Answer:
[tex]p2=0.667\\p1=0.167\\p3=0.167\\p=1[/tex]
Explanation:
Let's start writing the sample space for this exercise :
Let be ''M'' an abbreviation for Macrostate
Ω = { M1 , M2 , M3 }
Let be P(M1) the probability of Macrostate 1.
Reading the exercise, we know that ⇒
[tex]P(M1)=P(M3)[/tex]
Let's note this probability as ''p''.
[tex]P(M1)=P(M3)=p[/tex]
Macrostate 2 is four times more likely to occur than either of the other two macrostates ⇒
[tex]P(M2)=4p[/tex]
The sum of all probabilities must be equal to 1 for this sample space.Therefore,
[tex]p+p+4p=1[/tex]
[tex]6p=1[/tex]
[tex]p=\frac{1}{6}[/tex]
Finally :
[tex]P(M1)=\frac{1}{6}[/tex]
[tex]P(M2)=\frac{4}{6}[/tex]
[tex]P(M3)=\frac{1}{6}[/tex]
For Part A :
[tex]p2=\frac{4}{6}=0.667[/tex]
For Part B and C :
[tex]p1=p3=0.167[/tex]
For Part D :
The sum of the probabilities for all macrostates is equal to 1 :
[tex]p=p1+p2+p3=\frac{1}{6}+\frac{4}{6}+\frac{1}{6}=\frac{6}{6}=1[/tex]
Final answer:
The probability that equilibrium state 2 occurs is approximately 0.667. The probability that macro state 1 or macro state 3 occurs is each approximately 0.167. The sum of the probabilities for all macro states is exactly 1, as expected for a complete set of possible outcomes.
Explanation:
To find the probability of each macro state, we first recognize that the probabilities must sum to 1 for all macro states. Given that the equilibrium state, Macro state 2, is four times more likely than either of the other two Macro states 1 and 3, we can assign a probability 'p' to Macro states 1 and 3 and '4p' to Macro state 2.
Let's assume:
Probability of Macro state 1, p1 = p
Probability of Macro state 2, p2 = 4p
Probability of Macro state 3, p3 = p
Since the total probability must equal 1, we write the equation:
p1 + p2 + p3 = 1
Substituting the assumed probabilities gives:
p + 4p + p = 1
6p = 1
Therefore, p = 1/6
Now, we can calculate each probability:
p1 = 1/6 ≈ 0.167
p2 = 4/6 ≈ 0.667
p3 = 1/6 ≈ 0.167
The sum of all probabilities, p, is as expected:
p = p1 + p2 + p3 = 0.167 + 0.667 + 0.167 = 1
A museum curator moves artifacts into place on various different display surfaces. If the curator moves a 145 kg aluminum sculpture across a horizontal steel platform what is fs, fmax, and fk
Answer:
Static friction is, [tex]0\ N\leq f_s\leq 866.81\ N[/tex]
Maximum static friction is, [tex]f_{max}= 866.81\ N[/tex]
Kinetic friction is, [tex]f_k=667.87\ N[/tex]
Explanation:
The value of coefficient of static friction between Aluminium and Steel is [tex]\mu_s=0.61[/tex] and the value of coefficient of kinetic friction is [tex]\mu_k=0.47[/tex]
Given:
Mass of the sculpture is, [tex]m=145\ kg[/tex].
As the sculpture is moved along the horizontal direction only, there will no net force acting along the vertical direction. Therefore,
Normal force by the steel platform will be equal to the weight of the aluminium sculpture. That is,
[tex]N=mg[/tex]
Now, the static friction is given as the product of normal force and coefficient of static friction. The static friction varies from a value of 0 to limiting friction.
The limiting friction is the maximum static friction that acts on a body to resist the motion.
The value of the limiting friction or the maximum static friction is given as:
[tex]f_{max}=\mu_sN=\mu_smg=0.61\times 145\times 9.8=866.81\ N[/tex]
Thus, static friction is given as:
[tex]0\leq f_s\leq f_{max}\\0\leq f_s\leq \mu_smg\\0\ N\leq f_s\leq 866.81\ N[/tex]
Now, kinetic friction is the frictional force acting on a body when the body is moving under the action of forces. Thus kinetic friction is the product of normal force and coefficient of kinetic friction. Kinetic friction is a constant opposing force.
Thus, kinetic friction is given as:
[tex]f_k=\mu_kN\\f_k=\mu_kmg\\f_k=0.47\times 145\times 9.8=667.87\ N[/tex]
Consider a merry-go-round that has the form of a disc with radius 5.5 m and mass 155 kg. If five children, each of mass 20 kg, sit on the outer edge of the merry-go-round, what is the total moment of inertia?
