Explanation:
The given data is as follows.
Initial temperature ([tex]T_{1}[/tex]) = [tex]40^{o}F[/tex] = 499.67R
Initial pressure ([tex]P_{1}[/tex]) = 16.8 Psi
Volume (V) = 5 [tex]ft^{3}[/tex]
Work done = 4 Btu
Final temperature ([tex]T_{2}[/tex]) = 90 F
So, R of [tex]O_{2}[/tex] = 0.3353 Psi ft^{3}/lbm R
Therefore, we will calculate the mass of oxygen as follows.
M = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{16.8 \times 5}{0.3353 \times 499.67}[/tex]
= 0.50137 lbm
Therefore, mass of oxygen is 0.50137 lbm.
Now, we will calculate the change in internal energy as follows.
[tex]\Delta U = mC_{v} [T_{2} - T_{1}][/tex]
= [tex]0.50137 \times 0.157 \times (90 - 40)[/tex]
= 3.936 BTU
Relation between heat energy and internal energy is as follows.
[tex]\Delta Q = \Delta u + W.d[/tex]
= [tex]3.936 + (-4)[/tex]
= -0.0642 BTU
Therefore, amount of heat transfer in BTU is 0.0642 BTU.
A 10kg block is attached to a light cord that is wrapped around the pulley of an electric motor. What is the power output of the motor..?
when it is pulling the block upward with an instantaneous speed of 3m/s and an upward acceleration of 2 m/s?
a)300W b)360 c)600 d)240 e)480
Answer:
Option B is correct.
Power = 360 W
Explanation:
Power = Work done/time
Work done = Force × distance moved through by the force
Power = Force × (distance moved through by the force/time)
(Distance moved through by the force/time) = velocity = 3 m/s
Power = Force × velocity
Force = ma
But the acceleration in this case is this acceleration + acceleration due to gravity because the force has to be overcoming the force of gravity to now move the object upward at 2 m/s²
a = (2 + g) (assume acceleration due to gravity = 10 m/s²
a = 2 + 10 = 12 m/s²
F = ma = 10 × 12 = 120 N
Power = F × v = 120 × 3 = 360 W
Wayne exerts a force of 63 N to pull a 308 N sled along a snowy path using a rope that makes a 33° angle with the ground. The sled moves 11.3 m in 3.1 s. What is Wayne’s power? Answer in units of W.
Answer:
P = 192.6 Watt
Explanation:
given,
Force of Pull, F = 63 N
Weight of the sled, W = 308 N
Angle made with ground, θ = 33°
Movement of sled,s= 11.3 m
time, t = 3.1 s
Power of Wayne = ?
Work done = F .s cos θ
W = 63 x 11.3 x cos 33°
W = 597 J
We know.
[tex]Power = \dfrac{Work\ done}{time}[/tex]
[tex]P= \dfrac{597}{3.1}[/tex]
P = 192.6 Watt
The Wayne's Power is equal to P = 192.6 Watt
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above the top of the window was the rock dropped? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Answer:
h = 22.35 m
Explanation:
given,
initial speed of the rock,u = 0 m/s
length of the window,l = 2.7 m
time taken to cross the window,t = 0.129 s
Speed of the rock when it crosses the window
[tex]v = \dfrac{l}{t}[/tex]
[tex]v = \dfrac{2.7}{0.129}[/tex]
v = 20.93 m/s
height of the building above the window
using equation of motion
v² = u² + 2 g h
20.93² = 0² + 2 x 9.8 x h
h = 22.35 m
Hence, the height of the building above the top of window is equal to h = 22.35 m
A small crane has a motor that exerts 2.41 times 10^7 Pa of pressure on a fluid chamber. If the piston has an area of 168 cm2 how much force does the piston exert?
Answer:
[tex]F = 4.05\times 10^{9}\ N[/tex]
Explanation:
Given:
Exerts pressure
[tex]p = 2.41\times 10^{7}\ Pa[/tex]
Piston has an area
[tex]A = 168\ cm^{2][/tex]
We need to find the force exerted by piston.
Solution:
Formula of pascals law's is given as:
[tex]p=\frac{F}{A}[/tex]
Where:
F = Force applied
P = Pressure exerted
A = Cross sectional area
Substitute force and area in pascal's formula.
[tex]2.4\times 10^{7}=\frac{F}{168}[/tex]
Using cross multiplication.
