Answer:
[tex]331665750000\ m/s^2[/tex]
3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by
[tex]g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2[/tex]
The gravitational acceleration at the surface of such a star is [tex]331665750000\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s[/tex]
The velocity of the object would be 3257806.62409 m/s
(a) the acceleration due to gravity is 3.3 × 10¹¹ m/s²
(b) the final velocity of the object is 3.26 × 10⁶ m/s
Gravitational force:Given that the neutron star has a mass equal to Sun, M = 2 × 10³⁰ kg
Radius of the star is R = 20 km = 20 × 10³
The weight on the surface of a neutron star that has the same mass as our Sun and a diameter of 21.0 km will be 8.29 × 10¹³ N
(a) The gravitational force on the surface of the neutron star is given by:
F = GMm/R²
mg = GMm/R²
where G is the gravitational constant
M is the mass of the body
m is the mass of the person
and, R is the radius of the body
So,
g = GM/R²
g = (6.67 × 10⁻¹¹)( 2 × 10³⁰)/ (210 × 10³)²
g = 3.3 × 10¹¹ m/s²
(b) According to the third equation of motion:
v² = u² + 2gh
v² = 2gh , since u = 0 as the object is at rest initially
v = [tex]\sqrt{2\times3.3\times10^{11}\times16}[/tex]
v = 3.26 × 10⁶ m/s
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The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are __________.
Answer:
The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are due to water's partial charges.
Explanation:
The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.
During heavy rain, a section of a mountainside measuring 2.3 km horizontally (perpendicular to the slope), 0.75 km up along the slope, and 1.4 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.52 km x 0.52 km and that the mass of a cubic meter of mud is 1900 kg. What is the mass of the mud sitting above a 5.5 m2 area of the valley floor?
Answer:
93328 kg
Explanation:
[tex]\rho[/tex] = Density of mud = 1900 kg/m³
Volume of the cuboid is given by
[tex]V=2300\times 750\times 1.4\\\Rightarrow V=2415000\ m^3[/tex]
Volume is also given by
[tex]V=Al\\\Rightarrow l=\dfrac{V}{A}\\\Rightarrow l=\dfrac{2415000}{520\times 520}\\\Rightarrow l=8.9312\ m[/tex]
The length of the cuboid is 0.89312 m
For the small volume
[tex]V_a=al\\\Rightarrow V_a=5.5\times 8.9312\\\Rightarrow V_a=49.12\ m^3[/tex]
Mass is given by
[tex]m=\rho V_a\\\Rightarrow m=1900\times 49.12\\\Rightarrow m=93328\ kg[/tex]
The mass of the mud is 93328 kg
A bullet is fired through a board 13.0 cm thick in such a way that the bullet's line of motion is perpendicular to the face of the board. The initial speed of the bullet is 560 m/s and it emerges from the other side of the board with a speed of 460 m/s. (a) Find the acceleration of the bullet as it passes through the board.
Answer:
392307.6923 m/s²
Explanation:
t = Time taken
u = Initial velocity = 560 m/s
v = Final velocity = 460 m/s
s = Displacement = 13 cm
a = Acceleration
From the equation of motion we have
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{460^2-560^2}{2\times 13\times 10^{-2}}\\\Rightarrow a=-392307.6923\ m/s^2[/tex]
The acceleration of the bullet as it passes through the board is -392307.6923 m/s²
"The work done on an ideal gas system in an isothermal process is -400 J. What is the change in internal (thermal) energy of the gas?"
Answer:
zero
Explanation:
Work done = - 400 J
In an isothermal process, the temperature remains constant. So, according to the first law of thermodynamics
dQ = dU + dW
As the temperature remains constant and the change in internal energy is the function of change in temperature, here change is temperature is zero so the change in internal energy is also zero.
Final answer:
The change in internal energy of an ideal gas during an isothermal process is 0 J. This is because in an isothermal process for an ideal gas, any work done by the gas is invariably compensated by an equivalent amount of heat absorbed, resulting in no net change in internal energy.
