[tex]sp^3[/tex] hybridization and 1 lone pair is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen.
b. [tex]sp^3[/tex] hybridization and 1 lone pair
Explanation:
The Nitrogen particle is [tex]sp^3[/tex] hybridized with one crossover orbital involved by the solitary pair. Likewise, nitrogen is [tex]sp^3[/tex] hybridized which implies that it has four [tex]sp^3[/tex] half and half orbitals. The sub-atomic structure of water is predictable with a tetrahedral game plan of two solitary sets and two holding sets of electrons. Two of the [tex]sp^3[/tex] hybridized orbitals cover with s orbitals from hydrogens to frame the two N-H sigma bonds.
Nitrogen utilizes [tex]sp^3[/tex] orbitals to accomplish this geometry. Three of the mixtures are utilized to frame bonds to hydrogen and the fourth contains the solitary pair. Whereas lone pairs are the pairs of electron on atoms that don't take an interest in the holding bonding of two atoms. To distinguish solitary matches in a particle, make sense of the number of valence electrons of the molecule and subtract the number of electrons that have partaken in the holding.
Hybridization of nitrogen depends on its bonding context; sp hybridization leads to a linear geometry, whereas sp³ hybridization results in a trigonal pyramidal geometry with one lone pair, as seen in ammonia.
Explanation:The question is asking about the hybridization of nitrogen in various substances. Hybridization describes the mixing of atomic orbitals to form new hybrid orbitals that can accommodate bonding and lone pairs in molecules. For a nitrogen atom with sp hybridization, the molecule usually has a linear geometry, as in the case of hydrogen cyanide (HCN). Here, nitrogen has one sp hybrid orbital with a lone pair and one with a sigma bond, while the two p orbitals form pi bonds, resulting in a triple bond.
In cases where nitrogen is sp³ hybridized, the nitrogen atom can form three sigma bonds with its sp³ hybrid orbitals and retain one lone pair, giving a trigonal pyramidal geometry. An example of this is in ammonia (NH₃), where the nitrogen atom is bonded to three hydrogen atoms and has one lone pair.
Suzie is 2 meters away from the center of a see saw and she weighs 400 N. Where on the other side would Timmy, who weighs 800 N, have to sit on the see saw for it to be balanced?
Answer:
Timmy is 1 meters away from the center.
Explanation:
Given:
Distance of Suzie from center = 2 m
Weight of Suzie = 400 N
Weight of Timmy = 800 N
The see saw is balanced
To find the distance of Timmy from the center.
Solution:
Since the see saw is balanced,
So, the sum of clockwise moment = Sum of anticlockwise moments
Moment is given by:
⇒ [tex]Force\times Perpendicular\ distance\ from\ the\ center[/tex]
Let Suzies have moment in clockwise direction.
Moment of Suzie would be = [tex]400\ N \times 2\ m= 800\ Nm[/tex]
Timmy would have moment in anticlockwise direction.
Let Timmy be [tex]l[/tex] meters away from center
Moment of Timmy would be = [tex]800\ N\times l\ m=800l\ Nm[/tex]
Since, the see saw is balanced, the system is in equilibrium.
Thus, we have,
Anticlockwise Moment = Clockwise moment
[tex]800l=800[/tex]
Dividing both sides by 800.
[tex]\frac{800l}{800}=\frac{800}{800}[/tex]
[tex]l=1\ m[/tex]
Thus, Timmy is 1 meters away from the center.
What occurs in a nuclear power plant when Uranium-235 splits into two smaller isotopes? A nuclear fusion B nuclear fission C a chemical reaction D a neutralization reaction
B) Nuclear fission
Explanation:
Nuclear power plants work by using the process of nuclear fission.
Nuclear fission occurs when a heavy, unstable radioactive nuclei decays, breaking apart into two or more lighter nuclei, more stable. In the process, several neutrons are also released, alongside with energy.
In nuclear power plants, the nucleus used for the process is the Uranium-235. When an atom of uranium-235 absorbs a slow neutron, it becomes a very unstable nucleus of uranium-236, which quickly decays into a nucleus of Barium-141, Kripton-92 and 3 neutrons.
The uranium nuclei are located in the so-called fuel rods, which are placed in a moderator (usually water). The purpose of the moderator is to slow down the neutrons emitted in the reaction: this way, these neutrons can be absorbed by other nuclei of uranium-235, causing more fission reactions to occur.
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Stars produce heat by the process of 1. combustion. 2. implosion. 3. nuclear fission. 4. nuclear fusion.
Answer:
4. nuclear fusion.
Explanation:
The process that helps stars generate heat energy from atomic nuclei is called nuclear fusion. Due to gravity, pressure on the hydrogen gas in the center or core of the stars are enormous causing temperature to reach about 27 million°F (15 million°C). This hotness makes hydrogen atoms to fuse together to form helium atoms. This process is called nuclear fusion. Vast amount of energy is released which enables the star to shine.