[tex]F = (2.41\times 10^{7})\times 168[/tex]
[tex]F = 404.88\times 10^{7}[/tex]
[tex]F = 4.05\times 10^{9}\ N[/tex]
Therefore, the force exerted by piston [tex]F = 4.05\times 10^{9}\ N[/tex]
The piston exerts a force of 404,880 Newtons when the motor exerts a pressure of 2.41 x 10⁷ Pa on a fluid chamber with a piston area of 168 cm².
To determine the force exerted by the piston, we use the formula:
Force (F) = Pressure (P) × Area (A)
Given:
Pressure (P) = 2.41 x 10⁷ PaArea (A) = 168 cm²First, convert the area from cm² to m²:
1 m² = 10,000 cm²
A = 168 cm² / 10,000 = 0.0168 m²
Now, calculate the force:
F = 2.41 × 10⁷ Pa × 0.0168 m²
F = 404,880 N
Thus, the piston exerts a force of 404,880 Newtons.
The light-gathering power of a telescope is directly related to the area of the telescope's primary mirror. A mirror with four times the diameter of another mirror collects how many times more light as the smaller mirror does in the same amount of time?
Answer:
16 times
Explanation:
Generally, the light-gathering power of the mirror of a telescope is dependent on the area of the mirror. The area of the mirror is (π*d^2)/4. The variable 'd' is the diameter of the mirror. Therefore, if the diameter of A is four times the diameter of B, the light-gathering power of A will be (π*4^2)/4 while that of B will be (π*1^2)/4. This shows that A has 16 times that of B.
A mirror with four times the diameter of another mirror collects 16 times more light than the smaller mirror in the same amount of time.
The light-gathering power of a telescope is directly proportional to the area of its primary mirror. The area A of a circular mirror is given by the formula [tex]\( A = \pi \left(\frac{D}{2}\right)^2 \)[/tex], where D is the diameter of the mirror.
If one mirror has four times the diameter of another, we can denote the diameters as D for the smaller mirror and [tex]\( 4D \)[/tex] for the larger mirror.
The area of the smaller mirror is:
[tex]\[ A_{\text{small}} = \pi \left(\frac{D}{2}\right)^2 = \pi \frac{D^2}{4} \][/tex]
The area of the larger mirror is:
[tex]\[ A_{\text{large}} = \pi \left(\frac{4D}{2}\right)^2 = \pi \left(2D\right)^2 = \pi 4D^2 \][/tex]
To find out how many times more light the larger mirror collects compared to the smaller mirror, we divide the area of the larger mirror by the area of the smaller mirror:
[tex]\[ \frac{A_{\text{large}}}{A_{\text{small}}} = \frac{\pi 4D^2}{\pi \frac{D^2}{4}} = \frac{4D^2}{\frac{D^2}{4}} = 4 \times 4 = 16 \][/tex]
The Pentium 4 Prescott processor, released in 2004, had a clock rate of 3.6 GHz and voltage of 1.25 V. Assume that, on average, it consumed 10 W of static power and 90 W of dynamic power.The Core i5 Ivy Bridge, released in 2012, had a clock rate of 3.4 GHz and voltage of 0.9 V. Assume that, on average, it consumed 30 W of static power and 40 W of dynamic power.Find the percentage of the total dissipated power comprised by static power for the Pentium 4 Prescott. Round to a whole integer between 0-1
Answer:
For Pentium 4 Prescott:
% of Static Power = 10
For core i5 Ivy Bridge:
% of Static Power = 43
Given Information:
Static Power of P4 = 10 W
Dynamic Power of P4 = 90 W
Static Power of i5 = 30 W
Dynamic Power of i5 = 40 W
Required Information:
% of static power w.r.t total power dissipation = ?
Explanation:
For Pentium 4 Prescott:
% of static power = static power/total power * 100
% of static power = 10/(10 + 90) * 100
% of static power = 10/(100) * 100
% of static power = 10
For core i5 Ivy Bridge:
% of static power = static power/total power * 100
% of static power = 30/(30 + 40) * 100
% of static power = 30/(70) * 100
% of static power = 43 (rounded to nearest whole integer)
Final answer:
The static power comprises 10% of the total dissipated power for the Pentium 4 Prescott processor.