Explanation:
The question pertains to the change in internal energy of an ideal gas during an isothermal process in the context of thermodynamics, specifically according to the first law of thermodynamics. The first law states that the change in internal energy of a system (Triangle U) is equal to the heat added to the system (Q) minus the work done by the system (W). In an isothermal process, the temperature remains constant, and if the process is performed on an ideal gas, the internal energy also remains constant since it depends only on temperature for an ideal gas. Therefore, the change in internal energy ( triangle U) is zero.
In this specific scenario where the work done on the system is -400 J (-W), which means that the system has done 400 J of work on the surroundings. Since no heat transfer information is given, we can infer that in an isothermal process, whatever work is done by the gas is compensated by the heat received from the surroundings, keeping the internal energy unchanged. Thus, the change in internal energy of the gas is 0 J, as the energy spent as work would have been absorbed as heat from the surroundings.
At time t in seconds a particles distance s(t) in micrometers from a point is given by s(t)=e^t+2-sin(t) what is the average velocity of the particle from t=1 to t=5
Answer:
[tex]v_{avg}=29.499\ m.s^{-1}[/tex]
Explanation:
Given:
distance as a function of time is:
[tex]s=e^t+2-sin\ t[/tex]
Average velocity of the particle from time t=1s to t=5s can be given by the total distance covered divided by the total time consumed.
Position at t=1s:
[tex]s_1=e^1+2-sin\ 1[/tex]
[tex]s_1=3.8768\ m[/tex]
Position at t=5s:
[tex]s_5=e^5+2-sin\ 5[/tex]
[tex]s_5=151.3721\ m[/tex]
Therefore the distance covered during this time is:
[tex]\Delta s=s_5-s_1[/tex]
[tex]\Delta s=151.3721-3.8768[/tex]
[tex]\Delta s =147.4952\ m[/tex]
Now the average speed for the duration:
[tex]v_{avg}=\frac{\Delta s}{\Delta t}[/tex]
[tex]v_{avg}=\frac{147.4952}{5}[/tex]
[tex]v_{avg}=29.499\ m.s^{-1}[/tex]
The total magnification of a specimen being viewed with a 10X ocular lens and a 40X objective lens is _____.
Answer:
total magnification = 400 X
Explanation:
given data
ocular lens = 10 X
objective lens = 40 X
to find out
total magnification
solution
we know that total magnification is express as
total magnification = Objective magnification × ocular magnification .................1
put here value we get
total magnification = 10 × 40
total magnification = 400 X
The total magnification for a microscope with a 10X ocular lens and a 40X objective lens is 400X.
Explanation:When using a microscope, the total magnification can be found by multiplying the magnification of the ocular lens (also known as the eyepiece) by the magnification of the objective lens. Thus, in your case, the total magnification of the specimen being viewed with a 10X ocular lens and a 40X objective lens would be 400X. This is because 10 times 40 equals 400. Therefore, you are observing the specimen at 400 times its actual size.
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You throw a ball horizontally from the top of a building, with a speed of 3 m/s. In this problem, you can neglect the force of air resistance.
After 2 Seconds, what is the balls horizontal velocity?
A. Impossible to tell
B.Less than 3 m/s
C. 3 m/s
D. 0
E.More than 3 m/s
Answer:
Option C. 3 m/s
Explanation:
From the question it is stated that we can neglect the force of air resistance, therefore there is no force acting on the ball horizontally. From Newton's first law of motion an object will remain at rest or in uniform motion unless acted upon by an external force. thus the ball velocity will remain 3m/s.
A hockey puck is hit on a frozen lake and starts moving with a speed of 13.60 m/s. Exactly 6.2 s later, its speed is 7.20 m/s. (a) What is the puck's average acceleration?
Answer:
-1.03 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².
Mathematically, acceleration is expressed as
a = (v-u)/t ........................ Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time.
Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.
Substituting into equation 2
a = (7.20-13.60)/6.2
a = -6.4/6.2
a = -1.03 m/s²
Note: a is negative because, the hockey puck is decelerating.
Hence the average acceleration = -1.03 m/s²
5) The gravitational force between two objects that
are 2.1x10 m apart is 3.2x10N. If the mass of one
object is 55 kg what is the mass of the other object?
Answer:
F=G[tex]\frac{M1XM2}{R^{2} }[/tex]
Explanation:
F=F=G[tex]\frac{M1XM2}{R^{2} }[/tex]
Newton law define gravity as the product of two mass and the mean squired distance between the two bodies.