Answer:
[tex]I=5369.375[/tex]
Explanation:
Given:
mass of merry go round, [tex]M=155\ kg[/tex]radius of merry go round, [tex]r=5.5\ m[/tex]mass of child, [tex]m=20\ kg[/tex]Considering merry-go-round as a disk, its moment of inertia is given as:
[tex]I_d=\frac{1}{2} M.r^2[/tex]
[tex]I_d=0.5\times 155\times 5.5^2[/tex]
[tex]I_d=2344.375\ kg.m^2[/tex]
Considering children as point masses, their moment of inertia is given as:
[tex]I_C=5(m.r^2)[/tex]
since there are 5 children
[tex]I_C=5\times20\times 5.5^2[/tex]
[tex]I_C=3025\ kg.m^2[/tex]
Now, total moment of inertia:
[tex]I=I_C+I_d[/tex]
[tex]I=3025+2344.375[/tex]
[tex]I=5369.375[/tex]
A rubber ball and a lump of putty have equal mass. They are thrown with equal speed against a wall. The ball bounces back with nearly the same speed with which it hit. The putty sticks to the wall. Which objects experiences the greater momentum change?
Answer:
The rubber ball
Explanation:
In order to understand this, let's begin with the fact that momentum it's a vector quantity that has mass, sense, direction and a numerical value.
Now, we have the ball and the putty, in both hands. Both of them, has the same mass, let's say they have a mass of 20 g each.
In this point, you throw both balls against the wall, and they have a speed of 10 m/s (I'm assuming these values); from the moment that you let go the balls, they are both have a momentum, and as they have the same speed and mass, the momentum it's the same for both of them.
Now, they hit the wall. The putty sticked to the wall, so it's movement finished. At this point it's momentum becomes zero. Even though it still has mass, but it's not moving, so momentum equals zero here. However, inthe ball bounces back to you, at this point, the ball even with a reduced speed, it still has a momentum, so, it's greater than the one that the putty has because it becomes zero. Therefore, the ball has a greater change in momentum.
Star-forming clouds appear dark in visible-light photos because the light of stars behind them is absorbed by __________.
Answer:
Interstellar Dust
Explanation:
In the solar system Interstellar dust is a type of cosmic dust and unlike Interstellar gas they have large dust particles enabling them to block visible light. Unlike smoke and fog, they contain large number of closely packed clumps of atoms and molecules
Answer:
interstellar dust
Explanation:
Dust particles interact with light through scattering and absorption. Reducing the amount of starlight received
The colour of the sky during the day is blue, because the blue light is scattered by atmosphere particles. The increased path length causes the change of colour of sun light . This is due to dust grains in the atmosphere, a similar process happens with interstellar dust
How does a coal-fired power plant use the energy in coal to produce electricity?
Coal-fired power plants produce electricity by burning coal to boil water into steam, which drives a turbine connected to a generator. The efficiency of energy conversion is low, with significant heat loss to the environment and a large CO2 emission as one of the main environmental impacts.
Explanation:A coal-fired power plant converts the energy stored in coal into electricity through a multi-step process. First, coal is mined and processed to be suitable for burning. When coal is combusted in the plant, it heats water to turn it into steam. The steam at high pressure then drives a turbine, which is connected to a generator. As the turbine blades turn, they rotate the generator, which converts the kinetic energy into electricity. This process involves significant heat transfer to the surroundings, which is an inherent part of energy production from combustion.
During the energy conversion process, the efficiency of coal power stations is quite low, with only about 42% of the energy being used for electricity generation and the rest being lost as heat transfer to the environment. The chemical reaction during the combustion of coal is C + O2 → CO2, and a significant amount of CO2 is emitted into the atmosphere. This contributes to the warming of our planet, and coal power plants are known for being the least efficient and most CO2-emitting fossil fuel energy sources.
Who was Georges Lemaitre and how did he contribute to the Big Bang theory?
(15 points)
Final answer:
Georges Lemaître, a Belgian cosmologist, proposed the initial concept of the Big Bang, predicting the universe's expansion and Hubble's Law before they were empirically observed. His ideas on the universe starting from a 'primeval atom' and his work on particle interactions in the early universe laid the foundations of modern cosmology.
Explanation:
Georges Lemaître was a Belgian priest and cosmologist who made significant contributions to the Big Bang theory. Born in 1894, Lemaître studied theology as well as mathematics and physics. He was instrumental in exploring the concept of the expanding universe and was the first to propose a concrete model of the Big Bang. This model suggested that the universe started as a single 'primeval atom', which eventually fragmented into smaller pieces, leading to the formation of the current atoms in the universe through a process akin to nuclear fission. Lemaître's work predated and anticipated the empirical findings that became known as Hubble's Law, which observed that galaxies are moving away from each other, implying that the universe is expanding. His insights laid the groundwork for the understanding of the universe's beginnings and its initial hot, dense state.
Additionally, Lemaître, alongside collaborators such as George Gamow, further developed the Big Bang theory. They predicted that as the universe expands and cools, the interactions among particles would lead to the formation of protons, neutrons, and eventually the nuclei of light elements such as deuterium, helium, and lithium. This theoretical framework is bolstered by measurements such as the cosmic microwave background radiation and the abundance of deuterium, supporting the Big Bang theory as a robust model of the universe's inception.
A wire of length 25.0 cm carrying a current of 4.21 mA is to be formed into a circular coil and placed in a uniform magnetic field B with arrow of magnitude 5.55 mT. Suppose the torque on the coil from the field is maximized.
