Final answer:
To calculate the pressure after a temperature change, use the equation PV = nRT, and simplifying it to P1/T1 = P2/T2. Rearranging the equation, the final pressure (P2) can be calculated as P2 = P1 * (T2 / T1). In this case, the pressure after the temperature has risen to 35.0ºC is approximately 7.33 × 10^5 Pa.
Explanation:
To calculate the pressure after the temperature change, we can use the equation PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature. Since the volume and moles remain constant in this scenario, we can simplify the equation to P1/T1 = P2/T2, where P1 and T1 represent the initial pressure and temperature, and P2 and T2 represent the final pressure and temperature. Rearranging the equation, we get P2 = P1 * (T2 / T1). Therefore, the pressure after the temperature has risen to 35.0ºC can be calculated as:
P2 = 7.00 × 10^5 Pa * (35.0 + 273.15) / (21.0 + 273.15)
P2 = 7.00 × 10^5 Pa * 308.15 / 294.15
P2 = 7.33 × 10^5 Pa
Gas A has molecules with small mass. Gas B has molecules with larger mass. They are at the same temperature.
How do the gases compare with respect to the average translational kinetic energy?
a)A has a larger average kinetic energy.b)B has a larger average kinetic energy.c)The gases have the same average kinetic energy.
Answer:
c)The gases have the same average kinetic energy.
Explanation:
As we know that the kinetic energy of gas is given as
[tex]K = \frac{1}{2}mv^2[/tex]
here we know that
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
so we have
[tex]K = \frac{1}{2}m (\frac{3RT}{M})[/tex]
now we have
[tex]K = \frac{3}{2}n RT[/tex]
now mean kinetic energy per molecule is given as
[tex]K_{avg} = \frac{3}{2}KT[/tex]
so this is independent of the mass of the gas
so average kinetic energy will remain same for both the gas molecules
George Of The Jungle's wife, Mrs. Of The Jungle, has been pestering him to go on a diet. He should have listened. During his commute home last Thursday afternoon, the Number 8 vine upon which he was swinging (along a circular path) snapped.
At the time of the incident, George was at the bottom of his swing, moving at a peppy 14.1 m/s. Given that the maximum tension that the vine (length 7.3 m) was able to tolerate was 4150 N, determine George's mass.
A) 110 kg
B) 120 kg
C) 130 kg
D) 140 kg
E) 150 kg
Answer:
112.06 kg - Thats heavy !
Explanation:
Let's do force balance here. Let the object of our interest be George. The forces acting on him are the tension in the upward direction, his weight in the downward direction and the centrifugal force in the downward direction. Considering the upward and downward directions on the y-axis and f=given the fact that George doesn't move up or down, the forces are balanced along the y-axis. Hence doing force balance:
magnitude of forces upward =magnitude of forces downward
i.e., Tension(T) = Weight(mg) + Centrifugal force (mv²/r)
where: 'm' is the mass of George, g is the acceleration due to gravity (9.8 m/s²). v is the speed with which George moves (14.1 m/s) and r is the radius of the circle in which he's moving at the instant (Here since he's swinging on the rope, he moves in a circle with radius as the length of the rope and hence r=7.3m).
therefore, T = m (9.8 + (14.1)²/7.3) = 4150 N
Therefore, m = 112.06 kg
Part A What is a radio galaxy? How can radio galaxies affect the gas surrounding them? Drag the items on the left to the appropriate blanks on the right to complete the sentences. (Not all terms will be used.)
Explanation:
A radio galaxy is a galaxy with a powerful radio luminance relative to the stars' visible and infrared luminosity. Radio galaxies can emit strong and speedy particle jets. They inject in their surroundings high amounts of kinetic energy. Many radio galaxies have jets of plasma shooting out in opposite direction. Thus they ionize the gases surrounding them.
Assuming constant velocities, if a fastball pitch is thrown and travels at 40 m/s toward home plate, 18 m away, and the head of the bat is simultaneously traveling toward the ball at 18.0 m/s, how much time elapses before the bat hits the ball?a. About 0.3 sb. About 0.6 sc. About 0.9 sd. About 1.2 s
Explanation:
Here velocity of ball and bat are in opposite direction.