Where F is the gravitational pull and G is the gravitational constant and M1 and M2 are masses of two bodies while R is their mean distance
Replacing the value in the equation it becomes:
3.2x10N=G[tex]\frac{M1X55}{2.1X10^{2} }[/tex] Where G is the gravitation constant,[tex]6.67x10^{-11}[/tex]
M1=[tex]\frac{3.2X10X(2.1X10)^{2} }{55}[/tex]
M1=[tex]3.8468x10^{-11}[/tex]Kg
How long (in s) will it take an 875 kg car with a useful power output of 42.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.)
Answer:
The time taken by the car with a useful power output is 4.03 seconds.
Explanation:
Given that,
Mass of the car, m = 875 kg
Initial speed of the car, u = 0
Final speed of the car, v = 17 m/s
Power of the car, P = 42 hp = 31332 W
The power output of the car is given by the work done per unit time. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
Initial kinetic energy of the car will be 0 as it is at rest.
Final kinetic energy of the car is :
[tex]K=\dfrac{1}{2}mv^2[/tex]
[tex]K=\dfrac{1}{2}\times 875\times (17)^2[/tex]
K = 126437.5
As per the work energy theorem, the change in kinetic energy of the object is equal to the work done.
So,
[tex]P=\dfrac{W}{t}[/tex]
[tex]t=\dfrac{W}{P}[/tex]
[tex]t=\dfrac{126437.5}{31332}[/tex]
t = 4.03 seconds
So, the time taken by the car with a useful power output is 4.03 seconds. Hence, this is the required solution.
A certain compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD's information at a constant rate of 1.1 megabits per second. How many minutes does it take the player to read the entire CD?
Answer:
Time takes the player to read the entire CD is 94.9 minutes.
Explanation:
compact disc (CD) contains 783.216 megabytes of digital information.
Each byte consists of exactly 8 bits.
constant rate of 1.1 megabits per second
How many minutes does it take the player to read the entire CD?
1 Megabites = 1000000 bites
783.216 megabytes = 783216000 bites = 783216000 x 8 = 6265728000 bits
1.1 megabits/s = 1100000 bits/s
thus
t = 6265728000 bits/1100000 bits/s
= 5696 s
= 94.9 min
A truck is hoisted a certain distance in a garage and therefore has potential energy with respect to the floor. If it were lifted twice as high, it would have ____ as much potential energy.
Answer:
If the truck were lifted twice as high, it would have twice as much potential energy
Explanation:
Potential energy = mgh
where;
m is mass of the truck (kg)
g is acceleration due to gravity (m/s²)
h is the height above the floor.
Initial Potential Energy, E₁ = mgh₁
When the truck was lifted twice as high, New height (h₂) = 2h₁
New potential Energy, E₂ =mgh₂ = mg(2h₁)
E₂ = 2mgh₁
Recall, E₁ = mgh₁, then substitute in E₁ into E₂
E₂ = 2(mgh₁)
E₂ = 2E₁
Therefore, If the truck were lifted twice as high, it would have twice as much potential energy
A heavy-duty stapling gun uses a 0.179 kg metal rod that rams against the staple to eject it. The rod is attached and pushed by a stiff spring called a ram spring (k = 37107 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses. The ram spring is compressed by 3.20 10-2 m from its unstrained length and then releases from rest. Assuming that the ram spring is oriented vertically and is still compressed by 1.35 10-2 m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.
Final answer:
The speed of the ram at the instant of contact is 8.413 m/s.
Explanation:
To find the speed of the ram at the instant of contact, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is converted into the kinetic energy of the ram as it hits the staple. Since the mass of the spring is ignored, we can consider the system as consisting of only the ram.
Using the equation for the potential energy of the spring (PE = 0.5kx^2), we can calculate the initial potential energy. The compressed length of the spring is 3.20 x 10^-2 m, so the initial potential energy is 0.5(37107 N/m)(3.20 x 10^-2 m)^2 = 192.128 J.