Answer:
1.162 x 10^-7 Nm
Explanation:
length of wire, l = 25 cm
l = 2 π r
where, r is the radius of circular loop
25 = 2 x 3.14 x r
r = 3.98 cm
Magnetic field, B = 5.55 mT = 5.55 x 10^-3 T
Current, i = 4.121 mA = 4.21 x 10^-3 A
Torque, τ = i x A x B
τ = 4.21 x 10^-3 x 3.14 x 0.0398 x 0.0398 x 5.55 x 10^-3
τ = 1.162 x 10^-7 Nm
Thus, the maximum torque in the coil is 1.162 x 10^-7 Nm.
A manufacturer claims to have built a home stereo speaker that is 4 feet tall but uses an average of only 30 watts of power. A sample of 40 speakers from the manufacturer found that they used an average of 32 watts of power. The appropriate hypotheses to test the manufacturer’s claim are?
Answer:
Original claim is H_o: M = 30
Alternate claim is H_a: M > 30.
Explanation:
Given data:
4 ft tall speaker consume 30 watt power
a group of 40 speaker consume average power of 32 watt
calculation:
A home installed 4 ft tall speaker that use 30 watt Power
40 speakers used 32 watt Power
Original claim is H_o: M = 30
Alternate claim is H_a: M > 30.
A person is standing on a spring bathroom scale on the floor of an elevator which is moving up and slowing down at the rate of 3 m/s 2 . The acceleration of gravity is 9.8 m/s 2 . If the person’s mass is 93.3 kg, what does the scale read? Answer in units of n.
Answer:
[tex]F_N=1194.24\ N[/tex]
Explanation:
Given that,
The elevator is moving up and slowing down at the rate of, [tex]a=3\ m/s^2[/tex]
The acceleration due to gravity, [tex]g=9.8\ m/s^2[/tex]
Mass of the person, m = 93.3 kg
To find,
The reading of the scale.
Solution,
As the elevator is moving down with some acceleration. The net force acting on it is given by :
[tex]F_N=ma+mg[/tex]
[tex]F_N=m(a+g)[/tex]
[tex]F_N=93.3(3+9.8)[/tex]
[tex]F_N=1194.24\ N[/tex]
So, the scale will read 1194.24 N.
An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first started, it applies a constant torque of 1930 Nm to the propeller, which starts from rest.
a. What is the angular acceleration of the propeller? Treat the propeller as a slender rod.
b. What is the propeller's angular speed after making 5.00 rev?
c. How much work is done by the engine during the first 5.00 rev?
d. What is the average power output of the engine during the first 5.00 rev?
e. What is the instantaneous power output of the motor at the instant that the propeller has turned through 5.00 rev?
Answer:
a)30.14 rad/s2
b)43.5 rad/s
c)60633 J
d)42 kW
e)84 kW
Explanation:
If we treat the propeller is a slender rod, then its moments of inertia is
[tex] I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2[/tex]
a. The angular acceleration is Torque divided by moments of inertia:
[tex]\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2[/tex]
b. 5 revolution would be equals to [tex]10\pi[/tex] rad, or 31.4 rad. Since the engine just got started
[tex]\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5[/tex]
[tex]\omega = \sqrt{1893.5} = 43.5 rad/s[/tex]
c. Work done during the first 5 revolution would be torque times angular displacement:
[tex]W = T*\theta = 1930 * 31.4 = 60633 J[/tex]
d. The time it takes to spin the first 5 revolutions is
[tex]t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s[/tex]
The average power output is work per unit time
[tex]P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W[/tex] or 42 kW
e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:
[tex]P_i = T*\omega = 1930*43.5=83983 W[/tex] or 84 kW
A football is kicked straight up from a height of 5 feet with an initial speed of 55 feet per second. The formula h equals negative 16 t squared plus 55 t plus 5 describes the ball's height above the ground, h, in feet, t seconds after it was kicked. How long will it take for the football to hit the ground?
Answer:
3.53 second
Explanation:
The formula for the height is
[tex]h=-16t^{2}+55t+5[/tex]
When it hits the ground, the height is zero.
So, put h = 0 in the above equation
[tex]0=-16t^{2}+55t+5[/tex]
[tex]16t^{2}-55t-5=0[/tex]
[tex]t=\frac{+55\pm \sqrt{55^{2}+4\times 5\times 16}}{2\times 16}[/tex]
[tex]t=\frac{+55\pm 57.84}{2\times 16}[/tex]
Take positive sign
t = 3.53 second.
Thus, the time taken to hit the ground is 3.53 second.