Velocity of ball = 40 m/s
Velocity of bat = 18 m/s
Since they are in opposite direction relative velocity is given by,
Relative velocity = 40 + 18 = 58 m/s
Distance to home plate = 18 m
We have
Displacement = Velocity x Time
18 = 58 x Time
Time = 0.3 seconds
Option A is the correct answer.
A block of mass 3 kg, which has an initial
speed of 4 m/s at time t = 0, slides on a
horizontal surface.
Find the magnitude of the work that must
be done on the block to bring it to rest.Answer in units of J If a constant friction force of magnitude 2 Newtons is exerted on the block by the surface, find the magnitude of the acceleration of the block.
How far does the block slide before it comes to rest? units of m
Answer:
Explanation:
Kinetic energy of the block
= 1/2 m v²
= .5 x 3 x 4 x 4
= 24 J
Negative work of - 24 J is required to be done on this object to bring it to rest.
magnitude of acceleration due to frictional force
= force / mass
2 / 3
= 0 .67 m /s²
Let the body slide by distance d before coming to rest so work done by force = Kinetic energy
= 2 x d = 24
d = 12 m
The magnitude of the work done to bring the block to rest is 2 times the distance the block slides. The magnitude of the acceleration of the block is 2/3 m/s^2. The block slides a distance of 8/3 m before it comes to rest.
Explanation:The work done on an object is given by the equation:
Work = Force x Distance
In this case, the work done on the block to bring it to rest is equal to the force applied multiplied by the distance the block slides.
Given that the force exerted by the surface is 2 Newtons, we can calculate the magnitude of the work done:
Work = 2 N x Distance
To determine the distance the block slides, we need to calculate its deceleration using Newton's second law:
Force = mass x acceleration
Since the friction force is constant and in the opposite direction of motion, we can write:
2 N = 3 kg x acceleration
Solving for acceleration, we find:
acceleration = 2 N / 3 kg
With the acceleration calculated, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Since the final velocity is 0 (block comes to rest), the equation simplifies to:
0 = (4 m/s)^2 + 2(-acceleration)d
Solving for distance, we find:
d = (4 m/s)^2 / (2 x -acceleration)
Now, we can substitute the calculated acceleration into the equation to find the distance the block slides.
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A runner of mass 60.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.50m/s . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.190rad/s relative to the earth. The radius of the turntable is 3.60m , and its moment of inertia about the axis of rotation is 81.0kg*m2 .
A) Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat the runner as a particle.)
answer in rad/s please
To find the final angular velocity of the system, we need to apply the principle of conservation of angular momentum. The final angular momentum can be obtained by equating the initial angular momentum and the final angular momentum of the system. Solving for the final angular velocity gives us a value of approximately 38.54 rad/s.
Explanation:To find the final angular velocity of the system, we need to apply the principle of conservation of angular momentum. The initial angular momentum of the system is given by:
Li = Itωt + Irωr
Where It and Ir are the moments of inertia of the turntable and the runner respectively, and ωt and ωr are their respective angular velocities.
Since the runner comes to rest relative to the turntable, ωr = 0. Therefore, the final angular momentum of the system is:
Lf = Itωt
Using the conservation of angular momentum principle, we can set Li equal to Lf:
Itωt = Itωt
Substituting the given values:
81.0kg × m² × 0.190rad/s = It × ωt
Solving for ωt, we find that the final angular velocity of the system is approximately 38.54 rad/s.
To find the final angular velocity of the runner and turntable system, we apply conservation of angular momentum. The final angular velocity is calculated to be 0.611 rad/s. This involves determining the initial angular momenta of both the runner and the turntable, then using the total moment of inertia to find the final velocity.
To find the final angular velocity of the system when the runner comes to rest relative to the turntable, we need to apply the principle of conservation of angular momentum. The initial angular momentum of the system (runner plus turntable) must equal the final angular momentum since there are no external torques acting.