Explanation:
To calculate the percentage of the total dissipated power that is comprised of static power for the Pentium 4 Prescott, we need to add the static power to the dynamic power to get the total power and then find what percentage the static power is of the total power.
Total power for Pentium 4 Prescott = Static power + Dynamic power
Total power for Pentium 4 Prescott = 10 W + 90 W
Total power for Pentium 4 Prescott = 100 W
Percentage of static power = (Static power / Total power) imes 100
Percentage of static power = (10 W / 100 W) imes 100
Percentage of static power = 10%
So, the static power comprises 10% of the total dissipated power for the Pentium 4 Prescott.
A cylinder which is in a horizontal position contains an unknown noble gas at 54700 Pa 54700 Pa and is sealed with a massless piston. The piston is slowly, isobarically moved inward 0.150 m 0.150 m , while 16800 J 16800 J of heat is removed from the gas. If the piston has a radius of 0.272 m 0.272 m , calculate the change in internal energy Δ U ΔU of the system.
Answer:
-14892.93 J
Explanation:
given,
Pressure, P = 54700 Pa
heat removed, Q = 16800 J
radius, r = 0.272 m
distance, d = 0.150 m
internal energy ΔU of the system = ?
We know,
Force = Pressure x area
F = P x A
F = 54700 x π x r²
F = 54700 x π x 0.272²
F = 12713.79 N
Work done = F.d
W = 12713.79 x 0.15
W = 1907.07 J
Change in internal energy ΔU is
ΔU = W + Q
= 1907.07 + (-16800)
= -14892.93 J
Hence, the change in internal energy is equal to -14892.93 J
Aristotle said that a moving earthly or `mundane' object with nothing pushing or pulling on it will always A slow down and stop. B speed up. C keep moving at the same speed. D follow a circular path.
Answer:
A slow down and stop
Explanation:
When there is no force acting on something it automatically begins to slow down and then stops.Essentially, Aristotle's perspective of motion is that "it requires a force to move an object in an unnatural" way— or, plainly, that "movement involves strength." Indeed, if you propel a book, it keeps moving. Once you stop trying to push, it comes to a stop.
Recoil is noticeable if we throw a heavy ball while standing on a skateboard. If instead we go through the throwing motion but hold onto the ball, without friction effects our net recoil will be
A) zero.
B) the same as before.
C) small, but noticeable.
Answer:
A) zero.
Explanation:
If we go through the throwing motion but still hold onto the ball, then there will be 2 events:
1. you undergo the throwing motion, giving the ball's acceleration
2. you decelerate the ball so you can still hold onto it
Event 1 and 2 roughly need the same energy, but with the opposite vector. The sum of energy used for both events should be zero so the net recoil will also be zero too.
person walks in the following pattern: 2.3 km north, then 2.5 km west, and finally 5.4 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point
Answer:
a.3.86 Km
b.Direction=233.3 degree
Explanation:
We are given that
Person wale in North direction,DE=2.3 Km
Person walk in West direction==BD=2.5 Km
Person walk in South direction,EC=5.4 Km
a.We have to find the distance between initial and final position.
Vector AB=-2.5 Km
BD=-EA=-2.3 Km
CA=y=-(EC-EA)=-(5.4-2.3))=-3.1 m
AB=-DE=x=-2.5 Km
Magnitude of resultant vector=[tex]\sqrt{AB^2+CA^2}=\sqrt{(-2.5)^2+(-3.1)^2}[/tex]
Magnitude of resultant vector=3.86 Km
Hence, the distance between initial and final position=3.86 Km
b.Direction=[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
Direction=[tex]\theta=tan^{-1}(\frac{-3.1}{-2.3})\approx 53.3^{\circ}[/tex] S of W
x-coordinate and y-coordinate are negative therefore, the angle lies in third quadrant.
When we measured from East then
[tex]\theta=53.3+180=233.3^{\circ}[/tex]
Therefore, the direction=233.3 degree
An unknown weak acid with a concentration of 0.090 M has a pH of 1.80. What is the Ka of the weak acid?
Answer : The value of [tex]K_a[/tex] of the weak acid is, [tex]3.36\times 10^{-3}[/tex]
Explanation : Given,
Initial concentration = 0.090 M
pH = 1.80
First we have to calculate the hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
[tex]1.80=-\log [H^+][/tex]
[tex][H^+]=0.0158M[/tex]
Now we have to calculate the [tex]K_a[/tex] of the weak acid.
