Answer:
Boyle's Law
[tex]\therefore P_1.V_1=P_2.V_2[/tex]
Explanation:
Given that:
initially:
pressure of gas, [tex]= P_1[/tex]
volume of gas, [tex]= V_1[/tex]
finally:
pressure of gas, [tex]= P_2[/tex]
volume of gas, [tex]= V_2[/tex]
To solve for final volume [tex]V_2[/tex]
According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.
According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.
But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.
Here nothing is said about the temperature, so we consider the Boyle's Law which states that at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.
Mathematically:
[tex]P_1\propto \frac{1}{V_1}[/tex]
[tex]\Rightarrow P_1.V_1=k\ \rm(constant)[/tex]
[tex]\therefore P_1.V_1=P_2.V_2[/tex]
Final answer:
To calculate the new volume V2 in Boyle's Law, use the formula V2 = P1V1/P2, ensuring that P1 and P2 are in the same pressure units and that V1 and V2 are in the same volume units.
Explanation:
To determine the new volume V2 after a change in pressure of an ideal gas, one can utilize Boyle's Law, which states that at a constant temperature, the pressure of a gas is inversely proportional to its volume. The formula representing Boyle's Law is P1V1 = P2V2. In this equation, P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. To solve for V2, you simply rearrange the equation to V2 = P1V1/P2.
Remember that in order for Boyle's Law to be applied correctly, both volumes should be in the same unit. If you solve for V2 and find it to be greater than V1, it indicates that the pressure has decreased, consistent with the inverse proportionality.
You observe a spiral galaxy with a large central bulge and tightly wrapped arms. It would be classified a
Answer:
Sa
Explanation:
Spiral Galaxies -
It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .
According to Hubble , these galaxies are classified as Sa , Sb , Sc .
Where ,
Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .
Sb - have a lesser bulge and the spiral is looser .
Sc - It has very weak bulge with the open spiral structure .
Hence , from the question ,
The given information is about the Sa .
A sound can be _______ or ________.
A. Quiet; loud
B. Quiet; soft
C. Quiet; heavy
D. Loud; soft
Answer:
a
Explanation:
a sound cannot be soft
Answer:
Quite or loud
Explanation:
Beth exerts 14 Newton’s of force to propel a 4.5 kilogram bowling ball down the lane. Describe how the ball will travel.
The ball will accelerate at a rate of [tex]3.11 m/s^2[/tex]
Explanation:
We can describe the motion of the ball by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:
[tex]F=ma[/tex]
where
F is the net force
m is the mass
a is the acceleration
In this problem,
F = 14 N is the force exerted on the ball
m = 4.5 kg is the mass of the ball
Solving the equation, we find its acceleration:
[tex]a=\frac{F}{m}=\frac{14}{4.5}=3.11 m/s^2[/tex]
So, the ball will accelerate at a rate of [tex]3.11 m/s^2[/tex].
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You throw a 50.0g blob of clay directly at the wall with an initial velocity of -5.00 m/s i. The clay sticks to the wall, and the collision takes about 20.0 ms (2.00 x 10^-2 s). a) What is the change in momentum for the blob of clay?
Answer:0.25 kg-m/s
Explanation:
Given
mass of blob [tex]m=50 gm [/tex]
initial velocity [tex]u=-5 m/s\ \hat{i}[/tex]
time of collision [tex]t=20 ms[/tex]
we know Impulse is equal to change in momentum
initial momentum [tex]P_i=mu[/tex]
[tex]P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s[/tex]
Final momentum [tex]P_f=50\times 10^{-3}v[/tex]
[tex]P_f=0[/tex] as final velocity is zero
Impulse [tex]J=P_f-P_i[/tex]
[tex]J=0-(-0.25)[/tex]
[tex]J=0.25 kg-m/s[/tex]
The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a height 6450 km above Earth’s surface? Answer in units of N. What is the weight 33700 km above Earth’s surface? Answer in units of N.
Final answer:
The weight of a spacecraft at 6450 km above Earth's surface is 1767.5 N and at 33700 km above Earth's surface is 182.35 N, calculated using Newton's law of universal gravitation and considering the increased distance from the Earth's center.