Step-by-Step Calculation :
Determine the initial angular momentum of the runner:
The runner's linear velocity is 2.50 m/s, and they can be treated as a particle moving on a circular path with radius 3.60 m. So, the initial angular momentum (L_runner) is the product of the runner's mass (m), velocity (v), and radius (r):L_runner = m * v * r = 60.0 kg * 2.50 m/s * 3.60 m = 540 kg·m²/s.Determine the initial angular momentum of the turntable:
The initial angular velocity of the turntable (ω_t) is 0.190 rad/s, and its moment of inertia (I_t) is 81.0 kg·m². So, the initial angular momentum (L_turntable) is:L_turntable = I_t * ω_t = 81.0 kg·m² * 0.190 rad/s = 15.39 kg·m²/s.Calculate the total initial angular momentum:
The total initial angular momentum (L_initial) is:L_initial = L_runner - L_turntable = 540 kg·m²/s - 15.39 kg·m²/s = 524.61 kg·m²/s. (The minus sign indicates the turntable rotates in the opposite direction of the runner.)Find the final angular velocity:
When the runner comes to rest relative to the turntable, their combined angular momentum will be conserved. Let ω_final be the final angular velocity and I_total be the combined moment of inertia. The runner's moment of inertia can be treated as a point mass:I_runner = m * r² = 60.0 kg * (3.60 m)² = 777.60 kg·m².I_total = I_t + I_runner = 81.0 kg·m² + 777.60 kg·m² = 858.60 kg·m².Using conservation of angular momentum:L_initial = I_total * ω_finalω_final = L_initial / I_total = 524.61 kg·m²/s / 858.60 kg·m² = 0.611 rad/s.The final angular velocity of the system is 0.611 rad/s.
Tarzan swings on a 35.0 m long vine initially inclined at an angle of 44.0◦ with the vertical. The acceleration of gravity if 9.81 m/s2.
What is his speed at the bottom of the swing if he
a) starts from rest?
b) pushes off with a speed of 6.00 m/s?
Answer:
(A) Vf = 13.8 m/s
(B) Vf = 15.1 m/s
Explanation:
length of rope (L) = 35 m
angle to the vertical = 44 degrees
acceleration due to gravity (g) = 9.8 m/s^{2}
(A) from conservation of energy
final kinetic energy + final potential energy = initial kinetic energy + initial potential energy
0.5m(Vf)^{2} + mg(Hf) = 0.5m(Vi)^{2} + mg(Hi)
where
m = mass
Hi = initial height = 35 cos 44 = 25.17
Hf = final height = length of vine = 35 m
Vi = initial velocity = 0 since he starts from rest
Vf = final velocity
the equation now becomes
0.5m(Vf)^{2} + mg(Hf) = mg(Hi)
0.5m(Vf)^{2} = mg (Hi - Hf)
0.5(Vf)^{2} = g (Hi - Hf)
0.5(Vf)^{2} = 9.8 x (25.17 - 35)
0.5(Vf)^{2} = - 96.3 (the negative sign tells us the direction of motion is downwards)
Vf = 13.8 m/s
(B) when the initial velocity is 6 m/s the equation remains
0.5m(Vf)^{2} + mg(Hf) = 0.5m(Vi)^{2} + mg(Hi)
m(0.5(Vf)^{2} + g(Hf)) = m(0.5(Vi)^{2} + g(Hi))
0.5(Vf)^{2} + g(Hf) = 0.5(Vi)^{2} + g(Hi)
0.5(Vf)^{2} = 0.5(Vi)^{2} + g(Hi) - g(Hf)
0.5(Vf)^{2} = 0.5(6)^{2} + (9.8 x (25.17 - 35))
0.5(Vf)^{2} = -114.3 ( just as above, the negative sign tells us the direction of motion is downwards)
Vf = 15.1 m/s
Answer:
a) [tex]v_{f} \approx 0.328\,\frac{m}{s}[/tex], b) [tex]v_{f} \approx 6.009\,\frac{m}{s}[/tex]
Explanation:
Let consider that bottom has a height of zero. The motion of Tarzan can be modelled after the Principle of Energy Conservation:
[tex]U_{g,1} + K_{1} = U_{g,2} + K_{2}[/tex]
The final speed is:
[tex]K_{2} = U_{g,1} - U_{g,2} + K_{1}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v_{f}^{2} = m\cdot g \cdot L\cdot (\cos \theta_{2}-\cos \theta_{1}) + \frac{1}{2}\cdot m \cdot v_{o}^{2}[/tex]
[tex]v_{f}^{2} = 2 \cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1}) + v_{o}^{2}[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot L \cdot (\cos \theta_{2}-\cos \theta_{1})}[/tex]