The dissociation reaction of weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Initial conc. 0.090 0 0
At eqm. (0.090-x) x x
x = 0.0158 M
The expression for dissociation constant is:
[tex]K_a=\frac{(x)\times (x)}{(0.090-x)}[/tex]
Now put all the given values in this expression, we get:
[tex]K_a=\frac{(0.0158)\times (0.0158)}{(0.090-0.0158)}[/tex]
[tex]K_a=3.36\times 10^{-3}[/tex]
Thus, the value of [tex]K_a[/tex] of the weak acid is, [tex]3.36\times 10^{-3}[/tex]
The value of Ka of the weak acid with a concentration of 0.090 M and has a pH of 1.80 is 3.36 × 10-³.
How to calculate Ka of an acid?To calculate the Ka of an acid, we have to calculate the hydrogen ion concentration of the acid using the following expression:
pH = -log {H+}
1.80 = -log {H+}
{H+} = 0.0158M
The dissociation equation is given as follows:
HA ⇌ H+ + A-
Ka = 0.0158²/(0.090 - 0.0158)
Ka = 2.49 × 10-⁴/7.42 × 10-²
Ka = 0.336 × 10-²
Ka = 3.36 × 10-³
Therefore, the value of Ka of the weak acid with a concentration of 0.090 M and has a pH of 1.80 is 3.36 × 10-³.
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The development of solar panels for home installation has improved due to technological advances and subsidies. These factors shift the supply curve to the _____ and result in the equilibrium price of solar panel installations to _____.
right; fall
left; rise
left; fall
right; rise
Answer:
The correct answer is
right; fall
Explanation:
When there are changes in the costs of the factors of production, it can result in the supply curve shifting to the right or to the left. That means the quantity supplied at the given price has either increased or decreased, hence the supply curve shows the relationship between quantity supplied at a given price
Factors that can cause a shift in supply curve include, prices of imput materials, increased competition, technological advancement, social or natural factors and general expectations.
When the supply curve shifts right, that means increased in supply or more units are available at a given price, hence prices fall. However if the supply curve shifts left that means less goods are available at a given market price, hence prices of the commodities will rise
A startled armadillo leaps upward, rising 0.587 m in the first 0.193 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.587 m? (c) How much higher does it go? Use g=9.81 m/s2.
Answer:
a) Initial speed as it leaves the ground is 3.99 m/sb) Speed at the height of 0.587 m is 2.10 m/sc) Height reached is 0.81 mExplanation:
a) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = ?
Acceleration, a = -9.81 m/s²
Time, t = 0.193 s
Displacement, s = 0.587 m
Substituting
s = ut + 0.5 at²
0.587 = u x 0.193 + 0.5 x -9.81x 0.193²
u = 3.99 m/s
Initial speed as it leaves the ground is 3.99 m/s
b) We have equation of motion v = u + at
Initial velocity, u = 3.99 m/s
Final velocity, v = ?
Time, t = 0.193 s
Acceleration, a = -9.81 m/s²
Substituting
v = u + at
v = 3.99 + -9.81 x 0.193
v = 2.10 m/s
Speed at the height of 0.587 m is 2.10 m/s
c) We have equation of motion v² = u² + 2as
Initial velocity, u = 3.99 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = 0 m/s
Substituting
v² = u² + 2as
0² = 3.99² + 2 x -9.81 x s
s = 0.81 m
Height reached is 0.81 m
According to Coulomb’s Law, the force between two charged objects is related to _____.
a. the inverse of the square of the distance separating them
b. the distance separating them
c. the inverse of the charges of the objects
d. the mass of the objects
Answer:
A.) the inverse of the square of the distance separating them
Explanation:
Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."
Mathematically, F = kq1q2/r²
Where q1 and q2 are the charges
r is the distance between the charges.
According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.
Coulomb’s Law states that the force between two charged objects is directly related to the inverse of the square of the distance between them.
Explanation:According to Coulomb’s Law, the force between two charged objects is related to (a) the inverse of the square of the distance separating them. This law states that the force (F) between two charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. It's written as F = k * q1 * q2 / r^2, where k is Coulomb's constant. Thus, if the distance between the objects is doubled, the force between them is reduced to one fourth of its original value.