Explanation:
The weight of a spacecraft can be calculated using Newton's law of universal gravitation which states that every mass attracts every other mass with a force that is directly proportional to the product of their masses and inversely proportional to the distance squared between their centers, F = G * (m₁ * m₂) / r², where G is the gravitational constant, m₁ and m₂ are the masses involved, and r is the distance between the centers of the two masses. To find the weight of the spacecraft at a certain height, we need to use the spacecraft's mass and the new distance from the Earth's center, which includes both the Earth's radius and the altitude above the surface.
To answer the first part of the question, we calculate the weight at 6450 km above Earth's surface. Since the radius of the Earth is also 6450 km, the distance from the center of the Earth to the spacecraft is now 2 * 6450 km. Applying the law of gravitation, the gravitational force, and hence the weight, will be (6450 km / 2 * 6450 km)² = 1/4 of the original weight, which is 7070 N/4 = 1767.5 N.
For the second part of the question, at a height of 33700 km above the Earth's surface, the distance from the center is 6450 km + 33700 km = 40150 km. Repeating the calculation, the weight at this height will be (6450 km / 40150 km)² times the original weight, giving us a reduced weight of (7070 N * (1/6.23)²) ≈ 182.35 N.
The cheetah is one of the fastest-accelerating animals, because it can go from rest to 19.6 m/s (about 44 mi/h) in 2.9 s. If its mass is 108 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
(a) watts(b) horsepower.
Answer:
a)P =14288.4 W
b)P = 19.16 horsepower
Explanation:
Given that
m= 108 kg
Initial velocity ,u= 0 m/s
Final velocity ,v= 19.6 m/s
t= 2.9 s
Lets take acceleration of Cheetah is a m/s²
We know that
v= u + a t
19.6 = 0 + a x 2.9
a= 6.75 m/s²
Now force F
F= m a
F= 108 x 6.75 N
F= 729 N
Now the power P
P = F.v
P = 729 x 19.6 W
P =14288.4 W
We know that
1 W= 0.0013 horsepower
P = 19.16 horsepower
P =14288.4 W
Sound wave A delivers 2J of energy in 2s. Sound wave length B delivers 10J of energy in 5s. Sound wave C delivers 2mJ of energy in 1ms. Rank in order, from largest to smallest, the sound powers of Pa, Pb, Pc of these three waves.Explain. What equation would you use to determine this?
Answer:
[tex]P_c=P_b>P_a[/tex]
Explanation:
E = Energy
T = Time
Power is given by the equation
[tex]P=\frac{E}{T}[/tex]
For first case
[tex]P_a=\frac{2}{2}\\\Rightarrow P_a=1\ W[/tex]
For second case
[tex]P_b=\frac{10}{5}\\\Rightarrow P_b=2\ J[/tex]
For third case
[tex]P_c=\frac{2\times 10^{-3}}{1\times 10^{-3}}\\\Rightarrow P_b=2\ J[/tex]
The rank of power would be [tex]P_c=P_b>P_a[/tex]
Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its previous value. If the speed of the water in the larger section of the pipe was 32 m/s what is its speed in this smaller section if the water behaves like an ideal incompressible fluid?
Answer:
The speed in the smaller section is [tex]43.2\,\frac{m}{s}[/tex]
Explanation:
Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:
[tex] \Delta Q=0 [/tex] (1)
With Q the flux of water that is [tex] Av[/tex] with A the cross section area and v the velocity, so by (1):
[tex] A_{2}v_{2}-A_{1}v_{1}=0 [/tex]
subscript 2 is for the smaller section and 1 for the larger section, solving for [tex] v_{2} [/tex]:
[tex]v_{2}=\frac{A_{1}v_{1}}{A_{2}} [/tex] (2)
The cross section areas of the pipe are:
[tex] A_{1}=\frac{\pi}{4}d_{1}^{2} [/tex]
[tex] A_{2}=\frac{\pi}{4}d_{2}^{2} [/tex]
but the problem states that the diameter decreases 86% so [tex] d_{2}=0.86d_{1} [/tex], using this on (2):
[tex] v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1} [/tex]
[tex]v_{2}\approx(1.35)(32)\approx43.2\,\frac{m}{s} [/tex]
A molecule moves down its concentration gradient using a transport protein in the plasma membrane. This is an example of
Final answer:
Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane.