A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on the hoop to stop it
Answer:
The work done by the hoop is equal to 5.529 Joules.
Explanation:
Given that,
Mass of the hoop, m = 96 kg
The speed of the center of mass, v = 0.24 m/s
To find,
The work done by the hoop.
Solution,
The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,
[tex]K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]
I is the moment of inertia, [tex]I=mr^2[/tex]
Since, [tex]\omega=\dfrac{v}{r}[/tex]
[tex]K_i=mv^2[/tex]
[tex]K_i=96\times (0.24)^2=5.529\ J[/tex]
Finally it stops, so the final energy of the hoop will be, [tex]K_f=0[/tex]
The work done by the hoop is equal to the change in kinetic energy as :
[tex]W=K_f-K_i[/tex]
W = -5.529 Joules
So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble simultaneously at opposite ends of a diameter, and stick. W is the turntable's angular velocity, in rpm, just after this event
There are mistakes in the question.The correct question is here
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?
Answer:
w=50 rpm
Explanation:
Given data
The mass turntable M=2kg
Diameter of the turntable d=20 cm=0.2 m
Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s
Two blocks Mass m=500 g=0.5 kg
To find
Turntable angular velocity
Solution
We can find the angular velocity of the turntable as follow
Lets consider turntable to be disk shape and the blocks to be small as compared to turntable
[tex]I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}[/tex]
where I is moment of inertia
[tex]w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm[/tex]
The problem is a case of angular momentum conservation within the domain of rotational dynamics in physics. The turntable's initial angular momentum remains conserved despite the addition of the blocks. By accounting for the added moment of inertia from the blocks, the final angular velocity of the system can be calculated.
Explanation:The subject we're discussing here comes under the physics concept of rotational dynamics particularly focusing on the conservation of angular momentum.
Before the blocks hit the turntable, we know that the turntable is rotating with an angular velocity given in RPM (revolutions per minute), which we can convert to rad/s for our calculations. So, the initial angular momentum can be represented as Lim = (moment of inertia of the system) * (initial angular velocity).
Once the blocks fall onto the turntable, they contribute to the moment of inertia of the system, while the angular momentum of the system remains conserved. Thus resulting in a decreased angular velocity. The final angular momentum can be represented as Lfm = (moment of inertia including the blocks) * (final angular velocity).
Since the initial and final angular momenta need to be equal (Lfm = Lim), we can solve the resulting equation for the final angular velocity.
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A ball is kicked from a location 7, 0, −8 (on the ground) with initial velocity −11, 19, −5 m/s. The ball's speed is low enough that air resistance is negligible. (a) What is the velocity of the ball 0.4 seconds after being kicked? (Use the Momentum Principle!) v = Incorrect: Your answer is incorrect. m/s
Answer:
[tex]\vec{v} = <-11,15.08,-5>[/tex]
Explanation:
given,
location of the ball ⟨7,0,−8⟩
initial velocity of the ball ⟨-11,19,−5⟩
time = 0.4 s
speed of the ball = ?
using Momentum Principle
change in momentum = Force x time
[tex]m \vec{v} - m \vec{u}= \vec{F}\times \Delta t[/tex]
[tex]\vec{v} =\vec{u} + \dfrac{\vec{F}}{m}\times \Delta t[/tex]
Net force acting in this case will be equal to force due to gravity because air resistance is negligible.
F_net = F_g = ⟨0 ,-9.8 m , 0⟩
now,
[tex]\vec{v} = <-11,19,-5>+ \dfrac{<0 ,-9.8 m , 0>}{m}\times (0.4-0)[/tex]
[tex]\vec{v} = <-11,19,-5>+ <0 ,-3.92 , 0>[/tex]
[tex]\vec{v} = <-11,15.08,-5>[/tex]
hence, the velocity of the ball 0.4 s after being kicked is equal to [tex]\vec{v} = <-11,15.08,-5>[/tex]
The velocity of the ball 0.4 seconds after being kicked is; v = (-11, 15.08, -5) m/s
Velocity Vector
We are given;
Location of the ball = (7,0,−8)Initial velocity of the ball = (-11, 19, −5)Time = 0.4 sFrom the Momentum Principle, we know that;
Change in momentum = Impulse
Thus;
m(v - u) = F * Δt
Divide through by m and make v the subject to get;
v = u + [(F/m)Δt]
In this question, due to the fact that air resistance is negligible, the net force acting will be equal to force due to gravity. Thus;
F_net = F_g = mg
Since acceleration due to gravity is 9.8 m/s², then;
F = (0, -9.8m, 0)
Thus;
v = (-11, 19, −5) + [((0, -9.8m, 0)/m) * (0.4 - 0)]
v = (-11, 19, −5) + (0, -3.92m, 0)
v = (-11, 15.08, -5) m/s
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A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper needs to be drawn into a wire with a diameter of 9.50 mm. What length of wire, in meters, can be produced?