a) The final speed is:
[tex]v_{f} = \sqrt{(0\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (35\,m)\cdot (\cos 0^{\textdegree}-\cos 44^{\textdegree})}[/tex]
[tex]v_{f} \approx 0.328\,\frac{m}{s}[/tex]
b) The final speed is:
[tex]v_{f} = \sqrt{(6\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (35\,m)\cdot (\cos 0^{\textdegree}-\cos 44^{\textdegree})}[/tex]
[tex]v_{f} \approx 6.009\,\frac{m}{s}[/tex]
Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on the ground directly underneath it when flying at an altitude of 2.25 km?
Answer:71 dB
Explanation:
Given
sound Level [tex]\beta _1=124 dB[/tex]
distance [tex]r_1=5.01 m[/tex]
From sound Intensity
[tex]\beta =10dB\log (\frac{I_1}{I_0})[/tex]
[tex]124=10dB\log (\frac{I_1}{I_0})[/tex]
[tex]12.4=\log (\frac{I_1}{I_0})[/tex]
[tex]I_1=(1\times 10^{-12})\times 10^{12.4}[/tex]
[tex]I_1=2.51 W/m^2[/tex]
we know Intensity [tex]I\propto ^\frac{1}{r^2}[/tex]
[tex]I_1r_1^2=I_2r_2^2[/tex]
[tex]I_2=I_1(\frac{r_1}{r_2})^2[/tex]
[tex]I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2[/tex]
[tex]I_2=1.24\times 10^{-5} W/m^2[/tex]
Sound level corresponding to [tex]I_2[/tex]
[tex]\beta _2=10\log (\frac{I_2}{I_0})[/tex]
[tex]\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})[/tex]
[tex]\beta _2=70.93\approx 71 dB[/tex]
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats per second. The tuner increases the tension on the as-yet untuned string, and now when they are played together the beat frequency is
1.0s−1.
(a) What was the original frequency of the untuned string?
(b) By what percentage did the tuner increase the tension on that string?
Final answer:
To find the two possible frequencies of the untuned piano string, we can use the formula for beat frequency. From the given information, the original frequency of the untuned string can be either 266.0 Hz or 262.0 Hz. To find the percentage increase in tension on the untuned string, we can use the formula for calculating percentage increase.
Explanation:
To find the two possible frequencies of the untuned piano string, we can use the formula for beat frequency:
Beat frequency = |Frequency of the first string - Frequency of the second string|
In this case, the beat frequency is given as 2.00 s. The frequency of the first string is 264.0 Hz. Let's assume the frequency of the second string is x Hz.
So, we can set up the equation:
2.00 = |264.0 - x|
Solving for x, we get two possible frequencies: 266.0 Hz and 262.0 Hz.
To find the original frequency of the untuned string, we can use the formula:
Original frequency = Frequency of the first string ± Beat frequency
For positive beat frequencies, the original frequency would be:
Original frequency = 264.0 + 2.00 = 266.0 Hz
For negative beat frequencies, the original frequency would be:
Original frequency = 264.0 - 2.00 = 262.0 Hz
To find the percentage increase in tension on the untuned string, we can use the formula:
Percentage increase = (Change in tension / Original tension) x 100
Since the tension on the first string is unchanged (as it is the tuned string), the change in tension on the untuned string is equal to the change in frequency. Assuming the original frequency of the untuned string is 262.0 Hz:
Change in tension = |Original frequency - New frequency|
Change in tension = |262.0 - 264.0| = 2.0 Hz
Therefore, the percentage increase in tension on the untuned string is:
(2.0 / 262.0) x 100 = 0.763%
A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg person running tangentially to the rim of the merry-go-round at 3.51 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.