At the instant of contact, all of the potential energy has been converted into kinetic energy. Therefore, we can equate the initial potential energy to the final kinetic energy of the ram. The kinetic energy is given by KE = 0.5mv^2, where m is the mass of the ram (0.179 kg) and v is the speed of the ram at the instant of contact.
Solving for v, we have: 192.128 J = 0.5(0.179 kg)v^2. Rearranging the equation and solving for v, we find that v = 8.413 m/s. Therefore, the speed of the ram at the instant of contact is 8.413 m/s.
how do you describe and determine the direction of the magnetic field produced by an electric current?
Answer:
the magnet or electric current produces a magnetic field
Explanation:
this magnetic field can be visualized as a pattern of a circular field lines surrounding a wire. hall probes can determine the magnitude of the field
hope it helps you
The direction of a magnetic field produced by an electric current in a straight wire can be determined using the right-hand rule. The Biot-Savart law is used to calculate the magnetic field strength at a distance from the wire.
Explanation:To describe and determine the direction of the magnetic field produced by an electric current, one commonly uses the right-hand rule. For a straight, current-carrying wire, point your right-hand thumb in the direction of the current; then, the direction in which your fingers curl around the wire represents the direction of the magnetic field lines, which form concentric circles around the wire.
For example, if the current flows from right to left in a wire, and you are looking at the wire end-on from the left end, your thumb would point to the left, indicating the direction of the current. Your fingers would naturally curl in a counterclockwise direction, suggesting the magnetic field direction at that point.
For calculating the magnetic field strength, B, produced by a long straight wire carrying current I, the Biot-Savart law can be used. It states that the strength of the magnetic field is directly proportional to the magnitude of the current and inversely proportional to the distance from the wire (assuming the wire is very long compared to the distance).
A Sense of Proportion: The Sun is roughly 100 times the diameter of Earth. If you built a model solar system with a golf ball (diameter of 1.68 inches) for Earth, how big would the model Sun be?
Explanation:
Diameter of Sun = 100 x Diameter of Earth.
We are creating a model where Earth is same as golf ball,
Diameter of Earth in model = Diameter of golf ball
Diameter of Earth in model = 1.68 inches
Diameter of Sun in model = 100 x Diameter of Earth in model
Diameter of Sun in model = 100 x 1.68
Diameter of Sun in model = 168 inches = 14 feet
Diameter of Sun in model = 168 inches = 14 feet = 4.27 m
Which statements describe the Mercalli scale?
a. This scale measures seismic waves based on their size.
b. This scale rates an earthquake according to how much damage it causes.
c. This scale produces a single rating for earthquakes that reach the surface.
d. This scale uses Roman numerals to rank the damage caused by an earthquake.
e. This scale measures the magnitude of an earthquake based on the size of seismic waves.
Answer:
The correct options are b. This scale rates an earth quake according to how much damage it causes. and d. This scale uses roman numerals to rank the damage caused by an earth quake.
Explanation:
According to the definition of the Mercalli scale " it is a scale used for the measurement of the intensity of the earthquakes" and was first invented in 1884 by the Italian volcanologist Giuseppe Mercalli. This scale ranges from I to XII.
Answer:
b d
Explanation:
did the quiz
ill has just gotten out of her car in the grocery store parking lot. The parking lot is on a hill and is tilted 3 degrees. Fifty meters downhill from Jill, a little old lady lets go of a fully loaded shopping cart. The cart, with frictionless wheels, starts to roll straight downhill. Jill immediately starts to sprint after the cart with her top acceleration of 2.0m/s2.How far has the cart rolled before Jill catches it?
Answer:
[tex]t=25.6446\ s[/tex]
is the time after which Jill will be able to catch the cart.
Explanation:
Given:
angle of inclination, [tex]\theta=3\ ^{\circ}[/tex]height of the cart from the level ground, [tex]h=50\ m[/tex]acceleration of Jill, [tex]a=2\ m.s^{-2}[/tex]Now the component of gravity acting on the cart along the inclined plane:
[tex]g'=g.sin\ \theta[/tex]
[tex]g'=9.8\times sin\ 3^{\circ}[/tex]
[tex]g'=0.5129\ m.s^{-2}[/tex]
Time taken by Jill to reach this speed:
[tex]v=u+a.t[/tex]
where:
t = time taken
u = initial velocity = 0
[tex]51.289=0+2\times t[/tex]
[tex]t=25.6446\ s[/tex] is the time after which Jill will be able to catch the cart.