Answer:
120.125 m
Explanation:
Density = Mass/volume
D = m/v .............................. Equation 1.
Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.
Making v the subject of the equation,
v = m/D............................... Equation 2
Given: m = 76.5 kg,
Constant: D = 8960 kg/m .
Substituting into equation 2
v = 76.5/8960
v = 0.0085379 m³
Since the copper sphere is to be drawn into wire,
Volume of the copper sphere = volume of the wire
v = volume of the wire
Volume of wire = πd²L/4
Where d = diameter of the wire, L = length of the wire.
Note: A wire takes the shape of a cylinder.
v = πd²L/4 ........................ equation 3.
making L the subject of the equation,
L = 4v/πd²..................... Equation 4
Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095 and π = 3.14
Substitute into equation 4
L = 4×0.0085379/(3.15×0.0095²)
L = 0.0341516/0.0002843
L = 120.125 m.
L = 120.125 m
Thus the length of the wire produced = 120.125 m
For a particular scientific experiment, it is important to be completely isolated from any magnetic field, including the earth's field. The earth's field is approximately 50 μT, but at any particular location it may be a bit more or less than this. A 1.00-m-diameter current loop with 200 turns of wire is adjusted to carry a current of 0.203 A ; at this current, the coil's field at the center is exactly equal to the earth's field in magnitude but opposite in direction, so that the total field at the center of the coil is zero.
Answer:
51.019 μT
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
i = Current in wire = 0.203 A
N = Number of turns = 200
d = Diameter = 1 m
Magnetic field is given by
[tex]B=\dfrac{N\mu_0i}{d}\\\Rightarrow B=\dfrac{200\times 4\pi \times 10^{-7}\times 0.203}{1}\\\Rightarrow B=0.000051019\ T[/tex]
The strength of the magnetic field at this location would be 51.019 μT
The magnetic field at the given point is 51.019 μT.
Magnetic field:
It is given that the current in the coil is I = 0.203A
the diameter of the loop carrying the current is d = 2r = 1m
and the number of turns is N = 200.
The magnetic field to a current-carrying loop having N number of turns and carrying a current I with the radius of the loop being r is given by:
[tex]B=\frac{\mu_oNI}{2r} \\\\B=\frac{4\pi\times10^{-7}\times200\times0.203}{1}\\\\B=51.019\;\mu T[/tex]
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What soil conditions favor the use of belled caissons? What soil conditions favor piles over caissons? What type of piles are especially well suited to repair or improvement of existing foundations and why? List and explain some cost thresholds frequently encountered in foundation design.
Answer:
What soil conditions favor the use of belled caissons?
Answer:
- where the bell can be unearthed from a solid surface.
- where the supporting stratum below the bottom of the caisson is impermeable to water movement.
What soil conditions favor piles over caissons?
Answer:
- non-cosheal soils
- subterranean water or excessive depth of bearing strata make caisson unworkable
What type of piles are especially well suited to repair or improvement of existing foundations ?
Answer:
Without hammering, minipiles or helical piles are placed which escapes much of the vibration and noise associated with traditional pile installation. for working close to existing buildings or for improving the exiting foundations where excessive vibration could damage exiting structures or noise may interfere with ongoing activities these piles are good options.
Why?
Their slenderness involves little or no displacement of the soil, thus minimizing the risk of disturbance to nearby foundations.
List and explain some cost thresholds frequently encountered in foundation design.
Answer:
building below the water table- site dewatering must occur, strengthening of slopes supper system must be done and waterproofing of the foundation all of which entails money
building near existing building - this requires underpinning(The process of reinforcing the base of an existing building or other structure underpins it.)
increase in column/wall load- building height determines the foundation depth
The answer discusses the use of belled caissons, piles, and helical piles in different soil conditions, along with common cost thresholds in foundation design.
Belles caissons are favored in soil conditions with the potential for liquefaction, as driving deep piles or piers can strengthen the soil and reduce liquefaction risk. On the other hand, piles are preferred over caissons in soil conditions that are unsuitable for drilling due to hard rock or boulders. Helical piles are especially well-suited for repairing or improving existing foundations as they can be installed quickly with minimal noise and vibration, making them ideal for retrofitting projects. Some common cost thresholds in foundation design include budgets for materials, labor, and specialized equipment, all of which can impact the overall project cost.
You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.
(a) What are the two possibilities for its x-component?
(b) Assuming the x-component is known to be positive, specify the magnitude of the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the negative x-direction.
(c) Specify the direction of the vector.