Calculate the final kinetic energy for this system.
Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the surface of the water.
You want to load bricks onto the floating block and then push it horizontally through the water to an island where you are building an outdoor grill.
a. What is the volume of the block? Express your answer with the appropriate units
b. What is the maximum mass of bricks that you can place on the block without causing
it to sink below the water surface? Express your answer with the appropriate units
Answer:
a) V = 0.085 m^3
b) m = 17 kg
Explanation:
1) Data given
mb = 68 kg (mass for the block)
20% of the block volume is floating
100-20= 80% of the block volume is submerged
2) Notation
mb= mass of the block
Vw= volume submerged
mw = mass water displaced
V= total volume for the block
3) Forces involved (part a)
For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)
Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :
B = (mass water displaced) g = (mw) g (1)
The definition of density is :
[tex] \rho_w = \frac{m_w}{V_w} [/tex]
If we solve for mw we got [tex]m_w = \rho_w V_w [/tex] (2)
Replacing equation (2) into equation (1) we got:
[tex] B = \rho_w V_w g [/tex] (3)
On this case Vw represent the volume of water displaced = 0.8 V
If we replace the values into equation (3) we have
0.8 ρ_w V g = mg (4)
And solving for V we have
V = (mg)/(0.8 ρ_w g )
We cancel the g in the numerator and the denominator we got
V = (m)/(0.8 ρ_w)
V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3
4) Forces involved (part b)
For this case we have bricks above the block, and we want the maximum mass for the bricks without causing it to sink below the water surface.
We can begin finding the weight of the water displaced when the block is just about to sink (W1)
W1 = ρ_w V g
W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N
After this we can calculate the weight of water displaced before putting the bricks above (W2)
W2 = 0.8 x 833 N = 666.4 N
So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)
W3 = W1 -W2 = 833-666.4 N = 166.6 N
And finding the mass fro the definition of weight we have
m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg
Final answer:
The volume of the block of wood and the maximum mass of bricks it can carry without sinking are determined using principles of buoyancy and uniform density.
Explanation:
(a) To determine the volume of the block of wood, we use the fact that 20.0% of its volume is above the water. Given a uniform density, this means the submerged volume is 80% of the total volume. Hence, the volume of the block is 0.8 x (68.0 kg / 920 kg/m³) = 0.472 m³.
(b) The maximum mass of bricks that can be loaded without sinking the block can be calculated using the concept of buoyancy. The buoyant force equals the weight of the water displaced, so the mass of the bricks should not exceed the mass of the water displaced by the volume of the submerged part of the block. Therefore, the maximum mass of bricks is 0.472 m³ x 1100 kg/m³ x 9.8 m/s² = 5148.64 kg.
Soil tilth refers to ________. Select one:
A. ratio of bulk density to particle density
B. the moisture content at which a soil is best suited for tillage
C. the physical suitability of a soil for plant growth
D. the bearing strength of a soil under a given downward force
E. micro-aggregates produced as a by-product of tillage
Answer:
Option c
Explanation:
Soil tilth refers to the physical condition of the soil, particularly with regards to the suitability of the soil for the growth of crop.
The determinants of the tilth in the soil incorporates the arrangement and steadiness of aggregated particles of the soil, pace of water penetration, level of air circulation, dampness content and seepage.
Thus option C follows the definition of the soil tilth.
Answer:
Option (C)
Explanation:
The soil tilth is defined as one of the physical property that determines the necessary condition for the better growth of plants and crops.
This property depends upon various factors such as-
fertility of the soil amount of moisture content in the soil rate and type of drainage pattern rate of percolation of water into the deeper zone of the soil porosity and permeability of the soil.These are the main features that characterize the soil tilth, and fulfilling the above required necessary condition, a plant grows in a much better way.