A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact with the top of the beam. Then it drives the beam 12.0 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest. a) m=2100 kg b) Xi=5.00m c) Xf=12.0 cm =.12m
Answer:
f = 878,080 N
Explanation:
mass of pile driver (m) = 2100 kg
distance of pile driver to steel beam (s) = 5 m
depth of steel driven (d) = 12 cm = 0.12 m
acceleration due to gravity (g0 = 9.8 m/s^{2}
calculate the average force exerted on the pile driver by the beam.
from work done = force x distance work done = change in potential energy of the pile driverequating the two equations above we haveforce x distance = m x g x (s - d)
f x 0.12 = 2100 x 9.8 x (5- (-0.12))
d = - 0.12 because the steel beam went down at we are taking its
initial position to be an origin point which is 0
f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12
f = 878,080 N
The average force that the pile driver exerted on the steel beam can be calculated using energy considerations, specifically by using the principle of conservation of energy. The potential energy of the pile driver is converted into the work done to drive the beam into the ground. This results in an average force of approximately 857,500 Newtons.
Explanation:To solve this problem, we can first consider the principle of conservation of energy. The energy of the pile driver, when it starts falling, is purely potential energy, and when it has driven the steel beam into the ground, it's all been converted to work done against the resistance of the ground.
Firstly, calculate the potential energy of the pile driver as it begins to fall. The formula for potential energy (P.E.) is mass (m) times the acceleration due to gravity (g), which is about 9.8 m/s², times the height (h, the distance fallen): P.E. = m * g * h = 2100 kg * 9.8 m/s² * 5m = 102,900 Joules.
Secondly, the work done (W) in driving the steel beam into the ground can be calculated using this energy. Since this work was done to overcome the force of the beam as it went into the ground, we can also write W = F * d, where F is the average force and d is the distance it drove the beam down (0.12m).
Lastly, solve for the average force (F) by rearranging the equation to F = W / d = 102,900 Joules / 0.12 m = approx. 857,500 Newtons. Assuming all the energy was used in driving the steel beam into the ground, the pile driver would have had to exert an average force of around 857,500 N.
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A pressure vessel at rest at the origin of an xy coordinate system explodes into three pieces that remain in the xy plane. Just after the explosion, one piece, of mass m, moves with velocity (−30 �/�)� and a second piece, also of mass m, moves with velocity (−30 �/�)�. The third piece has mass 3m. Just after the explosion, what is the velocity of the third piece? Use unit vector notation.
Answer:
v₃ = (10) i + (10) j
v₃ = 10√2
Explanation:
Given info
Before the explosion
ux = 0
uy = 0
After the explosion
v₁x = -30
v₁y = 0
v₂x = 0
v₂y = -30
We can use the Principle of Conservation of Momentum as follows
pi = pf
where
pix = M*ux = M*0 = 0
piy = M*uy = M*0 = 0
pfx = p₁x + p₂x + p₃x = (m*v₁x + m*v₂x + 3m*v₃x) = m*(-30) + m*(0) + 3m*v₃x
⇒ pfx = -30m + 3m*v₃x
if
pix = pfx ⇒ 0 = -30m + 3m*v₃x ⇒ v₃x = 10
pfy = p₁y + p₂y + p₃y = (m*v₁y + m*v₂y + 3m*v₃y) = m*(0) + m*(-30) + 3m*v₃y
⇒ pfy = -30m + 3m*v₃y
if
piy = pfy ⇒ 0 = -30m + 3m*v₃y ⇒ v₃y = 10
then we have
v₃ = (10) i + (10) j
and its module can be obtained as follows
v₃ = √(v₃x² + v₃y²) = √(10² + 10²) = 10√2
During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by 2.00ºC . What is the rate of heat transfer from this forced convection alone, assuming blood has the same specific heat as water and its density is 1050 kg/m
Answer:
-293 W
Explanation:
mass = density * volume = 1050 kg/m^3 * 0.002 m^3/min = 2.1 kg/min
Heat transferred = mass * specific heat capacity * change in temperature
= mcΔT
= 2.1 kg/min * 4186 J/kg-°C * -2 °C
= -17 581.2 kJ/min
= -17 581.2 kJ/60s
= -293 J/s
= -293 W
The negative sign shows us that the heat is being given off the blood
What are the characteristics of the atmospheres of venus
Answer:
Explanation:
The atmosphere is mainly composed of Carbon Dioxide (96%), 3.5% nitrogen, and less than 1% is composed of carbon monoxide, argon, sulfur dioxide and water vapor.Venus's atmosphere is very hot and denseThe Venusian atmosphere is crossed by occasional electrical phenomena of remarkable energyThe presence of a dense atmosphere and particularly strong winds causes the surface temperature to be approximately 710-740 K throughout the planetThe atmospheric pressure on the surface reaches the value of 90 atmospheres.An Object is moving due south at 16 m/s, when a force accelerates it at 4m/s in a southerly direction. How long will it take for the object to reach a velocity of 48 m/s south. Please help me out.
Answer:
Time taken to accelerate to 48 m/s = 8 seconds
Explanation:
Initially the object is moving south at = [tex]16 m/s[/tex]
So, initial velocity of the object = [tex]16 m/s[/tex]
Acceleration caused by the force = 4 [tex]m/s^2[/tex]
Final velocity towards south = [tex]48 m/s[/tex]
Using equation of motions:
[tex]v_f=v_i+at[/tex]
where
[tex]v_f\rightarrow[/tex] final velocity
[tex]v_i\rightarrow[/tex] initial velocity
[tex]a\rightarrow[/tex] acceleration
[tex]t\rightarrow[/tex] time
Plugging in values.