Answer:
Explanation:
a)Magnitude = [tex]\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }[/tex]
84=[tex]\sqrt{(0- (-67))^{2} + (x-0)^{2} }[/tex]
x= +50.67 or -50.67 units
b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.
To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.
Magnitude = [tex]\sqrt{(0- (67))^{2} + (-130.67)^{2} }[/tex] = 146.85 units
c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - [tex]Tan^{-1}(67/130.67)degrees[/tex] i.e 152.85 degrees from the +ve x-axis.
A) The two possibilities for the x-component are; +50.67 units or -50.67 units
B) The magnitude of the vector added to the original one is; 146.85 units
C) The direction of the vector is; θ = 207.15°
A) We are given;
Magnitude of vector = 84 units
Y-component of the vector = -67 units
We know that the formula for for 2 vectors like this in x and y direction is;
A = xi^ + yj^
Where A is the magnitude of the resultant
x is the value of the x-component
y is the value of the y-component
Thus;
A = √(x² + y²)
84 = √(x² + (-67)²)
84² = x² + 4489
7056 = x² + 4489
x = ±√(7056 - 4489)
x = ±50.67 units
B) From A above, let us take the positive value of the x-component and as such our original vector will be;
A = 50.67i^ - 67j^
We want to add another vector to this that would make the resultant to be -80 units in the x direction. Thus, A = -80i and if the new additional vector is V^, then we have;
-80i^ = (50.67i^ - 67j^) + V^
V^ = -(80 + 50.67)i^ + 67j^
V = -130.67i^ + 67j^
The magnitude of vector V is;
V = √(x² + y²)
V = √(-130.67)² + 67²)
V = 146.85 units
C) The direction of the vector V is;
θ = tan^(-1) (y/x)
θ = tan^(-1) (67/-130.67)
θ = -27.15°
Since it points entirely in the negative x axis, then the angle is;
180 - (-27.15) = 207.15°
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As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed (0.516)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.413)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)
The mass (M) of the target and the final speed (V) of the bullet and target combined after an inelastic collision can be determined by setting the total momentum before the collision equal to the total momentum after the collision and solving for M and V using the given information.
Explanation:The question deals with the concept of conservation of momentum in an inelastic collision. Since the bullet and the target paper are involved in an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. This can be defined as m*v=(m+M)*V, where m and v are the mass and velocity of the bullet respectively, M is the mass of the target, and V is the final velocity of the bullet and target combined.
Given in the question is that after collision, the speed of the bullet becomes (0.516)*v, therefore the velocity V of the bullet and the target combined, the moment after the collision would be V = 0.516 * v. Solving these equations will give the required values for M and V in terms of the initial mass and velocity of the bullet.
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The question involves an inelastic collision between a bullet and a target. Using conservation of momentum, the target's mass is found to be 0.937 times the bullet's, and the target's speed immediately after collision is approximately 0.484 times the bullet's initial speed.
Explanation:Given that the collision is inelastic, the quantity of momentum before and immediately after the collision is conserved. Hence, we can express this conservation of momentum as: m*v = (m+M)V. From this we get the mass M of the target as: M = [m*(1 - 0.516)]/0.516 = 0.937m which means the mass of the target is approximately 0.937 times the mass of the bullet.
Subsequently, we can solve for the speed V of the target using the above conservation of momentum giving: V = (m*v)/(m+M) = v/(1+1.064) = 0.484v approximately.
In conclusion, the mass M of the target the instant after the collision is 0.937m and the speed V of the target is 0.484v in terms of the mass m of the bullet and initial speed v of the bullet.
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A ________ is a device that converts digital signals from a computer into analog signals so that telephone lines may be used as a transmission medium to send and receive electronic data.
Answer:
analog-to-digital
Explanation:
An analog-to-digital converter, or ADC as it is more commonly called, is a device that converts analog signals into digital signals.
A person has been exposed to a particular antigen and now experiences a repeat exposure. What stimulates a quicker immune response?a) Memory T cellsb) immunityc) antibodiesd) macrophages
Answer:
a. Memory T cells
Explanation:
Memory T cells are actually the antigen-specific T cells that remain long-term after an infection has been eliminated. These memory T cells are quickly converted into large numbers of effector T cells upon reexposure to the specific invading antigen, thus providing a rapid response to past infection that has been experienced before
The brakes on your automobile are capable of creating a deceleration of 4.9 m/s2. If you are going 149 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 100 km/h speed limit?
Answer:
You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.
Explanation:
Hi there!
The equation of velocity for an object moving in a straight line is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time.
