Answer:
V2=42.93cm3
Explanation:
According to Gay-Lussac's Law: "The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. With an increase in temperature, the pressure will go up."
P1V1=P2V2
Given data
P1=4.25atm
V1=10.1cm³
P2=1atm
V2=?
V2=P1V1/P2
V2=(4.25*10.1)/1
V2=42.93cm3
A proton in a cyclotron has a velocity of 2x108 m/s, which is nearly the speed of light. The uncertainty in the velocity is 1%. What is the minimum uncertainty in the position of the proton in meters
Answer:
The minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m
Explanation:
Given;
velocity of the proton, v = 2 x 10⁸ m/s
uncertainty in the velocity, = 1 % = 0.01 v = 0.01 x 2 x 10⁸ m/s = 2 x 10⁶ m/s
The momentum of the proton is given by;
ΔP = mv = (1.67 x 10⁻²⁷)(2 x 10⁶ ) = 3.34 x 10⁻²¹ kgm/s
The minimum uncertainty in the position of the proton is given by;
[tex]\delta x = \frac{h}{4\pi \delta P}\\\\ \delta x =\frac{6.626*10^{-34}}{4\pi *3.34*10^{-21}}\\\\ \delta x = 1.578*10^{-14} \ m[/tex]
Therefore, the minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m
The log left the cliff with a horizontal velocity of 7.5 m/s. When it reached the water, it had a vertical velocity of -8.9 m/s. What was the log's speed (two dimensional) before it entered the water
Answer:
v = 11.6 m/s
Explanation:
Assuming no other forces acting upon the log than gravity (which acts only in the vertical direction) in the horizontal direction, velocity must be constant at any point.So, before it entered the water, its velocity has two components, as follows:vₓ = 7.5 m/svy = -8.9 m/sWe can find the magnitude of the velocity vector, just applying Pythagorean Theorem to both components, as follows:[tex]v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{((7.5m/s)^{2} + (-8.9m/s)^{2}[/tex]
⇒ v = 11.6 m/s
3. A 0.14-kg ball traveling with a speed of 43 m/s is brought to rest in a catcher's mitt.
Determine the magnitude of the impulse exerted by the mitt on the ball.
Answer:
change of momentum
= mass * change in velocity
= 0.14 * (36 - 0)
= 5.04 kg m/s ---answer
Explanation:
What is the potential energy of a 20-kg safe sitting on a shelf 0.5 meters
above the ground?
A-20j
B-10j
C-98j
D-196j
Answer:
C - 98j
Explanation:
Answer:
c- 98j
hope it helped
If you pour liquid into a tall, narrow glass, you may hear sound with a steadily rising pitch. What is the source of the sound
Answer:
the vibration of the glass tube creating sound waves within itself. reason is the rising pitch is that the liquid rise shortens the length of the vibrating area within the tube.
Explanation:
Two equal masses travel in opposite directions with equal speeds. They collide in a collision that is between elastic and inelastic. Just after the collision, their velocities are
Answer:
less in magnitude and opposite in direction to their original velocities
Explanation:
The nature of collisions can be determined by the coefficient of restitution.
Coefficient of restitution is given by
[tex]e=\dfrac{\text{Kinetic energy after collision}}{\text{Kinetic energy before collision}}[/tex]
In the case described here the value of [tex]e[/tex] is in between 0 and 1.
This means that the kinetic energy after collision is less than the kinetic energy before collision.
If the kinetic energy after collision is less this means that the velocity is not balanced with the velocity before the collision however the direction of the masses are opposite Some amount of energy is dissipated as heat or through other forms of energy.
So, just after the collision, their velocities are less in magnitude and opposite in direction to their original velocities.