Explanation:
Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane. This process does not require the input of energy and allows substances to diffuse across the membrane more easily. For example, glucose is transported into cells using glucose transporters that utilize facilitated transport. This process is important for the movement of larger or charged molecules that cannot freely diffuse across the cell membrane.
A 0.095-kg aluminium sphere is dropped from the roof of a 55-m-high building. The specific heat of aluminium is 900 J/kg⋅C∘ .
If 65 % of the thermal energy produced when it hits the ground is absorbed by the sphere, what is its temperature increase?
Answer:
Increase in temperature will be [tex]0.389^{\circ}C[/tex]
Explanation:
We have given mass of the aluminium m = 0.095 kg
Height h = 55 m
Specific heat of aluminium c = 900 J/kg°C
We know that potential energy is given as
[tex]PE=mgh=0.095\times 9.8\times 55=51.205[/tex]
Now 65 % of potential energy [tex]=\frac{51.205\times 65}{100}=33.28[/tex]
Now this energy is used to increase the temperature
So [tex]mc\Delta T=33.28[/tex]
[tex]0.095\times 900\times \Delta T=33.28[/tex]
[tex]0.095\times 900\times \Delta T=33=0.389^{\circ}C[/tex]
As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an axis through its center. Since you were a good physics student, you know what to do. You measure the diameter of the wheel to be 0.88 m and find that it weighs 280 N . You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang a 6.32 kg mass from the free end of the rope. You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has speed 4.0 m/s after it has descended 2.5 m .(a) What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center?
Final answer:
The moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.
Explanation:
To calculate the moment of inertia of the wheel, we can use the principle of conservation of energy. The initial gravitational potential energy of the mass is equal to the final rotational kinetic energy of the wheel. This can be represented by the equation:
mg * h = 1/2 * Iω^2
Where m is the mass, g is the acceleration due to gravity, h is the distance the mass has descended, I is the moment of inertia of the wheel, and ω is the angular velocity of the wheel. Rearranging the equation:
I = 2mg * h / ω^2
Substituting the given values:
I = 2 * 6.32 kg * 9.8 m/s^2 * 2.5 m / (4.0 m/s)^2
I = 0.964 kg * m^2
Therefore, the moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.
A ledge on a building is 23 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up over a pulley and straight down to a 3.0-kg can of nails on the ground. If the can of paint is accidentally knocked off the ledge, what time interval does a carpenter have to catch the can of paint before it smashes on the ground?
Answer:
The time can catch before it smashes on the ground is [tex]t=5.73 s[/tex]
Explanation:
Using the force equation
[tex]F=m*a[/tex]
[tex]F_{net}=m*a[/tex]
So replacing and solving to find the acceleration
[tex]a = (m_1*g-m_2*g) / m_1+m_2[/tex]
Finding the factor
[tex]a = g *( m_1-m_2)/m_1+m_2[/tex]
[tex]a=9.8m/s^2 *( 4.0 kg- 3.0 kg) / (4.0 + 3.0) kg[/tex]
[tex]a=1.4 m/s^2[/tex]
Now replacing in Newtons law to find the time before can catch so:
[tex]d= \frac{1}{2}*a*t^2[/tex]
[tex]t=\sqrt{\frac{2*d}{a}}=\sqrt{\frac{2* 23m}{1.4 m/s^2}}[/tex]
[tex]t=5.73 s[/tex]
Calculate the work required to move a planet’s satellite of mass 571 kg from a circular orbit of radius 2R to one of radius 3R, where 8.8 × 106 m is the radius of the planet. The mass of the planet is 7.76 × 1024 kg. Answer in units of J]
Final answer:
The work required to move a satellite from an orbit of radius 2R to 3R around a planet is calculated using the gravitational potential energy formula and is found to be 3.897×1010 J.