We have to find the time at which the velocity is 100 km/h with a decceleration of 4.9 m/s² and an initial velocity of 149 km/h. Let´s first convert km/h into m/s:
149 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 41.4 m/s
100 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 27.8 m/s
Now, let´s solve the equation of velocity for the time:
v = v0 + a · t
(v - v0) / a = t
Replacing with the data:
(27.8 m/s - 41.4 m/s) / -4.9 m/s² = t
Notice that the acceleration is negative because you are slowing down.
t = 2.8 s
You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.
To get the car under the speed limit, it would take approximately 4.34 seconds.
Explanation:To calculate the minimum time it takes for the car to get under the 100 km/h speed limit, we need to find the deceleration required to slow down from 149 km/h to 100 km/h.
First, let's convert the speeds to m/s. 149 km/h is equal to 41.4 m/s and 100 km/h is equal to 27.8 m/s.
Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can rearrange the equation to solve for acceleration:
a = (v² - u²) / (2s)
Plugging in the values, we get a = (27.8² - 41.4²) / (2(-30.5)).
Solving this equation, we find that the minimum deceleration required is approximately -3.13 m/s².
Finally, we can use the formula a = Δv / t to find the minimum time:
t = Δv / a = (41.4 - 27.8) / 3.13 = 4.34 seconds.
An unknown solid with a mass of 2.00 kilograms remains in the solid state while it absorbs 32.0 kilojoules of heat. Its temperature rises 4.00 degrees Celsius. What is the specific heat of the unknown solid?
Answer: The specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]
Explanation:
As we know that,
[tex]q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex] (1)
where,
q = heat absorbed = 32.0 kJ = [tex]32.0\times 10^3J[/tex] J (1kg=1000g)
[tex]m[/tex] = mass of unknown solid= 2.00 kg = [tex]2.00\times 10^3g[/tex] (1kg=1000g)
[tex]T_{final}[/tex] = final temperature
[tex]T_{initial}[/tex] = initial temperature
[tex]\Delta T[/tex] =[tex]4.00^0C[/tex]
[tex]c[/tex] = specific heat of unknown solid = ?
Now put all the given values in equation (1), we get
[tex]32.0\times 10^3J=2.00\times 10^3g\times c\times (4.00^0C)][/tex]
[tex]c=4.00J/g^0C[/tex]
Therefore, the specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]
The scientific heat of the unknown solid will be "4.00 J/g°C".
Specific heat:Given values are:
Heat absorbed, q = 32.0 kJ or, [tex]32.0\times 10^3[/tex] J
Mass, m = 2.00 kg or, [tex]2.00\times 10^3[/tex] g
Rise in temperature, ΔT = 4.00°C
We know the relation,
→ q = m×c×ΔT
or,
→ = m×c×([tex]T_{final} - T_{initial}[/tex])
By substituting the values,
[tex]32.0\times 10^3=2.00\times 10^3\times c\times 4.00[/tex]
[tex]c = 4.00[/tex] J/g°C
Thus the above answer is appropriate.
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Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint of the distance is marked to make it easier to see how the locations compare. Assume the spaceship has the same mass throughout the trip (that is, it is not burning any fuel). Rank the five positions of the spaceship from left to right based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest.
The five positions of the spaceship from left to right are based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest is [tex]5, 1, 2, 4, 3[/tex]
Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another.
The gravitation force that Earth exerts on the spaceship will be:[tex]F_{ES}=(Gm_1m_E)/r^2[/tex]
Where [tex]F_{ES}[/tex] the force exerted on the spaceship by Earth [tex]m_1\\\\[/tex] is the mass of the spaceship and r is the distance between the.
[tex]F_{ES}\ \alpha\ 1/r^2[/tex]
This indicates larger the distance smaller will be the force. The correct order is [tex]5, 1, 2, 4, 3[/tex].
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The strength of the gravitational force that Earth exerts on a spaceship varies depending on the distance between them. The force is strongest when the spaceship is closest to Earth and weakest when it is closest to the Moon.
Explanation:Position 1: The spaceship is closest to Earth, so the gravitational force is strongest here. Position 2: The spaceship is moving away from Earth, so the gravitational force is slightly weaker than at Position 1 but stronger than at the other positions. Position 3: The spaceship is at the midpoint between Earth and the Moon, so the gravitational force is weaker than at Positions 1 and 2 but still stronger than at Positions 4 and 5. Position 4: The spaceship is closer to the Moon than to Earth, so the gravitational force from the Moon is stronger than the force from Earth. Position 5: The spaceship is closest to the Moon, so the gravitational force from the Moon is strongest here, and the force from Earth is weakest. Learn more about Gravitational force here:
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To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride, F−. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
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A ball is tossed straight up and later returns to the point from which it was launched. If the ball is subject to air resistance as well as gravity, which of the following statements is correct?
a. The net work done by gravity on the ball during its flight is greater than zero.
b. The speed at which the ball returns to the point of launch is less than its speed when it was initially launched.
c. The net work done by air resistance on the ball during its flight is zero.
d. The force of air resistance is directed down-ward both when the ball is rising and when it is falling.