[tex]48=16+(4)t[/tex]
[tex]48=16+4t[/tex]
Subtracting both sides by 16
[tex]48-16=16+4t-16[/tex]
[tex]32=4t[/tex]
Dividing both side by 4.
[tex]\frac{32}{4}=\frac{4t}{4}[/tex]
[tex]8=t[/tex]
∴ [tex]t=8[/tex]
Time taken to accelerate to 48 m/s = 8 seconds
A 66.0−kg short-track ice skater is racing at a speed of 10.0 m/s when he falls down and slides across the ice into a padded wall that brings him to rest. Assuming that he doesn't lose any speed during the fall or while sliding across the ice, how much work is done by the wall while stopping the ice skater?
Answer:
3300J
Explanation:
Work done is the energy that is lost by the skater
Formula for workdone = 1/2*mV^2
m = 66kg
V = 10m/s
Work done = 1/2 * 66 * 10^2
= 3300J
The work done by the wall to stop a 66.0-kg ice skater moving at 10.0 m/s is calculated using the work-energy theorem and is found to be 3300 joules.
The student is asking how much work is done by the wall to stop a 66.0-kg ice skater who is moving at a speed of 10.0 m/s. To solve this, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Since the skater is coming to rest, the final kinetic energy is 0. The initial kinetic energy can be calculated using the equation KE = 0.5 × m × v^2, where m is the mass and v is the velocity. After plugging in the values, we get KE = 0.5 × 66.0 kg × (10.0 m/s)^2 = 3300 J. Therefore, the padded wall does 3300 joules of work to bring the skater to rest.
** You pull a rope oriented at a 37° angle above the horizontal. The other end of the rope is attached to the front of the first of two wagons that have the same 30-kg mass. The rope exerts a force of magnitude T1 on the first wagon. The wagons are connected by a second horizontal rope that exerts a force of magnitude T2 on the second wagon. Determine the magnitudes of T1 and T2 if the acceleration of the wagons is 2.0 m⁄s2.
Answer:
T2= 60 N and T1= 150,25 N
Explanation:
a free body diagram has to be done
The magnitude of the forces T1 and T2 if the acceleration is 2m/s² is 60N and 150.26N respectively.
Find the free body diagram attached. According to newton's second law;
[tex]\sum F_x = ma_x[/tex]
∑Fx is the sum of applied force in the horizontal direction
m is the mass of the object
ax is the acceleration of the object
For the body of mass 30kg
∑T = ma
T2 = ma
T2 = 30 * 2
T2 = 60N
For the sum of force acting on the second body;
T1 cos θ - T2 = ma
T1 cos 37 - 60 = 30(2)
T1 cos 37 = 120
T1 = 120/cos37
T1 = 120/0.7986
T1 = 150.26N
This shows that the magnitude of the forces T1 and T2 if the acceleration is 2m/s² is 60N and 150.26N respectively.
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A person exerts a horizontal force of 190 n in the test apparatus shown in the figure. Find the horizontal force that his flexor muscle exerts on his forearm.
Answer:
Check explanation.
Explanation:
From the question, we know that The person exerted 190N, force on the flexor is unknown. Since, we don't have access to our diagram, we have to make one or two assumption; (1) that ba= 0.3 m, ac= 0.05.
Therefore, the horizontal force that his flexor muscle exerts on his forearm,F(flexor) = (190N) × (0.3m) / 0.05m.
The horizontal force that his flexor muscle exerts on his forearm,F(flexor) = 57 Nm/ 0.05 m.
The horizontal force that his flexor muscle exerts on his forearm,F(flexor) = 1140N.
Using principles of physics, specifically equilibrium, we know the force exerted by muscles can be greater than the load they support, especially when the load is distant from the joint. In the case of a 190 N horizontal force, the exact force of the flexor muscle can't be calculated without more information, but it's likely larger than 190 N.
Explanation:To answer the student's question, we must refer to concepts in physics, specifically the principle of equilibrium and the analysis of free-body diagrams. Given that the student is trying to find the force that a person's flexor muscle exerts horizontally when the person exerts a horizontal force of 190 N.
As mentioned in the provided information, forces and tensions in muscles and joints can be quite substantial. In this scenario, the flexor muscle in the forearm is acting against the applied force to maintain balance or equilibrium. This can be seen in the free-body diagram where forces are broken down into their x- and y-components.
It's important to note that the force exerted by muscles can be far greater than the load they support, especially when the load is a considerable distance from the joint, as stated in the provided text snippet. This may be relevant when analyzing the muscle force in this scenario.
Unfortunately, without more specific details about the angles involved and the exact configuration of the arm and forearm in the test apparatus, we can't calculate the exact force exerted by the flexor muscle. However, we can say that it's likely to be substantially larger than the 190 N applied force, considering the principles discussed above.