Explanation:
To calculate the work required to move a satellite from one circular orbit to another around a planet, we must consider the gravitational potential energy differences in the two orbits.
The gravitational potential energy (U) of an object of mass m in orbit around a planet of mass M at a distance r is given by U = -GmM/r, where G is the gravitational constant (6.67×10-11 N m2/kg2).
For the initial orbit at radius 2R, the potential energy is U1 = -GmM/(2R), and for the final orbit at radius 3R, the potential energy is U2 = -GmM/(3R). The work done (W) in moving the satellite is the difference in gravitational potential energy, W = U2 - U1. Substituting the values, we get:
W = (-GmM/3R) - (-GmM/2R) = (GmM/6R)
Let's calculate the work required using the given values: G = 6.67×10-11 N m2/kg2, m = 571 kg, M = 7.76×1024 kg, R = 8.8×106 m.
W = (6.67×10-11 N m2/kg2 × 571 kg × 7.76×1024 kg) / (6 × 8.8×106 m)
W = 3.897×1010 J
Therefore, the work required to move the satellite from a circular orbit of radius 2R to one of radius 3R is 3.897×1010 J.
A tennis ball is thrown upward from the top of a 680 foot high building at a speed of 56 feet per second. The tennis ball's height above ground can be modeled by the equation . When does the tennis ball hit the ground?
Answer:
t = 8.5 s
Explanation:
Kinematic equation of the movement of the tennis ball that is thrown upwards :
y = y₀ + v₀*t -½ g*t² Equation (1)
Where :
y : position of the ball as a function of time
y₀ : Initial position of the ball
t: time
g: acceleration due to gravity in m/s²
Known data
g = 32 ft/s²
y₀ = 680 ft
v₀ = 56 ft/s
Calculation of the time it takes for the ball to thit the ground
We replace data en the equation (1)
y = y₀ + v₀*t -½ g*t²
0 = 680+(56)*t -½( 32) *t²
16*t²-(56)*t- 680 = 0 equation (2)
solving equation (2) quadratic:
t₁ = 8.5 s
t₁ = -5 s
Time cannot be negative so the time it takes for the ball to hit the ground is t = 8.5 s
A Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia In one set of experiments, they studied the maximum speed that quolls could run around a curved path without slipping. One quoll was running at 2.4 m/s around a curve with a radius of 1.6 m when it started to slip.
What was the coefficient of static friction between the quoll's feet and the ground in this trial?
Answer:
Coefficient of static friction = 0.37
Explanation:
At the point the the quoll slides, quoll attains its maximum velocity.
So Ne = (mv^2)/r ....equa 1
And N =mg....equ 2
Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s
Sub equ 2 into equ 1
Mge= (mv^2)/r ...equa3
Simplfy equ3
e = v^2/(gr)...equ 4
Sub figures above
e = 5.76/(9.8*1.6)
e = 0.37
Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height h above the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground.
Answer:
The final velocities of all the six balls will be same.
Explanation:
According to law of conservation of energy:
Gain in K.E = Loss in potential energy
½ mv^2 = mgh
Where “m” and “g” are constant. The interchange in energies will occur only with the change in velocity and height. Since, balls are thrown from the same hight with the same initial velocity so, their final velocities will also be same just before striking the ground.
The six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.
During a downward motion of an object, the speed of the object increases as the object moves downwards and becomes maximum before the object hits the ground.
The equation for estimating the final velocity of the six balls is given as;
[tex]v_f = v_i + gt[/tex]
If air resistance is negligible, the six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.
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The end point of a spring vibrates with a period of 2.1 seconds when a mass m is attached to it. When this mass is increased by 6.810×101 kg, the period is found to be 3.4 seconds. Find the value of m.