The correct answer is that the speed of the ball when it returns to the point of launch is less than its speed when it was initially launched, due to the effects of air resistance.
Explanation:The subject of this question is Physics, specially mechanics that is a part of Physics dealing with motion and the forces that produce motion.
The correct answer is option b. 'The speed at which the ball returns to the point of launch is less than its speed when it was initially launched'. This is due to the presence of air resistance, which counteracts the force of gravity and slows the ball down both on its way up and on its way down. Therefore, the speed of the ball when it returns to the hand will be less than the speed at which it was launched.
Option a is incorrect because the net work done by gravity is zero since the ball returns to its original position. Option c is also incorrect because air resistance always does work on the ball as it changes its velocity. Lastly, option d is incorrect because the direction of air resistance changes; it is upward when the ball is rising and downward when the ball is falling.
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The correct answer is b. The ball's return speed is less than its launch speed due to the opposing forces of air resistance during its upward and downward motion.
Explanation:
The correct statement is b. The speed at which the ball returns to the point of launch is less than its speed when it was initially launched. This is because air resistance, a form of friction, works against the motion of the ball. When the ball is launched upward, air resistance acts downwards, slowing it down. Similarly, when the ball falls downwards, air resistance acts upwards, again slowing it down.
As a result, the return speed of the ball is less than the launch speed. It’s important to remember that, in an ideal scenario without air resistance, the return speed would be equal to the launch speed because of the conservation of energy. However, in real-world situations like this one, air resistance, a form of energy dissipation, reduces the speed.
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High power lasers in factories are used to cut through cloth and metal. One such laser has a beam diameter of 0.863 mm and generates an electric field at the target having an amplitude 0.955 MV/m. The speed of light is 2.99792 × 10⁸ m/s the permeability of free space is 4π × 10⁻⁷ T· N/A. What is the amplitude of the magnetic field produced? Answer in units of T.
Answer:
B_m = 3.186 x 10⁻³ T
Explanation:
given,
diameter of the beam = 0.863 mm
Amplitude = 0.955 MV/m
speed of light = 2.99792 × 10⁸ m/s
he permeability of free space= 4π × 10⁻⁷ T· N/A
Amplitude of the magnetic field is given by
[tex]B_m = \dfrac{E_m}{c}[/tex]
E_m is amplitude of the electric field.
[tex]B_m = \dfrac{0.955\times 10^{6}}{2.99792\times 10^8}[/tex]
B_m = 0.31855 x 10⁻² T
B_m = 3.186 x 10⁻³ T
The amplitude of magnetic field is equal to 3.186 x 10⁻³ T
Answer:
I= 12.09×10^8 W/m^2
Explanation:
Em = amplitude of electric field = 0.955 MV/m = 0.955 x 10^6 V/m
B_m = amplitude of magnetic field = ?
c = speed of light = 2.99792 x 10^8 m/s
amplitude of magnetic field is given as
B_m = E_m /c
B_m = (0.955 x 10^6)/(2.99792 x 10^8 )
B_m = 0.00318 T
b)
intensity is given as
[tex] I=(0.5)\epsilon\times E_m^2 c [/tex]
[tex]I=0.5(8.85\times10^{-12})(0.955\times10^6)^2(2.99792\times10^8)[/tex]
I= 1209873759.69
I= 12.09×10^8 W/m^2
A Ping-Pong ball has a diameter of 1.99 cm and average density of 0.121 g/cm3 . What force would be required to hold it completely submerged under water? The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.
Answer:
0.035 N
Explanation:
Density = mass/volume
D = m/v
m = D× v ..................... Equation 1
Where D = density of the ping-pong, v = volume of the ping-pong, m = mass of the ping-pong
Note: The ping-pong is spherical in shape.
v = 4/3πr²
Where r = radius, π = pie
d = 1.99 cm, π = 3.14
v = 4/3(1.99/2)²(3.14)
v = 4.12 cm³
Also D = 0.121 g/cm³
Therefore,
m = 0.121(4.12)
m = 0.499 g
W = mg
Where W = weight of the ping-pong
W = (0.499/1000)×9.81
W = 0.005 N.
From Archimedes principle,
Upthrust = density of water × volume of water displaced × acceleration due to gravity.
U = D'vg/1000...................... Equation 2
Note: The volume of water displaced is equal to the volume of the ping-pong.
given: v = 4.12 cm³, g = 9.81 m/s², D' = 1 g/cm³
Substitute into equation 2
U = 1(4.12)(9.81)/1000
U = 0.04 N
The force required to hold the ball completely submerged under water = U-W
= 0.04-0.005 = 0.035 N
When 1.60 × 10 5 J 1.60×105 J of heat transfer occurs into a meat pie initially at 17.5 °C , 17.5 °C, its entropy increases by 485 J / K . 485 J/K. Estimate the final temperature of the pie.