Radio Station: What is the wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz? Remember that λ = c/f, where λ is the wavelength, f is the frequency and c is the speed of light (c = 300000 km/s)
Answer: The wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.
Explanation:
To calculate the wavelength of light, we use the equation:
[tex]\lambda=\frac{c}{\nu}[/tex]
where,
= wavelength of the light = ?
c = speed of light = 300000 km/s = [tex]3\times 10^8m/s[/tex] (1km=1000m)
[tex]\nu[/tex] = frequency of light = 100kHz = 100000Hz or (1kHz=1000Hz)
[tex]\lambda=\frac{3\times 10^8m/s}{100000s^{-1}}[/tex]
[tex]\lambda=3000m[/tex]
Thus the wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.
Baby Susie's pediatrician notices that one of her eyes rotates outward and that she does not appear to be using it for vision. What is her condition and what does the pediatrician recommend?
Answer:
Susie has strabismus.
Explanation:
Strabismus is a medical condition in which the patient can not keep his/her eyes aligned under the normal conditions. In strabismus one or both of eyes rolls into up, down, in or out direction. Physicians usually recommend the exercises for eyes or placing a patch on the unaffected eye or they also do surgery where it's required like in severe cases.
Two balls collide elastically the balls have masses of 3.4 and 5.7, if the 3.4 ball is moving at 10.5 m/s toward the other ball and the 5.7 kg ball is moving with a speed of 6.7 what are the final speeds of the balls?
Answer: The final speed of the balls will be 8.12m/s
Explanation:
For least collision of bodies, both momentum and kinetic energy is conserved.
According to Law of conservation of energy,
The sum of momentum of the objects before collision is equal to the sum of their momentum after collision. Note that the objects will move with a common velocity after impact.
Let m1 and m2 be the masses of the objects
u1 and u2 be their velocities before impact
v be their common velocity after impact
Since momentum is mass × velocity
Mathematically,
m1u1 + m2u2 = (m1+m2)v
Given m1= 3.4kg m2 = 5.7kg u1 = 10.5m/s u2 = 6.7m/s v = ?
Substituting this values in the formula, we have
3.4(10.5)+5.7(6.7) = (3.4+5.7)v
35.7+38.19 = 9.1v
73.89 = 9.1v
V = 73.89/9.1
V= 8.12m/s
A 6.0-kg box slides down an inclined plane that makes an angle of 39o with the horizontal. If the coefficient of kinetic friction is 0.40, at what rate does the box accelerate down the slope?3.1m/s23.4m/s23.7m/s24.1m/s2
Answer:
A. 3.1m/s²
Explanation:
Since the body is sliding down the plane, the frictional force (Ff) being a force of opposition will be acting upwards.
The forces acting on the body in the vertical direction will be the weight (W) and the normal reaction(R) acting in the opposite direction.
Taking the sum of the forces along the plane,
Fm - Ff = mass × acceleration (newton's law)
Since Fm = Wsin(theta)
Ff = nR where n is the coefficient of friction we will have;
Wsin(theta) - nR = ma... (1)
W = mg = 6×10 = 60N
n = 0.40
R = Wcos (theta) {resolving weight to the vertical)
R = 60cos39°
R = 46.6N
Substituting this values into eqn 1 to the get the acceleration,
60sin39° - 0.4(46.6) = 6a
37.8 - 18.64 = 6a
19.16 = 6a
a = 19.16/6
a = 3.19m/s²
The acceleration of the body down the plane will be 3.1m/s²
A demented scientist creates a new temperature scale, the "Z scale." He decides to call the boiling point of nitrogen 0°Z and the melting point of iron 1000°Z.
A) What is the boiling point of water on the Z scale?
B) Convert 100°Z to the Celsius scale.
C) Convert 100°Z to the Kelvin scale.
Answer:
A) [tex] Z = 0.577 C +112.931[/tex]
[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]
B) [tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]
C) [tex] K = C +273.15[/tex]
[tex] K = -22.41 +273.15 =250.739 K[/tex]
Explanation:
For this case we want to create a function like this:
[tex] Z = a C + b[/tex]
Where Z represent the degrees for the Z scale C the Celsius grades and tha valus a and b parameters for the model.