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A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.718 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.34 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. (a) Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round? stay the same increase decrease (b) Calculate the initial and final kinetic energies for this system.Ki = kJKf = kJ
Answer:
The kinetic energy of the system decrease
Ki = 5.78 KJ
Kf = 4.55 KJ
Explanation:
For answer this question we will use the law of the conservation of the angular momentum so,
Li = Lf
Where Li is the inicial momentum of all the system, and Lf is the final momentum of the system.
also, the angular momentum L can be calculated in two ways
L = IW
where I is the momentum of inertia and the W is the Angular velocity.
or,
L = MVD
where M is the mass, V is the lineal velocity and the D is the lever arm.
Therefore,
Li = Ld ( merry-go-round) + Lp ( person )
Lf = Ls
Where Ld is the angular momentum of the merry go round, Lp is the angular momentum of the person and Ls is the angular momentum of the sistem (merry-go-round + person)
so,
[tex]L_d=I_dW_d[/tex]
Ld = [tex]\frac{1}{2}M_dR^{2}W_d[/tex]
Ld = [tex]\frac{1}{2}(155) (2.63)^{2}(0.718*2\pi)[/tex]
Ld = 2418.43
and,
[tex]L_p=M_pV_pD[/tex]
Lp = (59.4)(3.34)(2.63)
Lp = 521.78
then,
Lf = Ls
L_d=I_sW_s
Lf = [tex](\frac{1}{2}(155)(2.63)^{2}+(59.4)(2.63^2))(W_s)[/tex]
[tex]Lf = 946.92W_s[/tex]
so, solving for Ws
Lf = Li
[tex]946.92W_s = 521,78 + 2418.43[/tex]
Ws = 3.1 rad/s
Finally, the inicial and the final Kinetic energy
Ki = [tex]\frac{1}{2}I_d(W_d)^2 + \frac{1}{2}M_p(V_p)^2[/tex]
Ki = 5786.284 J = 5.78 KJ
Kf = [tex]\frac{1}{2}I_s(W_s)^2[/tex]
Kf = 4549.97 J = 4.55 KJ
Then, The kinetic energy of the system decrease because Kf < Ki
Wearing a seat belt protects you from possible harm by a certain force. Which force?
Explain the difference between weight and mass.
What force causes objects to move in a circular path?
Answer:
push
Explanation:
it pushes you back when you lean forward really fast
The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. ... Weight usually is denoted by W. Weight is mass multiplied by the acceleration of gravity (g).
GRAVITY
THE UNBALANCED FORCE THAT CAUSES OBJECTS TO MOVE IN A CIRCULAR PATH IS CALLED A CENTRIPETAL FORCE. GRAVITY PROVIDES THE CENTRIPETAL FORCE THAT KEEPS OBJECTS IN ORBIT. THE WORD CENTRIPETAL MEANS "TOWARD THE CENTER."
Determine the magnitude of the gravitational force Mars would exert on man if he was on the surface of Mars. The mass of the man is 68.0 kg . The mass of the Mars is 6.42×1023kg and its radius is 3396 km
The gravitational force that Mars would exert on a man with a mass of 68.0 kg standing on its surface is approximately 252.28 N (Newtons).
Explanation:To calculate the gravitational force (Weight) that Mars would exert on a man standing on its surface, we can use the formula for the weight which is 'W = mg', where 'm' is the mass of the man and 'g' is the acceleration due to gravity. However, on Mars, the value of 'g' (acceleration due to gravity) is different than on Earth. On Mars, 'g' is approximately 3.71 m/s².
Therefore, by substituting the given and calculated values into the formula we get:
W = mg = 68.0 kg x 3.71 m/s² = 252.28 N.
So, the gravitational force that Mars would exert on the person would be approximately 252.28 N (Newtons).
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a 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what is his recoil velocity? PLEASE HELP ME
Answer:
His recoil velocity is 0.269 m/s to the East
Explanation:
This question is easily solved by using the law of conservation of linear momentum.
The formula for the momentum is
[tex]Mo = mv[/tex]
, where m is the mass and v its speed.
The person + the book are at rest which means their momentum is
[tex]Mo_0=0[/tex]
After the book is released, they both start to move and their combined momentum is
[tex]Mo_f=m_pv_p+m_bv_b[/tex]
Where [tex]m_p, v_p[/tex] are the mass and speed of the person respectively and [tex]m_b, v_b[/tex] are the mass and speed of the book
Knowing that
[tex]m_p=74.9 Kg, m_b=2.44 Kg, v_b=-8.25m/s[/tex] (positive speed is assumed to the right or East), and the total momentum is zero:
[tex]m_pv_p+m_bv_b=0[/tex] =>
[tex]v_p=-\frac{m_bv_b}{m_p} =-\frac{2.44 (-8.25)}{74.9}[/tex]
[tex]v_p=0.269 m/s[/tex]
Since the sign of [tex]v_p[/tex] is positive, it's directed to the East
The recoil velocity of the person after throwing the book is approximately 0.27 m/s to the east, which is the opposite direction of the thrown book.
The recoil velocity of a person after throwing an object can be found by using the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum afterwards, assuming no external forces act on the system. In this case, the person and the book together form a closed system with no external forces since they are on an icy pond, which we can consider frictionless.