The John Hancock Center in Chicago is the tallest building in the United States in which there are residential apartments. The Hancock Center is 343 m tall. Suppose a resident accidentally causes a chunk of ice to fall from the roof. What would be the velocity of the ice as it hits the ground
Answer:
Explanation:
The chunk of ice will fall with acceleration of 9.8 m /s²
initial velocity u = 0 ,
v² = u² + 2 gH
H = 343 m
g = 9.8 m /s²
v is final velocity
v² = 0 + 2 x 9.8 x 343
v² = 6723
v = 82 m /s approx .
The velocity of the the chunk of ice as it hits the ground is 82m/s.
Given the data in the question;
Height of John Hancock Center; [tex]h = 343m[/tex]Before the chunk of ice to fell off the roof, its was initially at rest, so
Initial velocity; [tex]u = 0[/tex]Final velocity as the ice hits the ground; [tex]v = \ ?[/tex]
To determine the final velocity, we use the third equation of motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is final velocity, u is the initial velocity, s is distance or height of the building and a is acceleration due to gravity{ since the ice will be under gravity as it falls, ([tex]a = g = 9.8m/s^2[/tex]) }
We substitute our values into the equation
[tex]v^2 = [0^ 2] + [ 2\ *\ 9.8m/s^2\ *\ 343m]\\\\v^2 = 2\ *\ 9.8m/s^2\ *\ 343m\\\\v^2 = 6722.8m^2/s^2\\\\v = \sqrt{6722.8m^2/s^2}\\\\v = 81.99m/s\\\\v = 82m/s[/tex]
Therefore, the velocity of the the chunk of ice as it hits the ground is 82m/s.
Learn more: https://brainly.com/question/24679384
i. What is the gravitational potential energy added to the steel beam when the beam reaches a height of 18.4 m? (1 point)
Answer:
59,566.32JoulesExplanation:
The question is incomplete. Here is the complete question.
At a construction site, a crane lifts a steel beam with a mass of 330 kg.
i. What is the gravitational potential energy added to the steel beam when the beam reaches a height of 18.4 m? (1 point)
Gravitational potential energy GPE = mgh
m is the mass of the steel beam = 330kg
g is the acceleration due to gravity = 9.81m/s²
h is the height the beam reaches = 18.4m
Substitute
GPE = 330 * 9.81 * 18.4
GPE = 3237.3 * 18.4
GPE = 59,566.32Joules
Hence the gravitational potential energy is 59,566.32Joules
Answer:
The increase in gravitational potential energy is approximately 60,000 J.
Explanation:
GPE=mgh
m=330
g=9.8^2
h=18.4
GPE=330*9.81*18.4
GPE=59,566.32 J
Hope I could help! :)
Calculate the magnitude of the internal normal force, NB, shear force, VB, and moment, MB, at point B.
Answer:
The answer is "[tex]\bold{MB= \frac{L^2 \ WB}{12}}[/tex]"
Explanation:
please find the complete question in the attached file:
From A to B
[tex]reqAB = \int^{\frac{1}{2}}_{0} \frac{WB}{\frac{L}{2}} x \ dx = \frac{L\ WB}{4}[/tex]
From symmetry [tex]Ay = Cy = FeqAB[/tex]
The FeqAB position is found on a table of common centres:
[tex]\frac{2}{3}( \frac{L}{2})[/tex]
[tex]\to \Sigma_{NB=0} FX = NB=0\\\\\to \Sigma_{VB=0} Fy = \frac{LWB}{4} - \frac{LWB}{4} -VB =0 \\\\\to \Sigma M_{.CCW \ A} = \frac{L}{3} - \frac{LWB}{4} - \frac{L}{2}VB+MB =0 \\\\MB= \frac{L^2 \ WB}{12}[/tex]
The most distant galaxy we have observed is more than
13.2 billion light-years away.
What does this evidence indicate?
e
A. That the universe is less than 13.2 billion years old
B. That the universe will exist for at least 13.2 billion more years
C. That the universe is at least 13.2 billion years old
D. That the universe is infinitely large and has always existed
Answer:
That the universe is at least 13.2 billion years old
Explanation:
The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old
What is the universe?
