a lower level; the difference between these two altitudes is referred to as the head. The falling water triggers the rotation of turbines at the bottom of its descent through the pipes.
To know more about Hydroelectricity:
https://brainly.com/question/15228003
#SPJ2
Riley, a student, notices that the protractor tool does not measure the angle just as the ball leaves the surface. She sees that the ball must travel some distance before it crosses the protractor, so the direction of travel may have changed as the ball moves upwards. She says that this is the cause of the discrepancy between her predicted angle and the measured angle. Does this reasoning explain the discrepancy between your predicted angle and your measured angle. Use evidence to support your claim.
Answer:
Riley's reasoning is correct
Explanation:
Her reasoning is correct because as the ball moves upwards, the acceleration due to gravity would be vertical and in downward position. Therefore at all points as the ball moves, the velocity of the ball is going to change in magnitude as well as in direction. given that the direction keeps changing at certain points, the angle made by the initial velocity just as the ball left the surface would also have to continuously change.
If Riley has to wait for this ball to move some inches before she uses the protractor to measure the angle, the angle of travel would have to change.
Therefore there is going to be discrepancies between the measured angle and the predicted angle. The predicted is the angle of velocity with the horizontal just as this ball moves from the surface.
what is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s
Answer:
8m/s
Explanation:
Vavg= 16-0/2=8m/s
Write submultipels units of time
What
is the the improvetanse of water in our body and how do
Or
open Meet and enter this code
Answer:
Or open Meet and enter this code to the source of definition and how to use it to make
7. If a load of 300N is pulled along the inclined plane shown in the figure, answer the following. B 200 N 0.5m 2m 300 N А i. Calculate the VR and MA of the inclined plane. Calculate the input work and output work. ii.what efficiency of inclined plane?iv.what should be the length of inclined plane if same load has to be pulled with a50N effort for the same efficiency as above
Explanation:
700n I think friend .. if worng
The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-3 m3 of benzene [β = 1240 × 10-6 (C°)-1]. Both the copper and the benzene have the same temperature. By what amount ΔT should the temperature of the sphere and the benzene within it be increased, so that the liquid just begins to spill out?
Answer:
The answer is "[tex]60.74^{\circ}[/tex]".
Explanation:
Cavity and benzene should be extended in equal quantities.
[tex]\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\[/tex]
[tex]\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\[/tex]
[tex]\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\[/tex]
Suppose a van de Graaff generator builds a negative static charge, and a grounded conductor is placed near enough to it so that a 7.0 mu C of negative charge arcs to the conductor. Calculate the number of electrons that are transferred.
Answer:
# _electron = 4.375 10¹³ electrons
Explanation:
In this exercise it is indicated that 7.1 μC is transferred, let's use a direct ratio or rule of three. If an electron has a charge of 1.6 10⁻¹⁹ C, how many electrons have a charge of 7.0 10⁻⁶ C
# _electron = 7.0 10⁻⁶ C (1 electron / 1.6 10⁻¹⁹ C)
# _electron = 4.375 10¹³ electrons