The initial momentum of the system is zero because the person is at rest. After the person throws the book, the momentum of the book is mass of the book multiplied by velocity of the book. To find the recoil velocity of the person, we use the formula:
m1×u1 + m2×u2 = m1×v1 + m2×v2,
where m1 and m2 are the masses of the person and book, u1 and u2 are their initial velocities (which are zero), and v1 and v2 are their final velocities. The final velocity of the person (v1) is what we are looking for:
(74.9 kg)(0 m/s) + (2.44 kg)(0 m/s) = (74.9 kg)v1 + (2.44 kg)(-8.25 m/s),
v1 = -(2.44 kg)(-8.25 m/s) / 74.9 kg,
v1 = 0.27 m/s to the east (opposite direction to the book).
How many σ bonds and π bonds does the co2 molecule have?
Answer:
2 [tex]\sigma[/tex] bonds and 2 [tex]\pi[/tex] bonds
Explanation:
If we consider the the bonding in the [tex]CO_{2}[/tex] molecule:
[tex]O = 1s^{2}\ 2s^{2}\ 2p^{4}[/tex]
[tex]C = 1s^{2}\ 2s^{2}\ 2p^{2}[/tex]
Thus carbon forms double bonds with oxygen:
O = C = O
Now,
We know that double bond comprises of a [tex]\sigma\ bond[/tex] and a [tex]\pi \ bond[/tex]
Since, in the [tex]CO_{2}[/tex], there are 2 double bonds thus there are 2 [tex]\sigma[/tex] bonds and 2 [tex]\pi[/tex] bonds in the molecules.
"The CO₂ molecule has 2 σƒ (sigma) bonds and 2 π (pi) bonds.
To determine the number of sigma and pi bonds in CO₂, we need to consider its Lewis structure. Carbon dioxide has a linear molecular geometry with carbon at the center and two oxygen atoms double-bonded to it.
In CO₂:
- The carbon atom forms two double bonds with the two oxygen atoms.
- Each double bond consists of one sigma bond and one pi bond.
Therefore, for each carbon-oxygen double bond, there is:
- One sigma bond (σƒ), which is the head-on overlap of atomic orbitals.
- One pi bond (π), which is the side-to-side overlap of p-orbitals.
Since there are two carbon-oxygen double bonds in CO₂, we have:
- A total of 2 sigma bonds from the double bonds.
- Additionally, each double bond includes a sigma bond from the overlap of the sp hybrid orbital of carbon with the sp2 hybrid orbital of oxygen, contributing another 2 sigma bonds.
In summary, CO₂ has 4 sigma bonds in total:
- 2 sigma bonds from the sp-sp² hybrid orbital overlaps.
- 2 sigma bonds from the head-on overlap of p-orbitals that form part of the double bonds.
And CO₂ has 2 pi bonds in total:
- 2 pi bonds from the side-to-side overlap of p-orbitals in the double bonds.
Thus, the final count is 4 σƒ bonds and 2π bonds in the CO₂ molecule. However, it is important to note that the question specifically asks for the number of sigma and pi bonds, and the correct answer should reflect the number of each type of bond individually, not the total number of bonds. Therefore, the correct answer is 2σ bonds and 2π bonds.
An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is 16.5 N. When completely submerged in water, its apparent weight is 13.3 N. What is the volume of the object?
Answer:
V = 1.7 x 10^{-3} m^{3}
Explanation:
apparent weight in ethyl alcohol = 16.5 N
apparent weight in water = 13.3 N
apparent weight = mg - ρgV
where
m = mass
g = acceleration due to gravity = 9.8 m/s^{2}
ρ = density
V = volume
for ethyl alcohol 16.5 = mg - ρ₁gV .....equation 1for water 13.3 = mg - ρ₂gV .......equation 2we can solve the two equations above as simultaneous equations by subtracting equation 2 from equation 1
we have
16.5 - 13.3 = (mg - mg) - ρ₁gV - (-ρ₂gV)
3.2 = gV(ρ₂ - ρ₁)
V = \frac{3.2}{g(ρ₂-ρ₁)}
where
ρ₂ = density of water = 1000 kg/m^{3}
ρ₁ = density of ethyl alcohol = 806 kg/m^{3}
V = \frac{3.2}{9.8(1000 - 806)}
V = 1.7 x 10^{-3} m^{3}
The volume of the object can be found using Archimedes' principle and the given apparent weights in water and ethyl alcohol. The volume is computed from these values using the densities of the two fluids, yielding a volume of about 0.013 cubic meters.
Explanation:To answer the question about the object's volume when it's submerged in ethyl alcohol and water, we'll need to use Archimedes' principle. This principle tells us that the apparent weight loss of the object in a liquid equals the weight of the fluid displaced by the object. In this case, the difference in apparent weights given for the object when in water and when in ethyl alcohol represents the weight of the fluids displaced.
We have to remember that weight can be calculated as the product of volume, density, and gravity. If you rearrange this equation, you get the volume as the quotient of weight and the product of density and gravity. Note that gravity cancels out in this problem since it remains constant for both fluids.