The universe is the sum total of all existent matter and space. The universe is estimated to be at least 10 billion light-years across and contains an enormous number of galaxies;
It has been expanding since the Big Bang some 13 billion years ago.
The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old
Hence option c is correct.
To learn more about the universe refer to the link;
https://brainly.com/question/11987268
The tendon from Lissa’s knee extensor muscles attaches to the tibia bone 1.5 in. (4 cm) below the center of her knee joint, and her foot is 15 in. (38 cm) away from her knee joint. What arc length does Lissa’s foot move through when her knee extensor muscles contract and their point of insertion on the tibia moves through an arc length of 2 in. (5 cm)?
Answer:
s₂ = 0.475 m = 47.5 cm
Explanation:
The arc length and the angle of rotation are related through the formula:
s = rθ
where,
s = arc length
r = radius of curvature
θ = angle of rotation
First, we consider the arc length covered by the point of insertion of extensor muscles.
s₁ = r₁θ
where,
s₁ = arc length covered by insertion of extensor muscle = 5 cm
r₁ = length of insertion from knee = 4 cm
θ = Angle of Rotation = ?
Therefore,
5 cm = (4 cm)(θ)
θ = (5 cm)/(4 cm)
θ = 1.25 rad
Now, we consider the arc length covered by the foot.
s₂ = r₂θ
where,
s₂ = arc length covered by the foot = ?
r₂ = distance from knee to foot = 38 cm = 0.38 m
The angle of rotation will be the same for the foot as the insertion.
Therefore,
s₂ = (0.38 m)(1.25 rad)
s₂ = 0.475 m = 47.5 cm
When trying to stop the train, Spiderman applies a 4000 N force over 200 seconds. What is his impulse on the train?
Answer:
Initial speed of train = 35.76 m/s (80 mph)
I figured out the speed of the train (which equals 80mph) by viewing the movie. In one of the scenes, it is shown that the train reaches 80mph after Doc Ock desroys the break on the train.
1mph = 0.44704mps
80(0.44704) = 35.7632
= 35.76m/s
vf = 0m/s
vi = 35.76m/s
Mass of train = approximately 137,100 kg (302,254 lbs) (found in research)
Time taken to stop train = approximately 46 seconds
a = vf - vi
t
= 0 - 35.76
46
= -.7773913043
= -.78 m/s^2
(The number is negative since the accelerarion of the train is slowing down)
According to Newton's Second Law, a = f/m. When rearranged, the formula reads f = m(a). As shown in the free body diagram, the weight of the train equals the force due to gravity.
(force = tension)
F = m(a)
= 137,100(-.78)
= -106938 N
106938 N is the calculated tension of Spider-Man's webs
I also calculated the distance it took for Spider-Man to stop the train:
Approximate distance traveled by train:
d = 1/2(a)(t)^2
= 1/2(-.78)(46)^2
= 825.24 m
In order for Spider-Man to have been able to completely stop the train before falling off the edge of the tracks, I found that the tension of his webs must have equaled 106938N.
In order to determine the tension of Spider-Man's webs, I needed to know the mass of the train. Since the movie was unclear on what type of trian was used, I researched online to find the mass of the train. Knowing the mass of the train, i was then able to calculate the tension of Spider-Man's webs and the distance the train took to stop.
Explanation:
Suppose a small compact car with a mass of 1000 kg is traveling north on Morewood Avenue at a speed of 15 m/s. At the intersection of Morewood and Fifth Avenues, it collides with a truck with a mass of 2000 kg that is traveling east on Fifth Avenue at 10 m/s. Treating each vehicle as a particle, find the total momentum (magnitude and direction) just before the collision.