Answer:
Mass attached to the spring is 41.95 kg
Explanation:
We have given time period of the spring T = 2.1 sec
Let the mass attached is m
And spring constant is k
We know that time period is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]2.1=2\pi \sqrt{\frac{m}{k}}[/tex]---------eqn 1
Now if the mass is increased by 68.10 kg then time period become 3.4 sec
So [tex]3.4=2\pi \sqrt{\frac{m+68.10}{k}}[/tex]------eqn 2
Now dividing eqn 1 by eqn 2
[tex]\frac{2.1}{3.4}=\sqrt{\frac{m}{m+68.10}}[/tex]
[tex]0.381=\frac{m}{m+68.10}[/tex]
[tex]m=41.95 kg[/tex]
So mass attached to the spring is 41.95 kg
Final answer:
To find the value of mass m, use the formula for the period of a mass-spring system.
Explanation:
In order to find the value of mass m, we can use the formula for the period of a mass-spring system:
T = 2π√(m/k)
Where T is the period, m is the mass, and k is the spring constant.
For the initial system with period 2.1 seconds, we have:
2.1 = 2π√(m/k)
For the system with mass increased by 6.810×10^1 kg and period 3.4 seconds, we have:
3.4 = 2π√((m + 6.810×10^1)/k)
Using these two equations, we can solve for the value of m.
Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2. Gun #1 weighs three times as much as gun #2. The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is:
a) 2:3
b) 3:2
c) 2:1
d) 3:1
e) 6:1
Answer:
option C
Explanation:
Let mass of the bullet be m and velocity be v
mass of gun be M and bullet be V
now,
using conservation of momentum for gun 1
(M+m) V' = 2 mv + 3 MV
V' = 0
3 M V = - 2 mv
momentum of gun 1 =- 2 mv---------(1)
now for gun 2
(M+m) V' = mv + MV
V' = 0
M V = - mv
momentum of gun 1 = -mv-----------(2)
dividing equation (1) by (2)
[tex]\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}[/tex]
[tex]\dfrac{P_m1}{P_m2} = \dfrac{2}{1}[/tex]
the correct answer is option C
The correct option is Option C (2:1).The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is 2:1. This conclusion follows from the principle of conservation of momentum. Thus, the correct answer is option c) 2:1.
When the bullets are fired, each gun experiences a momentum change due to the bullet's ejection. By the conservation of momentum, the momentum imparted to each gun should equal the momentum of its corresponding bullet in magnitude but in the opposite direction.Define the mass of the bullet from gun #2 as m. The bullet from gun #1 then has a mass of 2m since it is twice as heavy. Denote the velocity of both bullets as v:
The momentum of the bullet from gun #2 is p2 = m * v.The momentum of the bullet from gun #1 is p1 = 2m * v.So, the momentum imparted to gun #1 is 2m * v, and for gun #2 it is m * v.
To find the ratio of the momentum imparted to gun #1 to that imparted to gun #2, we calculate:
Ratio = (momentum imparted to gun #1) / (momentum imparted to gun #2) = (2m * v) / (m * v) = 2/1 = 2:1Therefore, the correct answer is option c) 2:1
A father racing his son has half the kinetic energy of the son, whohas three-fifths the mass of the father. The father speeds up by2.5 m/s and then has the same kinetic energy as the son.a) What is the original speed of the father?b) What is the original speed of the son?
Answer:
a) 6.04 m/s
b) 11.02 m/s
Explanation:
a) Let the father mass be M, and his speed be V. His son mass is m = 3M/5. Since his kinetic energy initially is half of after he increases his speed by 2.5m/s
[tex]E_2 = 2E_1[/tex]
[tex]\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}[/tex]
[tex]V^2 + 5V + 6.25 = 2V^2[/tex]
[tex]V^2 - 5V - 6.25 = 0[/tex]
[tex]V \approx 6.04m/s[/tex]
b) The son kinetic energy initially is:
[tex]E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J[/tex]
We can solve for the son speed by the following formula
[tex]E_s = \frac{mv^2}{2}[/tex]
[tex]v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s[/tex]
[tex]v = \sqrt{121.4} = 11.02 m/s[/tex]
A crate is lifted vertically 1.5 m and then heldat rest. The crate has weight 100 N (i.e., it issupported by an upward force of 100 N).How much work was done in lifting thecrate from the ground to its final position?1. More than 150 J2. 150 J3. A bit less than 150 J4. No work was done.5. None of these
Answer:
option 1
Explanation:
given,
Weight of crate = 100 N
Crate is lifted up to height = 1.5 m
Work done =?
work = Force x distance
work = 100 N x 1.5 m
work = 150 J
However, work would be more than the 150 J to lift the crate slightly higher to its final position.