The boiling point of nitrogen is -195,8 °C
The melting point of iron is 1538 °C
We know the following equivalences:
-195.8 °C = 0 °Z
1538 °C = 1000 °Z
Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)
So then we can calculate the slope for the linear model like this:
[tex] a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}[/tex]
And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:
[tex] 0 = 0.577 (-195.8) + b[/tex]
And if we solve for b we got:
[tex] b = 0.577*195.8 =112.931 Z[/tex]
So then our lineal model would be:
[tex] Z = 0.577 C +112.931[/tex]
Part A
The boiling point of water is 100C so we just need to replace in the model and see what we got:
[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]
Part B
For this case we have Z =100 and we want to solve for C, so we can do this:
[tex] Z-112.931 = 0.577 C[/tex]
[tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]
Part C
For this case we know that [tex] K = C +273.15[/tex]
And we can use the result from part B to solve for K like this:
[tex] K = -22.41 +273.15 =250.739 K[/tex]
Final answer:
The boiling point of water on the Z scale is 500°Z. To convert 100°Z to the Celsius scale, the equivalent temperature is 10°C. To convert 100°Z to the Kelvin scale, the equivalent temperature is 283.15 K.
Explanation:
The Z scale is a new temperature scale created by the demented scientist. According to the Z scale, the boiling point of nitrogen is 0°Z and the melting point of iron is 1000°Z.
A) To find the boiling point of water on the Z scale, we need to look at the temperature range between the boiling point of nitrogen and the melting point of iron. Since the boiling point of nitrogen is 0°Z and the melting point of iron is 1000°Z, the range between them is 1000-0 = 1000°Z. Therefore, the boiling point of water on the Z scale is 500°Z, which is halfway between 0°Z and 1000°Z.
B) To convert 100°Z to the Celsius scale, we can use the formula: Celsius = (Z - 0) * (100 - 0) / (1000 - 0). Plugging in the values, we get Celsius = (100 - 0) * (100 - 0) / (1000 - 0) = 10000 / 1000 = 10°C.
C) To convert 100°Z to the Kelvin scale, we use the formula: Kelvin = Celsius + 273.15. Plugging in the value, we get Kelvin = 10 + 273.15 = 283.15 K.
For every factor of 10 difference in Keq' (e.g., going from 102 to 10, or from 10–3 to 10–4), what is the difference in standard free energy change, ΔG°' (kJ/mol)?
Answer:
The standard free energy is the difference between that of products and reactants. The value of G has a greater influence on the reaction been considered.
Explanation:
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system. The spontaneity of a reaction is explained by the standard gibb's free energy.
If Delta-G = -ve ( the reaction is spontaneous)if Delta -G = +ve ( the reaction is non-spontaneous)if Delta-G = 0 ( the reaction is at equilibrium)Use this hints for any reaction involving the Gibb's free, Enthalpy and entropy.
For each factor of 10 change in Keq, there is a corresponding change of approximately 5.7 kJ/mol in standard free energy change, ΔG°', at standard biochemical conditions (298.15 K).
Explanation:For every factor of 10 difference in the equilibrium constant, Keq, there is a corresponding change in the standard free energy change, ΔG°', as calculated using the relationship between ΔG°' and Keq:
ΔG°' = -RT ln Keq, where R is the universal gas constant and T is the temperature in Kelvin.
Given that R = 8.314 J/mol·K and at standard biochemical conditions T = 298.15 K, a tenfold change in Keq results in a change of (8.314 J/mol·K x 298.15 K x ln(10)), which equates to approximately 5.7 kJ/mol. Therefore, for example, increasing Keq from 102 to 103 would decrease ΔG°' by 5.7 kJ/mol, making the reaction more product-favored under standard conditions. Conversely, decreasing Keq from 10-3 to 10-4 would increase ΔG°' by 5.7 kJ/mol, making the reaction less product-favored under standard conditions.
a child stands with each foot on a different scale. The left scale reads 200N and the right scale reads 150N. what is her mass in kg?