Explanation:
We are expected to solve for the total momentum before collision
the expression for momentum is given as
p = mv
p =momentum
m=mass
v=velocity
For small compact truck
m1=1000kg
v1=15m/s
For small truck
m1=2000kg
v1=10m/s
the total momentum can also be expressed as
Ptotal=m1v1+m2v2
Ptotal=1000*15+2000*10
Ptotal=15000+20000
Ptotal=35000kg•m/s
The direction is east because the momentum of the truck is more than the car
did he wash the shoe.change it to passive voice
Answer:
Did he wash the shoes?-Were the shoes washed by him?
Part A What electron current is driven through a 2.3-mm-diameter iron wire by a 0.062 V/m electric field?
We know, Current is given by :
[tex]I = \dfrac{EA}{\rho}[/tex]
Here, [tex]\rho[/tex] is resistivity and for iron.
[tex]\rho=9.71\times 10^{-8}\ \Omega\ m[/tex]
Putting [tex]\rho[/tex] in above equation, we get :
[tex]I = \dfrac{0.062\times \pi\times (2.3\times 10^{-3})^2}{4\times 9.71\times 10^{-8}}[/tex]
[tex]I=2.65\ A[/tex]
Therefore, the electron current is 2.65 A.
Hence, this is the required solution.
A uniform string of length 2.5 m and mass 0.01 kg is placed under a tension of 10 N. (5 pts) A) What is the frequency of the fundamental mode
Answer:
The fundamental frequency is 10 Hz
Explanation:
The frequency of the fundamental mode of the string is given by;
[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }[/tex]
where;
L is length of the string = 2.5 m
T is tension of the string, = 10 N
μ is mass per unit length = m / L = 0.01 / 2.5 = 0.004 kg/m
Substitute the given values and solve for the frequency;
[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }}\\\\ f_o =\frac{1}{2*2.5}\sqrt{\frac{10}{0.004}}\\\\ f_o = 0.2\sqrt{2500}\\\\ f_o = 0.2 *50\\\\f_o = 10 \ Hz[/tex]
Therefore, the fundamental frequency is 10 Hz
A spherical balloon is inflated with gas at a rate of 800 cubic centimeters per minute. (a) How fast is the radius of the balloon changing at the instant the radius is 70 centimeters
Answer:
0.013 cm/minute
Explanation:
We are given;
Volume rate; dV/dt = 800 cm³/min
Now, volume of a sphere is given by the formula;
V = (4/3)πr³
We want to find the rate at which the radius of the balloon is changing. This is represented by dr/dt.
Now, dr/dt will be gotten from the relation;
dr/dt = dV/dt ÷ dV/dr
Now, dV/dr = 3 × (4/3)πr²
dV/dr = 4πr²
Thus;
dr/dt = 800/(4πr²)
At r = 70 cm
dr/dt = 800/(4π × 70²)
dr/dt = 0.013 cm/minute
Two masses, m_1m ​1 ​​ = 3.75 kg and m_2m ​2 ​​ = 6.34 kg, are connect by a string of negligible mass. The string passes over a frictionless pulley so that m_1m ​1 ​​ and m_2m ​2 ​​ hang down on opposite sides of the pulley. The whole system is released from rest. Calculate the tension in the string as the masses are moving.
Answer:
Explanation:
For the mass m1;
The sum of forces acting on the body is expressed according to Newton's second law as;
\sum Fx = m1ax
T - Ff = m1a .... 1
T is the tension
Ff is the frictional force acting on m1
For the mass m2:
\sumFy = m2a
W - T = m2a
m2g - T = m2a.... 2
W is the weight
g is the acceleration due to gravity
If the acceleration of the system is 0, the equation becomes;
From 2:
m2g - T = m2a.... 2
a = m2g-T/m2
From 1:
a = T-Ff/m1
Equate both accelerations
m2g-T/m2 = T-Ff/m1
Cross multiply
m1m2g - Tm1 = m2T-m2Ff
m1m2g+m2Ff = m2T+m1T
m1m2g+m2Ff = T(m2+m1)
T = m1m2g+m2Ff/m2+m1
Hence the tension in strings is expressed as
T = m1m2g+m2Ff/m2+m1
A soccer player kicks a 0.5 kg ball and accelerates it at 50 m/s2. What force does the player apply to the ball?