The correct answer is option 1
A car traveling 6.0 m/s is uninformly accelerating at a rate of 3.0 m/s^2 for 15 seconds. What is it’s final velocity?
Answer:
The answer to your question is 11.2 m/s
Explanation:
Data
Initial speed (vo) = 6.0 m/s
Acceleration (a) = 3.0 m/s²
time = 15 s
Final speed = ?
Formula
d = vot + [tex]\frac{1}{2} at^{2}[/tex]
vf² = vo² + 2ad
Process
d = (6)(15) + [tex]\frac{1}{2} (3)(15)^{2}[/tex]
d = 90 + 337.5
d = 427.5 m
vf² = (6)² + 2(3)(15)
vf² = 36 + 90
vf² = 126
vf = 11.2 m/s
Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.
Answer:
Qtot = 6C * deltaV
Explanation:
you can find the total capacitance from adding 1C+2C+3C=6C. and the total voltage is 1V. Capacitance = charge/voltage--> C = Q / V--> 6C = Q / deltaV. this makes Qtot = 6C* deltaV
The Total charge for the equivalent circuit is = [tex]Q_{tot}[/tex] = 6c * ΔV
Although your question is incomplete I found the missing part online and used it to resolve the question
Given data :
Total capacitance ( C ) = 6C ( 1 + 2 + 3 )
voltage = 1 V
Three capacitors having values of ; 1 C, 2 C, 3 C
Determine the total charge ( Qtot )
Applying the formula ; Q = CV ---- ( 1 )
where; Q = charge
C = capacitance
change in V = ΔV
∴ [tex]Q_{tot}[/tex] = 6c * ΔV
Hence the total charge Qtot for the equivalent capacitor = 6c * ΔV
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Models that explain the formation of the solar system through a series of gradual steps are considered evolutionary theories. What are models that explain the formation of certain objects in the solar system through other means considered?
Answer:
Catastrophic theories
Explanation:
The theory of catastrophe is a collection of methods used to analyze and describe the ways in which a system can experience sudden significant behavioral changes when one or more of the variables that govern it are continuously modified.
Georges Louis de Buffon suggested in 1745 the first destructive theory — that a comet pulled material from the Sun to form the planets.
Therefore the answer is -
Catastrophic theories
Assume the speed of light to be 299 792 458 m/s. If the frequency of an electromagnetic wave is 80,000 GHz (GHz = gigahertz = 109 Hz), what is the wavelength of that radiation? Express your answer in micrometres (μm)
Answer:
3.747 μm
Explanation:
To answer this question, the fundamental wave equation will be used. Light is an electromagnetic wave so we will use the speed of light for this electromagnetic wave.
v = fλ
299 792 458 m/s = 80,000 *10^9 * λ
λ = 3.747 *10^-6 = 3.747 μm
A figure skater is spinning slowly with arms outstretched. She brings her arms in close to her body and her moment of inertia decreases by 1/2. Her angular speed increases by a factor of
Answer:
the angular velocity increase by a factor of 2
Explanation:
using the law of the conservation of the angular momentum
[tex]L_i = L_f[/tex]
where [tex]L_i[/tex] is the inicial angular momentum and [tex]L_f[/tex] is the final angular momentum.
also, the angular momentum can be calculated by:
L = IW
where I is the inertia momentum and the W is the angular velocity.
so:
[tex]I_i W_i = I_fW_f[/tex]
we know that [tex]I_f = \frac{1}{2}I_i[/tex] then,
[tex]I_iW_i = \frac{1}{2}I_iW_f[/tex]
solving for [tex]W_f[/tex]:
[tex]W_f = 2W_i[/tex]
When a figure skater pulls her arms in while spinning, her moment of inertia decreases and her angular speed increases. In the provided case, with the moment of inertia decreasing by half, the angular speed will double.