Answer:
35.7 kg
Explanation:
We are given that
Left scale reads =200N
Right scale reads=150 N
We have to find the mass in kg.
We know that
Weight of child= Sum of two scales=200+150=350 N
Acceleration due to gravity=[tex]9.8 m/s^2[/tex]
We know that
Weight= mg
Using the formula
[tex]350=9.8 m[/tex]
[tex]m=\frac{350}{9.8}=35.7 kg[/tex]
Hence, the mass of child=35.7 kg
A very light cotton tape is wrapped around the outside surface of a uniform cylinder of mass M and radius R. The free end of the tape is attached to the ceiling. The cylinder is released form rest and as it descends it unravels from the tape without slipping. The moment of inertia of the cylinder about its center is I = 1/2 MR2
a. On the diagram above show all the forces applied on the cylinder. b. Find the acceleration of the center of the cylinder when it moves down. c. Find the tension force in the tape.
Answer:
Explanation:
Check attachment for answer.
The acceleration of the center of the cylinder and the tension force in the tape is mathematically given as
a=2g/3
T=ug/3
What are the forces applied on the cylinder, the acceleration of the center of the cylinder when it moves down, and the tension force in the tape?Generally, the equation for the Liner motion is mathematically given as
Mg-T=Ma
Therefore
Mg.R=(0.5MR^2+MR^2)\alpha
[tex]\alpha=2g/3R[/tex]
b)
Where
a=\alpha R
Hence
a=2g/3
c)
For the tension T
ug-T=Ma
T=u(g-a)
T=ug/3
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Assume that you land at an airport with the altimeter set to 29.92 instead of the current setting of 30.00. What will the altimeter read if the field elevation is 2,000 feet MSL?
When the altimeter is set to 29.92 inHg, it will read 4,000 feet MSL at a field elevation of 2,000 feet MSL.
To determine the altimeter reading when the field elevation is 2,000 feet MSL (Mean Sea Level) and the altimeter is set to 29.92 inches of mercury (inHg), we need to consider the difference in pressure settings.
The altimeter measures altitude based on the atmospheric pressure. When the altimeter setting (also known as the barometric pressure setting) is changed, it adjusts the reference point for sea level pressure. The standard atmospheric pressure at sea level is approximately 29.92 inHg.
Given:
Field elevation = 2,000 feet MSL
Altimeter setting = 29.92 inHg
We need to find the altimeter reading when the altimeter is set to 29.92 inHg.
To calculate the altimeter reading, we need to determine the difference in pressure between the current pressure (altimeter setting) and the pressure at the field elevation (2,000 feet above sea level).
The standard atmospheric pressure decreases as we go higher in altitude. A typical lapse rate for pressure is around 1 inHg per 1,000 feet of altitude gain. Since the field elevation is 2,000 feet above sea level, the difference in pressure would be approximately 2 inHg.
Therefore, the corrected altimeter reading would be 2,000 feet + 2,000 feet (for the 2 inHg difference) = 4,000 feet MSL.
Hence, when the altimeter is set to 29.92 inHg, it will read 4,000 feet MSL at a field elevation of 2,000 feet MSL.
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A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5 kg cart a speed of 26 cm/s to the left. What is the velocity of the 4.5-kg cart? a. 10 cm/s b. 4.56 cm/s c. 8.67 cm/s d. 7.3 cm/s
Answer:
c. 8.67 cm/s
Explanation:
From the law of conservation of momentum,
Total momentum before the thread was burned = Total momentum after was burned
mu + m'u' = mv + m'v'...................... Equation
Where m = mass of the heavier cart, m' = mass of the lighter cart, u = initial velocity of the bigger cart, u' = initial velocity of the smaller cart, v = final velocity of the bigger cart, v' = final velocity of the smaller cart.
Note: Both cart where momentarily at rest, as such u = u' = 0. i.e the total momentum before the thread was burn = 0
And assuming the left is positive,
We can rewrite equation 1 as
mv + m'v' = 0............................................ Equation 2
Given: m = 4.5 kg, m' = 1.5 kg, v' = 26 cm/s
Substitute into equation 2,
4.5v + 1.5(26) = 0
4.5v + 39 = 0
4.5v = -39
v = -39/4.5
v = -8.67 cm/s.