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A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions:
a. when the ropes are horizontal
b. when the ropes make a 30.0° angle with the vertical
c. at the bottom of the circular arc
Answer:
A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.
Gpe @ A=mgh=40*9.81*2=784.8J
B. At 30degree vertical angle the vertical displacement from lowest position is given by
2-2cos(30)=2-1.73=0.27m
Gpe @B= 40*9.81*0.27=106 J
C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.
(a) The gravitational potential energy when the ropes are horizontal is 784.8 J.
(b) The gravitational potential energy when the ropes make a 30 degree angle with vertical is 106 J.
(c) At the bottom of circular arc the gravitational potential energy is zero.
Given data:
The mass of child is, m = 40.0 kg.
The length of ropes is, L = 2 m.
The energy possessed by any object under the influence of gravity and by the virtue of position of object is known as gravitational potential energy.
(a)
As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position. Then the gravitational potential energy is,
[tex]PE = mgL\\\\PE = 40 \times 9.8 \times 2\\\\PE=784.8 \;\rm J[/tex]
Thus, the gravitational potential energy when the ropes are horizontal is 784.8 J.
(b)
At 30 degree vertical angle the vertical displacement from lowest position, the gravitational potential energy is given by,
[tex]PE = mgL(1-cos30^{\circ})\\\\PE = 40 \times 9.8 \times 2 \times (1-cos30^{\circ})\\\\PE = 106 \;\rm J[/tex]
Thus, the gravitational potential energy when the ropes make a 30 degree angle with vertical is 106 J.
(c)
At the bottom of circular arc it's gravitational potential energy is zero relative to lowest position as bottom of arc itself is lowest position.
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A 100 kg cart goes around the inside of a vertical loop of a roller coaster. The radius of the loop is 3 m and the cart moves at a speed of 6 m/s at the top. The force exerted by the track on the cart at the top of the loop is ________.
Final answer:
The force exerted by the track on the cart at the top of the loop is 220 N, calculated by the difference between the centripetal force required for circular motion and the gravitational force acting on the cart.
Explanation:
The student is asking for the force exerted by the track on a 100 kg cart at the top of a vertical loop, where the cart has a speed of 6 m/s and the loop's radius is 3 m. At the top of the loop, the cart's weight and the normal force from the track provide the centripetal force necessary to keep the cart moving in a circle. To calculate the total force exerted by the track on the cart, we need to account for both the cart's weight (gravitational force) and the centripetal force required to keep it in circular motion.
The gravitational force (Fg) can be calculated using the equation Fg = m × g, where m is the mass of the cart and g is the acceleration due to gravity. The centripetal force (Fc) required for circular motion at the top of the loop can be calculated with the formula Fc = m × v² / r, where v is the speed of the cart and r is the radius of the loop.
The force exerted by the track (Ft) is then the difference between the centripetal force and the gravitational force because, at the top of the loop, the track has to push down on the cart to provide the inward centripetal force while also supporting the cart's weight.
Let's calculate each force:
Gravitational Force (Fg): Fg = 100 kg × 9.8 m/s² = 980 N
Centripetal Force (Fc): Fc = (100 kg) × (6 m/s)² / (3 m) = 1200 N
Now, we calculate the total force exerted by the track at the top of the loop:
Total Force by Track (Ft):
Ft = Fc - Fg
Ft = 1200 N - 980 N = 220 N
Therefore, the force exerted by the track on the cart at the top of the loop is 220 N.
Find the distance between the cities. Assume that Earth is a sphere of radius 4000 miles and that the cities are on the same longitude (one city is due north of the other). (Round your answer to one decimal place.)
Answer: 20231.75km
Explanation: Assuming the earth is a sphere with a radius r, the formulae for calculating the distance between 2 points on the earth surface is given as
[tex]l=\frac{\pi r}{180} *\alpha[/tex]
l = distance between two points on the earth surface (km)
π = 22/7
r = radius of the earth = 4000miles =6437.376 km ( we will be using kilometer because that is the SI unit for distance)
α = angle between the two cities on the earth surface. For our question, α=180 because one of the city is north and the second city is in the opposite direction (due north of the first)
By slotting in the parameters into the foemulae, we have that
[tex]l = \frac{\frac{22}{7} * 6437.376 }{180} * 180 \\\\l={\frac{22}{7} * 6437.376\\\\\\l=20231.751km[/tex]
What evidence did Sir Isaac Newton use to support his particle explanation of light? Light reflects off metal surfaces. Light travels in a straight line, and it casts a shadow. Light produces an interference pattern. Light causes electrons to be emitted from a metal surface.