Answer:
Explanation:
Given
Heat transfer [tex]Q=1.6\times 10^5\ J[/tex]
initial Temperature [tex]T_i=17.5^{\circ}\approx 290.5\ K[/tex]
Entropy change [tex]dS=485\ J/K[/tex]
The expression for entropy is given by
[tex]dQ=TdS[/tex]
[tex]T=\frac{dQ}{dS}[/tex]
[tex]T=\frac{1.6\times 10^5}{485}[/tex]
[tex]T=329.89\ K[/tex]
Temperature can be written as average of initial and final temperature
[tex]T=\frac{T_i+T_f}{2}[/tex]
[tex]329.89=\frac{T_f+290.5}{2}[/tex]
[tex]T_f=659.78-290.5[/tex]
[tex]T_f=369.28\ K[/tex]
As you learned in Part B, a non-burning helium core surrounded by a shell of hydrogen-burning gas characterizes the subgiant stage of stellar evolution. As time goes on, the star continues to evolve, and eventually, it becomes a red giant. Rank the stages a star goes through as it evolves from a subgiant into a red giant, from latest to earliest.A) The shell of hydrogen surrounding the star's non-burning helium core ignites.B) Pressure in the star's core decreases.C) Pressure from the star's hydrogen-burning shell causes the non-burning envelope to expand.D) The star's non-burning helium core starts to contract and heat up.E) The star becomes a red giant.F) The surface of the star becomes brighter and cooler.
Answer:
E- The star becomes a red giant (LATEST STAGE)
F- The surface of the star becomes brighter and cooler
C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand
A- The shell of hydrogen surrounding the star's nonburning helium core ignites.
D- The star's non burning helium core starts to contract and heat up
B- Pressure in the star's core decreases (EARLIEST STAGE)
(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.
Final answer:
Explanation of the stages in stellar evolution from subgiant to red giant.
Explanation:
Ranking the stages of stellar evolution from subgiant to red giant:
The shell of hydrogen surrounding the star's non-burning helium core ignites.
The star's non-burning helium core starts to contract and heat up.
The surface of the star becomes brighter and cooler.
Pressure from the star's hydrogen-burning shell causes the non-burning envelope to expand.
Pressure in the star's core decreases.
The star becomes a red giant.
A spring with a force constant of 5.0 N/m has a relaxed length of 2.63 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.93 m. Calculate the elastic potential ene
Answer:
4.225 J
Explanation:
Elastic Potential Energy: This is the potential energy stored in an elastic material.This also the energy required to stretch an elastic material. The S.I unit is Joules.
Mathematically it is expressed as
E = 1/2ke²....................... Equation 1
Where E =elastic potential Energy, k = spring constant, e = extension.
Given: k = 5.0 N/m, e = 3.93-2.63 = 1.3 m.
Substitute into equation 1
E = 1/2(5)(1.3)²
E = 8.45/2
E = 4.225 J.
Thus the Elastic potential Energy = 4.225 J.
What conclusion can be derived by comparing the central tendencies of the two data sets?
A: {7, 6, 3, 1, 6, 2, 4, 6, 3, 5}
B: {2, 2, 2, 3, 4, 5, 2, 8, 7, 6}
A.
The mean of set A is smaller than the mean of set B.
B.
The median of set A is greater than the median of set B.
C.
The median and the mean of set B are greater than those of set A.
D.
The mode of set B is greater than the mode of set A.
Which layer in the Earth has a composition similar to the granite shown in this photograph?
Answer:
Earth crust and specifically the continental crust.
Explanation:
If we examine the earth crust there is mostly the granite and basalt and most of the granite is present in the continental crust part which is less thicker and denser. That's why we say that the continental crust has the composition similar to that of granite.
A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe slope? Neglect friction. The accelerationof gravity is 9.8 m/s2.Answer in units of N.
Final answer:
The force on the skier parallel to the slope is found by calculating the component of the skier's weight that acts along the slope. This force is the weight multiplied by the sine of the angle of the slope, which results in 534.8 Newtons for a 103 kg skier on a 32° slope.
Explanation:
To determine the force on the skier parallel to the slope, we can make use of the component of the gravitational force along the slope. Since we are neglecting friction, the only force acting on the skier in the direction parallel to the slope is the component of the skier's weight in that direction.
The weight of the skier can be calculated by multiplying the mass (m) by the acceleration due to gravity (g), which is W = m × g. The component of the weight parallel to the slope is Wparallel = W × sin(θ), where θ is the angle of the slope. Substituting the given values, we have W = 103 kg × 9.8 m/s² = 1009.4 N. The parallel component is then 1009.4 N × sin(32°).