Answer:
The answer is 25 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × accelerationFrom the question we have
force = 0.5 × 50
We have the final answer as
25 NHope this helps you
A 40.0 kg box is being pushed across the floor to the right with 200 N of applied force. There is 120 N of frictional force opposing the motion.
What is the net force acting on the box?
What is the acceleration of the box?
A playground merry-go-round has a radius R and a rotational inertia I. When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then:
Answer:
The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].
Explanation:
After a careful reading of the statement, we notice that merry-go-round-child system is a system with no external force exerting on it, such that the Principle of Angular Momentum Conservation can applied to analyze the system:
[tex]m_{C}\cdot \left(\frac{v}{R} \right) = (I+m_{C}\cdot R^{2})\cdot \omega[/tex] (Eq. 1)
Where:
[tex]m_{C}[/tex] - Mass of the child, measured in kilograms.
[tex]v[/tex] - Initial speed of the child, measured in meters per second.
[tex]R[/tex] - Radius of the playground merry-go-round, measured in meters.
[tex]I[/tex] - Moment of inertia, measured in kilograms per square meter.
[tex]\omega[/tex] - Angular velocity of the merry-go-round-child system, measured in radians per second.
Then, we clear the final angular speed:
[tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex]
The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].
A 227−lb fullback runs the 40−yd dash at a speed of 17.25 pm 0.10 mi/h. Enter your answers in scientific notation. (a) What is his de Broglie wavelength (in meters)? × 10 (select) m (b) What is the uncertainty in his position based on the de Broglie wavelength?
Answer:
His de Broglie wavelength is 8.35×10⁻³⁷ m
The uncertainty in his position is 1.15 × 10⁻³⁵ m
Explanation:
First, Convert 227-lb to kg and convert the unit of the speed from mi/h to m/s.
To convert 227-lb to kg,
1-lb = 0.453592 kg
∴ 227-lb = 227 × 0.453592 kg
227-lb = 102.97 kg
To convert 17.25 ± 0.10 mi/h to m/s
1 mi = 1609.34 m
and 1 h = 3600 s
Therefore,
17.25 mi/h = (17.25 ×1609.34)/3600 m/s = 7.71 m/s
and 0.10 mi/h = (0.10 ×1609.34)/3600 m/s = 0.044704 m/s
Hence, the speed 17.25 ± 0.10 mi/h = 7.71 ± 0.044704 m/s
Now
(a) To determine the de Broglie wavelength,
De Broglie wavelength is given by
λ = h/mv
Where λ is the de Broglie wavelength
h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)
m is the mass
and v is the speed (velocity)
From the question
m = 102.97 kg
v = 7.71 m/s
Therefore,
λ = 6.626×10⁻³⁴ / (102.97×7.71)
λ = 8.35×10⁻³⁷ m
Hence, his de Broglie wavelength is 8.35×10⁻³⁷ m
(b) To calculate the uncertainty in his position
From
Δx = h/(4πmΔv)
Where Δx is the uncertainty in the position
h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)
π is a constant ( π = 3.14)
m is the mass
Δv is the uncertainty in speed
From the question
m = 102.97 kg
Δv = 0.044704 m/s
Hence,
Δx = 6.626×10⁻³⁴ / (4×3.14×102.97×0.044704)
Δx = 1.15 × 10⁻³⁵ m
Hence, the uncertainty in his position is 1.15 × 10⁻³⁵ m.
What has the least potential energy in this world
Answer:
Anything that has the most kinetic has the least potential energy.
Explanation:
NEED BRAINLYEST PLZZ!!!!!!!!!!!!!!!