If we denote the volume of the object as V, the density of water as ρ_water (approximated to 1000 kg/m^3), the density of ethyl alcohol as ρ_alcohol (approximated to 789 kg/m^3), and the weight of the object in water and ethyl alcohol as W_water and W_alcohol respectively, we can write two equations:
V = W_water / ρ_waterV = W_alcohol / ρ_alcoholWe have the weights as the apparent weights from the problem, which are W_water = 13.3 N and W_alcohol = 16.5 N. If you solve these two equations, you should obtain the volume of the object, which should be around 0.013 m^3. This is the answer for the volume of the solid object.
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A person in a car during a sudden stop can experience potentially serious chest injuries if the combined force exerted by the seat belt and shoulder strap exceeds 16,000 N. Assume the mass of the passenger is 80 kg and the initial speed of the car is 16 m/s. Describe what it would take to avoid injury.
Answer:
minimum time interval to stop = 0.08 seconds
minimum stopping distance = 0.64 m
Explanation:
maximum force (F) = 16,000 N
mass (m) = 80 kg
initial velocity (U) = 16 m/s
what it would take for the passenger to avoid in this case refers to how long it would take the vehicle to come to a full stop and the stopping distance it would also take to come to a full stop. Therefore we are to find the time (t) and the distance (s)
from the impulse momentum equation,
impulse = change in momentum
Ft = m(V-U) (V-U is the change in velocity Δv)
where V is the final velocity = 0
and t = time
16000 x t = 80 (0 - 16)
16000t = -1,280 (he negative sign tell us there is a decrease in momentum, so we would not be using it further)
t = 0.08 seconds ( this is also the difference between the initial time when the vehicle started to come to a stop and the final time when it came to a full stop)
assuming the acceleration is constant, the stopping distance (s) would be given by the kinetic relation
change in distance (Δs) = \frac{(ΔV) x (Δt)}{2}
(Δ refers to change, that is final value - initial value)
Δs = \frac{16 x 0.08}{2}
Δs = 0.64 m
A management system that includes fire suppression will likely lead to
I. large quantities of biomass accumulating on the forest floor.
II. an increase in the likelihood of uncontrolled natural fires.
III. an increase in fire-dependent species.
Answer:
I and II
Explanation:
A management system that includes fire suppression will likely lead to
I. large quantities of biomass accumulating on the forest floor.
II. an increase in the likelihood of uncontrolled natural fires.
Biomass is the total quantity of weight of flora and fauna in a given area or volume. The recent fire in the Amazon rain forest is an example of uncontrolled natural forest fires.
IE is the energy required to remove an electron from an atom. As atomic radius increases, the valence electrons get farther from the nucleus. How do you think an atom’s size will affect its ability to hold on to its valence electrons? Why?
Answer:
The bigger the atom the lesser the ability of the atom to hold on to its valence electrons.
Explanation:
Atomic radius can be looked at as the distance between the nucleus and the outermost energy level. As an atom gets bigger, the outer shell gets further and further from the positive nucleus. this means that electrons that are in the outer energy level become less held (attracted) by the nucleus because of distance and shielding of the attractive forces by the electrons in the lower energy levels. This means that as an atom becomes bigger, its ability to hold on to its outer electrons lessens.
As an atom gets larger, it is unable to hold its valence electrons due to decreased electrostatic attraction between the nucleus and the valence electron.
An atom is composed of a nucleus that houses positive charges and an electron which is negatively charged and are found in orbits. Electrostatic attraction between the positively charged nucleus and electrons in orbits keep the atom together.
However, as an atom gets larger, the valence electrons are farther away from the nucleus. As the distance between the nucleus and the outermost electrons increases, the electrostatic interaction between electron and the nucleus is decreased according to Coulomb's law. Repulsion (screening) between inner and valence electrons further keep the valence electrons away from the attractive forces of the nucleus.
Therefore, as an atom gets larger, it is unable to hold its valence electrons due to decreased electrostatic attraction between the nucleus and the valence electrons.
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How much work is done by a force vector F = (4x N)i hat +(5 N)j, with x in meters, that moves a particle from a position vector r i = (5 m)i hat +(6 m)j, to a position vector r f = -(2 m)i hat -(5 m)j?
Answer:
-83
Explanation:
When light of wavelength 345 nm falls on a potassium surface, electrons are emitted that have a maximum kinetic energy of 1.67 eV. What is the work function of potassium? The speed of light is 3 × 108 m/s and Planck’s constant is 6.63 × 10−34 J · s.
Answer:
[tex]W=1.93eV[/tex]
Explanation:
The maximum kinetic energy of an ejected electron in the photoelectric effect is given by:
[tex]K_{max}=h\nu-W(1)[/tex]
Here h is the Planck's constant, [tex]\nu[/tex] the frequency of the light and W the work function of the element.
The frequency is equal to the speed of light, divided by the wavelength:
[tex]\nu=\frac{c}{\lambda}(2)[/tex]
Recall that [tex]1nm=10^{-9}m[/tex]. Replacing (2) in (1) and solving for W:
[tex]W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{345*10^{-9}m}-1.67eV\\W=1.93eV[/tex]