To find the sine of 32°, we use a calculator and get sin(32°) ≈ 0.5299. Multiplying this by the weight gives the parallel force on the skier, which is 1009.4 N × 0.5299 ≈ 534.8 N. Therefore, the force on the skier parallel to the slope is 534.8 Newtons.
A ladybug rests on the bottom of a tin can that is being whirled horizontally on the end of a string. Since the ladybug, like the can, moves in a circle, there must be a force on it. What exerts this force?
a. Gravityb. The stringc. Your handd. The can
Answer:uigu
Explanation:nvhv
The force that is experienced by the ladybug is being exerted by the can.
According to Newtons's third law of motion, action and reaction are equal and opposite. As a centripetal force is exerted on the can as it moves round the circle, the can also exerts a force back on the ladybug resting on its bottom.
Hence, the force that is experienced by the ladybug is being exerted by the can on which the ladybug sits.
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A 1200 Kg car travels at constant speed of 32 m/s/s. If the coefficient of friction between the road and the tires is 0.45 what force is applied by the car engine to run it?
Answer:
Fr= 5400 N
Explanation:
Given that
m = 1200 kg
v= 32 m/s
μ = 0.45
Given that car is moving with constant speed it means that acceleration of the car is zero or we can say that total force on the car is zero.That is why ,only force needed to over come the friction force.
Fr= μ m g
Fr= 0.45 x 1200 x 10 N
Fr= 5400 N
That is why the total force should be 5400 N.
Answer:
Explanation:
mass, m = 1200 kg
v = 32 m/s
coefficient of friction, μ = 0.45
As the car is moving with constant speed so the net force is zero, but the force applied by the car engine to run is the friction force.
F = μ mg = 0.45 x 1200 x 9.8 = 5292 N
Suppose two wagons, one with twice as much mass as the other, fly apart when a compressed spring that joins them is released. The heavier wagon rolls ____ as fast as the lighter wagon.
Answer:
The heavier wagon rolls 1/2 as fast as the lighter wagon.
Explanation:
When the compressed spring that joins them is released then the force acts on both wagons will be of equal magnitude but in the opposite direction. However as the mass of one wagon is twice that of other, so the acceleration will become half of the heavier wagon in comparison with lighter one.
A platform is rotating with an angular speed of 3.00 rad/s and an angular acceleration of 11.0 rad/s2. At a point on the platform that is 1.40 m from the axis of rotation, what is the magnitude of the total acceleration a?
Answer:
a=19.8977 m/s²
Explanation:
Given data
ω=3.00 rad/s
r=1.40 m
α=11.0 rad/s²
To find
Acceleration
Solution
As the object moves in a circle so it has tangential acceleration also due to circular motion is has centripetal acceleration
The total acceleration can be found by
[tex]a=\sqrt{(a_{c})^{2}+(a_{T})^{2}}[/tex]
where
at is tangential acceleration
ac is centripetal acceleration
First we need to find centripetal acceleration
so
[tex]a_{c}=rw^{2}[/tex]
put the values or r and ω
[tex]a_{c}=(1.40m)*(3.00rad/s)^{2}\\a_{c}=12.6 m/s^{2}[/tex]
Now for tangential acceleration
[tex]a_{t}=ra\\a_{t}=(1.40m)*(11.0rad/s^{2} )\\a_{t}=15.4 m/s^{2}[/tex]
Put values of ac and at to find total acceleration
So
[tex]a=\sqrt{(a_{t})^{2} +(a_{c})^{2} }\\ a=\sqrt{(15.4m/s^{2} )^{2}+(12.6m/s^{2} )^{2} }\\ a=19.8977m/s^{2}[/tex]
The magnitude of the total acceleration at a point on the rotating platform is approximately 19.79 m/s².
Explanation:To find the magnitude of the total acceleration at a point on the rotating platform, we need to consider both the tangential acceleration and the centripetal acceleration. The tangential acceleration is given by the product of the angular acceleration and the radius, while the centripetal acceleration is given by the square of the angular speed times the radius. Adding these two accelerations together will give us the magnitude of the total acceleration at that point.
First, calculate the tangential acceleration:
Tangential acceleration = angular acceleration × radius
Tangential acceleration = 11.0 rad/s² × 1.40 m = 15.4 m/s²
Then, calculate the centripetal acceleration:
Centripetal acceleration = (angular speed)² × radius
Centripetal acceleration = (3.00 rad/s)² × 1.40 m = 12.60 m/s²
Finally, find the vector sum of the tangential acceleration and the centripetal acceleration:
Total acceleration = √((tangential acceleration)² + (centripetal acceleration)²)
Total acceleration =√((15.4 m/s²)² + (12.60 m/s²)²)
Total acceleration ≈ 19.79 m/s²
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