In a class experiment to determine information about free-fall acceleration, a watermelon and a pumpkin are each set to fall from the back of the stands at your football stadium.
a. If the watermelon and the pumpkin are both dropped at the same time, which one will hit the ground first?
b. If the watermelon is thrown downward with an initial speed of 10 m/s and the pumpkin is dropped, which one will hit the ground first? Show calculation to support your answer.
c. If it takes the watermelon 1 second to reach the ground when it is thrown downward at 10 m/s, how tall are the stands?
d. How long does it take the pumpkin to reach the ground if it is dropped from this height you calculated in part c?
Answer:
a) They both hit the ground at the same time
b) Watermelon will hit first, since its speed is faster than the pumpkin speed
c) h = 29,6 m
d)t = 3,02 sec
Explanation:
Equations for fall free movement are:
vf = v₀ + g*t when v₀ = 0 (dropped case) vf = g*t
h = v₀*t + 1/2*g*t²
a) For both ( watermelon and pumpkin) the equation of speed is the same:
vf = g*t² Both will have the same speed second through second
They both hit the ground at the same time
b) Now is watermelon is thrown with v₀ = 10 m/s
Watermelon will hit first since its speed is faster than the pumpkin speed
vf(watermelon) = 10 + g*t
vf₂ (pumpkin) = g*t
c) h = v₀*t + (1/2)*g*t²
h = (10)*1 + (1/2)*9,8*1
h = 10 + 19,6
h = 29,6 m
d) h = g*t
t = 29,6/9,8
t = 3,02 sec
How does the temperature affect the amount of weathering
An 80 N box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and the table are 0.25 and 0.5 respectively. What is the friction force on this box if the pull is 25N?
A. 20 N
B. 40 N
C. 25 N
D. 0 N
Answer:
C. 25 N
Explanation:
w = Weight of box = 80 N
[tex]\mu_k[/tex] = Coefficient of kinetic friction = 0.25
[tex]\mu_s[/tex] = Coefficient of static friction = 0.5
F = Pulling force = 25 N
Static friction
[tex]f_s=\mu_s w\\\Rightarrow f_s=0.5\times 80=40\ \text{N}[/tex]
[tex]F<f_s[/tex]
The force of pulling is less than the static frictional force so this makes the friction force equal to the force of pulling.
So, the friction force on the box is 25 N.
What change to the atomic model helped solve the problem seen in Rutherford’s model?
Object A has a mass of 12 g and a density of 3 g/cm3. Object B has a mass of 12 g and a density of 5 g/cm3. Which object has a greater volume and by how much? (4 points)
Object A, by 2 6 cubic centimeters
Object A, by 1.6 cubic centimeters
Object B, by 2 6 cubic centimeters
Object B, by 1.6 cubic centimeters
Answer:
object B has a greater volume and by2 g/cm3
During the 3.7 min a 7.4 A current is set up in a wire, how many (a) coulombs and (b) electrons pass through any cross section of the wire's width?
Answer:
1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.
Explanation:
Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).
Then, the intensity of electric current is expressed as:
[tex]I=\frac{Q}{t}[/tex]
Where:
I is the intensity expressed in Amps (A) Q is the electric charge expressed in Coulombs (C) t is the time expressed in seconds (s)In this case:
I= 7.4 AQ= ?t= 3.7 min= 222 s (being 1 min= 60 s)Replacing:
[tex]7.4 A=\frac{Q}{222 s}[/tex]
Solving:
Q= 7.4 A* 222 s
Q= 1,642.8 C
A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?
[tex]electrons=\frac{222 C*6.24*10^{18} }{1C}[/tex]
electrons= 1.39*10²¹
1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.
what is hubbles law? please put in simple terms
Answer:
the law that the velocity of recession of distant galaxies from our own is proportional to their distance from us.
Explanation:
Hubble's law, also known as the Hubble–Lemaître law, is the observation in physical cosmology that galaxies are moving away from the Earth at speeds proportional to their distance. In other words, the farther they are the faster they are moving away from Earth.