Answer:
Isaac Newton supported his particle explanation of light by stating that light causes electrons to be emitted from a metal surface.
Explanation: The emission of electrons from metal surfaces when light of sufficient frequency falls on them is known as photoelecteic effect. He discovered photoelectric effect in the 19th century.
His other theories of light like light travels in a straight line could not explain diffraction and reflection. It needed the support of light as a wave to proof it.
Answer:
light travels in a straight line, and it casts a shadow.
Explanation:
Your textbook discusses the cosmic calendar, a model of the history of the universe scaled to a single year. The length of time represented by one month on this cosmic calendar is therefore closest to__________
Answer:
1.15 Billion Years
Explanation:
If the average age of the universe is 13.772 billion years and that equals to a year in cosmic calendar, then the length of time in a month will be 13.772 / 12 = 1.14766666667 ~ 1.15 Billion Years
The ratio of girls to boys in a an astronomy club was 1 : 8. There were 5 girls. How many total members were there in the club?
Answer:
45
Explanation:
If the ratio is 1 : 8 and there was 5 girls you would need to find the ratio of 5 girls to x amount of boys then add both genders together to find the total amount. You would need to multiply 8x5 to find the amount of boys.
8x5=40
Then add the five girls
40+5=45
Process in which permanent deformation of metals occurs due to applied stress and results in breaking of bonds and then reforming of bonds with new neighbors. After removing of the stress, the metal does not return to its initial form.
Answer:
This process involves the motion of dislocations and is termed slip (or glide in some textbooks)
Explanation:
Plastic deformation of metals (and other crystalline materials) usually occurs by slip, which is the sliding of planes of atoms over one another by dislocation movements.
On a microscopic scale, stress causes planes of crystalline objects to leave their original position and slide over other planes into new positions, these microscopic movements manifest as a slip on a macroscopic scale. And the planes do not return back to their original position after the removal of the dislocation-causing stress.
Skid is floating in space 3 m from Mitch. Skid and his backpack have a combined mass 50 kg. Skid releases his 10 kg backpack, which reduces his mass. If the same gravitational force is to be maintained between them, what will be the new distance between Skid and Mitch.?
Answer:
Explanation:
attached below is the working to the question asked. I equated the two formulas because of the statement "If the same gravitational force is to be maintained between them" mentioned in the question. The new distance between them is 2.68m
A statue is to be 'scaled down.' It will have its size changed without changing its shape. It starts with an initial volume of 3.25 m3 and ends up with a final volume of 1.00 m3 .
Answer:
The solution are in the explanation
Explanation:
Volume varies with the cube of the linear dimension
Ratio of volumes is 3:1, therefore the ratio of the linear dimensions is 3^(1/3):1 = 1.442:1
Smaller statue is 350/1.442 cm = 242.7 cm
C = pi*D, D is a linear dimension, so Ds = 50/1.442 = 34.67 cm
Area varies with the square of the linear dimension; so the ratio of the areas will be (1/3)^(2/3) = 0.48; the surface area of the smaller statue will be 0.48 that of the original
The mass of the model would vary directly with the volume, so the mass of the model will be 1/3 that of the original
If the masses of the models are the same, then the model must be made of a material 3x the density of that used for the original.
In mathematics, scaling down a shape involves reducing its size while keeping its shape unchanged. To scale down a statue, you need to determine the scale factor by dividing the final volume by the initial volume. Then, apply the scale factor to each dimension of the statue.
Explanation:In mathematics, scaling down a shape means reducing its size while keeping its shape unchanged. The process involves multiplying the dimensions of the shape by a scale factor less than 1.
In this case, the initial volume of the statue is 3.25 m3 and it needs to be scaled down to a final volume of 1.00 m3. To determine the scale factor, divide the final volume by the initial volume: 1.00/3.25 = 0.307.
Next, apply the scale factor to each dimension of the statue. If the dimensions of the statue were 2m x 2m x 2m, the scaled-down dimensions would be 2m x 2m x 2m x 0.307 = 0.614m x 0.614m x 0.614m.