Note: v is negative because it moves to right.
Hence the velocity of the 4.5 kg cart = 8.67 cm/s.
The right option is c. 8.67 cm/s
The velocity of the 4.5 kg cart after the thread is burned is 10 cm/s.
Explanation:When the thread is burned, the compressed spring pushes the carts apart. Since the 1.5 kg cart moves to the left with a speed of 26 cm/s, the 4.5 kg cart would also move to the left but with a slower speed. To determine the velocity of the 4.5 kg cart, we can use the principle of conservation of momentum:
Momentum before = Momentum after
(Mass of 1.5 kg cart) x (Initial velocity) + (Mass of 4.5 kg cart) x (Initial velocity) = (Mass of 1.5 kg cart + Mass of 4.5 kg cart) x (Final velocity)
Solving for the final velocity of the 4.5 kg cart:
(1.5 kg x 26 cm/s + 4.5 kg x 0 cm/s) / (1.5 kg + 4.5 kg) = (6.9 kg) x (Final velocity)
Final velocity = 10 cm/s
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Radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 million years. Assuming a constant rate, seafloor spreading in this area occurs at a rate of __________.
A. 5 cm per year
B. 1,012.5 km per year
C. 20,000 cm per year
D. 50 km per year
The rate of seafloor spreading in this area, based on the given radiometric dating results and calculated assuming a constant rate, is 25 km/million years.
Explanation:
To calculate the rate of seafloor spreading, we divide the distance that the seafloor has spread by the amount of time it took. Given the radiometric dating results, we know that this distance (225 km) was covered in 9 million years.
So, the calculation would be: 225 km / 9 million years = 0.025 km/year. Or, to convert this figure to kilometers per million years, you'd multiply by 1 million, giving you a seafloor spreading rate of 25 km/million years.
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What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?
Answer:
The frequency is 302.05 Hz.
Explanation:
Given that,
Speed = 18.0 m/s
Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .
We need to calculate the frequency
Using formula of frequency
[tex]f'=f(\dfrac{v+v_{p}}{v-v_{s}})[/tex]
Where, f = frequency
v = speed of sound
[tex]v_{p}[/tex] = speed of passenger
[tex]v_{s}[/tex] = speed of source
Put the value into the formula
[tex]f'=262\times(\dfrac{344+18}{344-30})[/tex]
[tex]f'=302.05\ Hz[/tex]
Hence, The frequency is 302.05 Hz.
A 1.25 in. by 3 in. rectangular steel bar is used as a diagonal tension member in a bridge truss. the diagonal member is 20 ft long, and its modulus of elasticity is 30,000,000 psi. if the strain in the diagonal member is measured as 0.001200 in./in., determine:
Answer:
axial stress in the diagonal bar =36,000 psi
Explanation:
Assuming we have to find axial stress
Given:
width of steel bar: 1.25 in.
height of the steel bar: 3 in
Length of the diagonal member = 20ft
modulus of elasticity E= 30,000,000 psi
strain in the diagonal member ε = 0.001200 in/in
Therefore, axial stress in the diagonal bar σ = E×ε
= 30,000,000 psi× 0.001200 in/in =36,000 psi
The stress in the bridge truss diagonal tension member is computed to be 36000 psi, and the elongation of the member under this stress is calculated to be approximately 0.384 inches.
Explanation:The subject of the question is in the field of Engineering, particularly dealing with the stress and strain effects in a steel structural member of a bridge.
The stress in the member can be calculated using stress and strain relationship along with Young's modulus, given by the formula Stress = Strain x Young's Modulus. Given that strain is 0.001200 in./in. and Young's Modulus (E) is 30,000,000 psi, stress (σ) in the member is σ = E x Strain = 30,000,000 psi x 0.001200 in./in. = 36000 psi.
The steel bar's cross-sectional area can also be calculated as Area = Width x Depth = 1.25 in. x 3 in. = 3.75 in^2. The change in length (∆L) of the member under this stress can be calculated using the formula ∆L = (Stress x Original length) / (Young's Modulus x Area) = (36000 psi x 20 ft) / (30,000,000 psi x 3.75 in^2) = 0.032 ft or 0.384 in.
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