Explanation:When a figure skater is spinning with her arms outstretched, and she pulls her arms in close to her body, her moment of inertia decreases. According to the law of conservation of angular momentum, if the moment of inertia of a spinning object decreases, its angular speed must increase to keep the angular momentum constant. In this case, since the skater's moment of inertia decreases by half, her angular speed will correspondingly double, or increase by a factor of 2.
Learn more about Conservation of Angular Momentum here:https://brainly.com/question/33713153
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Listed following are four models for the long-term expansion (and possible contraction) of the universe. Rank the models from left to right based on their predictions for the average distance between galaxies five billion years from now, from smallest to largest.
a. recollapsing universe
b. accelerating universe
c. coasting universe
d. critical universe
Answer:
gdsz
Explanation:
dsgzz cxvzdgctfgdsvftgdsftrdsfdtsardtgasfd5t6sgftsfdrstfdtgsv6cr5vsd5rw5
A human being can be electrocuted if a current as small as 48 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding. If his resistance is 2100 Ω, what might the fatal voltage in volts be?
Answer:
V = 100.8 V
Explanation:
given,
hum being can be electrocuted with current = 48 mA = 0.048 A
Resistance of the man = 2100 Ω
Fatal voltage = ?
we know,
V = I R
V is the fatal voltage in Volts
R is the resistance provided by the human body
I is current
V = I R
V = 0.048 x 2100
V = 100.8 V
the voltage which can be considered as fatal is equal to V = 100.8 V
Describe what happens, at a microscopic level, when an object is charged by rubbing. For instance, what happens when a plastic pipe is rubbed with a cloth? Describe the specific case where the rod becomes negatively charged
Answer:
Explanation:
The static charges are generated due to excess or deficiency of electrons, because these are the smallest quanta of charge available at the molecular level which can get transferred with minimal energy requirement.
These charges are usually generated by friction between the two surfaces leading to the transfer of electron from one to another.
When a plastic pipe is rubbed with a cloth then due to friction the surface of the cloth loses electron which gets stuck at the surface of the pipe making it negatively charged.
In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter contained 75.0 g H2O initially at 23.2°C. The final temperature of the solution was 31.8°C. What was the change in enthalpy for the dissolution of this compound?
Answer:
The enthalpy for dissolution is - 305.558 J/g
Solution:
Mass of the ionic compound, m = 10.00 g
Mass of water, m' = 75.0 g
Initial temperature, T = [tex]23.2^{\circ}C[/tex]
Final Temperature, T' = [tex]31.8^{\circ}C[/tex]
Now,
To calculate the change in enthalpy:
We know that the specific heat of water is 4.18 [tex]J/g^{\circ}C[/tex]
Total mass of the solution, M = m + m' = 10.00 + 75.0 = 85.0 g
Temperature, difference, [tex]\Delta T = T' - T = 31.8 - 23.2 = 8.6^{\circ}C[/tex]
Thus
The heat absorbed by the solution is given by:
[tex]Q = MC_{w}\Delta T = 85.0\times 4.18\times 8.6 = 3055.58\ J[/tex]
Enthalpy, [tex]\Delta H = -\frac{Q}{m} = - \frac{3055.58}{10} = - 305.558\ J/g[/tex]
The melting of glaciers and the polar ice caps poses a devastating threat of potential flooding for low-lying areas. Scientists predict that a sea level rise of ________ would inundate areas such as the delta regions of africa and asia
Answer:
4-6 millimeters
Explanation:
Global warming is causing devastating consequences for the planet such as rising sea levels and temperature in the oceans.
The melting of glaciers is one of the main causes of sea level rise.
Undoubtedly, the most affected and vulnerable areas correspond to areas of Asia, Africa and South America. Specifically, the study highlights that in six Asian countries such as China, Bangladesh, India, Vietnam, Indonesia and Thailand, there are approximately 237 million people who will suffer these floods if defense mechanisms are not activated.