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a 800 kg roller coaster cart is accelerated by a constant net force over a distance of 10 meters, as shown in the graph below: determine the speed of the cart after being accelerated for 10.0 meters.
Answer:
vf = 22.36[m/s]
Explanation:
First we must understand the data given in the problem:
m = mass = 800 [kg]
F = force = 20000[N]
dx = displacement = 10[m]
From newton's second we know that the sum of forces must be equal to the product of mass by acceleration.
[tex]F = m*a\\20000 = 800*a\\a = 20000/800\\a = 25 [m/s^2][/tex]
With the calculated acceleration, we can use the kinematics equations.
[tex]v_{f} ^{2} =v_{o} ^{2}+2*a*dx\\ v_{o} = initial velocity = 0\\a = acceleration = 25[m/s^2]\\dx= displacement = 10[m]\\[/tex]
The key to using this equation is to clarify that the initial velocity is zero since the body is at rest, otherwise the initial velocity would be an initial data.
[tex]v_{f} =\sqrt{2*25*10} \\v_{f} =22.36[m/s][/tex]
Another way of solving this problem is by means of the definition of work and kinetic energy, where work is defined as the product of the force by the distance.
W =F*d
W = 20000*10
W = 200000[J]
Kinetic energy is equal to work, therefore the value calculated above is equal to:
[tex]E_{k}=W =0.5*m*v_{f}^{2} \\200000=0.5*800*v_{f}^{2}\\v_{f}=\sqrt{\frac{200000}{0.5*800} } \\v_{f}=22.36[m/s][/tex]
Answer:
(A) 22.4 m/s
Explanation: hope this helps !
"Compared to infrared radiation, does ultraviolet radiation have longer or shorter wavelengths? Does ultraviolet radiation have higher or lower energy per photon?"
Answer:Ultraviolet radiation has shorter wavelengths and higher energy than infrared radiation.
Explanation: Electromagnetic radiation radiations which have both electrical and magnetic properties,they can be transmitted through space or through a medium.
It includes Gamma radiation, infra-red, visible light, Ultraviolet radiation etc they occur with different wavelength, the lower the wavelength the higher the Energy dissipated per photon. According to their order of decreasing wavelength and increased energy they are classified as follows.
RADIO WAVE, MICRO WAVE, INFRA-RED, VISIBLE LIGHT, ULTRAVIOLET RAY, X-RAY, GAMMA RAYS.
Ultraviolet radiation has shorter wavelengths and higher energy per photon compared to infrared radiation.
Explanation:Ultraviolet radiation has shorter wavelengths compared to infrared radiation. While infrared radiation has wavelengths between 700 nm and 1 mm, ultraviolet radiation has wavelengths between 100 nm and 400 nm.
Ultraviolet radiation also has higher energy per photon compared to infrared radiation. This is because energy is inversely proportional to wavelength, so shorter wavelengths have higher energy.
For example, ultraviolet radiation is responsible for causing sunburns and skin damage, which is evidence of its higher energy compared to infrared radiation.
Learn more about Ultraviolet radiation here:https://brainly.com/question/34609430
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When point charges q1 = +1.2 μC and q2 = +4.1 μC are brought near each other, each experiences a repulsive force of magnitude 0.67 N. Determine the distance between the charges.?
Answer:
2.57 cm or 2.57×10⁻² m
Explanation:
From coulomb's law,
F = kqq'/r²................. Equation 1
Where F = force between the charges, q and q' = The first and second charge respectively, k = constant of proportionality, r = distance between the charges.
Making r the subject of equation 1
r = √(kqq'/F)................ Equation 2
Given: F = 0.67 N, q = 1.2 μC = 1.2×10⁻⁶ C, q' = 4.1×10⁻⁶ C
Constant: k = 9×10⁹ Nm²/C².
Substitute into equation 2
r = √( 1.2×10⁻⁶×4.1×10⁻⁶×9×10⁹/0.67)
r = √(66.09×10⁻³)
r = √(6.609×10⁻⁴)
r = 2.57×10⁻² m
r = 2.57 cm or 2.57×10⁻² m
Hence the distance between the charge = 2.57 cm or 2.57×10⁻² m
In which situations can you conclude that the object is undergoing a net interaction with one or more other objects? (Select all that apply.
1)A car travels at constant speed around a circular race track.
2)A book slides across the table and comes to a stop.
3)A hydrogen atom remains at rest in outer space.
4)A proton in a particle accelerator moves faster and faster.
5)A spacecraft travels at a constant speed toward a distant star.
Answer:
4, A proton in a particle accelerator moves faster and faster.
Explanation:
A particle accelerator according to Wikipedia.com is a machine that uses electromagnetic fields to propel charged particles to very high speeds and energies, and to contain them in well-defined beams. Large accelerators are used for basic research in particle physics.
As the proton moves around, it collides with other proton and the wall of the accelerator thereby undergoing net interaction.
If the speed of the wave on the guitar string is 600 m/s. Determine the period of that wave. Show all of your work including the initial equations and include the units in your answer
Answer:
The period is (0.0017×wavelength) seconds
Explanation:
Speed = frequency × wavelength
Frequency = 1/period
Therefore, speed = 1/period × wavelength
Speed = wavelength/period
Period = wavelength/speed = wavelength/600 = (0.0017×wavelength) seconds
A 150 g baseball pitched at a speed of 45 m/s is hit straight back to the pitcher at a speed of 60 m/s. What is the magnitude of the average force on the ball from the bat if the bat is in contact with the ball for 9.5 ms?
Answer:
The average force is 1578.94 N.
Explanation:
Given that,
Mass of baseball = 150 g
Speed = 45 m/s
Speed of pitcher = 60 m/s
Time = 9.5 ms
We need to calculate the average force
Using formula of impulse
[tex]J=\Delta p[/tex]
[tex]J=m\Delta v[/tex]...(I)
[tex]J=F\Delta t[/tex]....(II)
From equation (I) and (II)
[tex]F=\dfrac{m(v_{f}-v_{i})}{\Delta t}[/tex]
Where, m = mass of baseball
[tex]v_{f}[/tex] = final velocity
[tex]v_{i}[/tex] = Initial velocity
[tex]\Delta t[/tex] = time
Put the value into the formula
[tex]F=\dfrac{150\times10^{-3}\times(60-(-40))}{9.5\times10^{-3}}[/tex]
[tex]F=1578.94\ N[/tex]
Hence, The average force is 1578.94 N.
Answer:
F = 1657.89 N
Explanation:
given,
mass of the baseball, m = 150 g
initial speed, u = 45 m/s
final speed, v = 60 m/s
time of contact,t = 9.5 ms
we know,
impulse is equal to change in momentum
J = m (v - u)
J = 0.15 x (60-(-45))
J = 0.15 x 105
J = 15.75 Kg.m/s
We also know that impulse
J = F x Δ t
F x 9.5 x 10⁻³ = 15.75
F = 1657.89 N
The magnitude of the average force is equal to F = 1657.89 N
A circular plastic disk with radius R = 1.80 cm has a uniformly distributed charge of Q = +(2.05 ✕ 106)e on one face. A circular ring of width 30 μm is centered on that face, with the center of the ring at radius r = 0.50 cm. What charge is contained within the width of the ring?
The charge contained within the width of the ring on the plastic disk is approximately [tex]\(1.89 \times 10^6\)[/tex] elementary charges (e).
To find the charge contained within the width of the ring, we'll first calculate the total charge of the disk and then subtract the charge within the inner circle defined by the radius of 0.5 cm.
1. Calculate the total charge of the disk:
Total charge, Q_total = Q = +(2.05 ✕ 10^6)e
2. Calculate the area of the entire disk:
Area_disk = π * R^2
= π * (1.80 cm)^2
= 10.17 cm^2
3. Calculate the area of the inner circle:
Area_inner_circle = π * (0.50 cm)^2
= π * 0.25 cm^2
= 0.785 cm^2
4. Calculate the area of the ring:
Area_ring = Area_disk - Area_inner_circle
= 10.17 cm^2 - 0.785 cm^2
= 9.385 cm^2
5. Calculate the charge density of the disk:
Charge density, σ = Q_total / Area_disk
= (2.05 ✕ 10^6)e / 10.17 cm^2
≈ 201,183.43 e/cm^2
6. Calculate the charge within the width of the ring:
Charge_ring = σ * Area_ring
≈ 201,183.43 e/cm^2 * 9.385 cm^2
≈ 1.89 ✕ 10^6 e
So, the charge contained within the width of the ring is approximately 1.89 ✕ 10^6 e.