Thus, the correct answer is option (C).
Which of the following statements are true as a block of ice melts? Check all that apply.
a. The temperature of the ice/water system remains constant.
b. Heat energy leaves the ice/water system.
c. Heat energy enters the ice/water system.
d. The temperature of the ice/water system decreases.
e. The temperature of the ice/water system increases.
Answer:
a. The temperature of the ice/water system remains constant.
c. Heat energy enters the ice/water system.
Explanation:
As we know that when ice coverts into the water then it is known as phase transfer. In the phase transfer process temperature and the pressure remains constant but the heat enters into the system. This heat is responsible for melting the ice. The heat is taken by ice is known as latent heat.
Therefore the option "a" and "c" are correct.
The true statements as a block of ice melts are that the temperature of the ice/water system remains constant, and heat energy enters the system. The temperature does not change, increase, or decrease during the melting process until all the ice is melted. So the correct options are a and c.
Explanation:As a block of ice melts, several key processes occur within the ice/water system:
Therefore, from the options provided:
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The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one-third of the wooden block is submerged. Is the unknown liquid more or less dense than water?
Final answer:
A wooden block that floats in an unknown liquid with only one-third submerged indicates that the unknown liquid is more dense than water.
Explanation:
When the wooden block is placed in an unknown liquid and only one-third of it is submerged, it suggests that the block is less dense than the unknown liquid. This is due to the principle of buoyancy, which states that an object will displace a volume of fluid that is equal to its own weight. Since the wooden block floats higher in the unknown liquid compared to how it floats in water, where it must submerge more to displace enough water to equal its weight, we can conclude that the unknown liquid is more dense than water.
A time-varying horizontal force F(t) = At4 + Bt2 acts for 0.500 s on a 12.25-kg object, starting at time . In the SI system, A has the numerical value 4.50 and B has the numerical value 8.75. (a) What are the SI units of A and B? (b) What impulse does this force impart to the object? 5) (a) A: N/s4 = kg • m/s6, B: N/s2 = kg • m/s4 (b) 12.9 N • s, horizontally
Answer:
Part a)
[tex]A = \frac{N}{s^4}[/tex]
[tex]B = \frac{N}{s^2}[/tex]
PART B)
[tex]I = 0.393 Ns[/tex]
Explanation:
PART A)
As we know that the force is given as
[tex]F = At^4 + B t^2[/tex]
here we know that each term of the equation must have same dimensions
so we will have
[tex]At^4 = N[/tex]
[tex]A = \frac{N}{s^4}[/tex]
similarly for other term
[tex]Bt^2 = N[/tex]
[tex]B = \frac{N}{s^2}[/tex]
PART B)
Impulse given by the force is given as
[tex]impulse = \int Fdt[/tex]
now we have
[tex]I = \int (At^4 + Bt^2)dt[/tex]
[tex]I = \int (4.50 t^4 + 8.75 t^2) dt[/tex]
[tex]I = \frac{4.50(0.5)^5}{5} + \frac{8.75(0.5)^3}{3}[/tex]
[tex]I = 0.028 + 0.36[/tex]
[tex]I = 0.393 Ns[/tex]
The SI units of the constants A and B are kg·m/s·6 and kg·m/s·4 respectively, essential for ensuring dimensional consistency in the force equation. The calculated impulse imparted to the object by this varying force over 0.500 s is 12.9 N·s.
The question asks two parts: (a) to determine the SI units of constants A and B in the equation F(t) = At4 + Bt2, and (b) to calculate the impulse imparted to the object by this force during 0.500 s. To address part (a), we recognize that force (F) has SI units of kg·m/s2, known as newtons (N). To ensure dimensional consistency, the units of A must be N/s4 = kg·m/s6, and the units of B must be N/s2 = kg·m/s4, as these adjustments yield a force measurement when applied to time (t) in seconds. For part (b), impulse, which is the integral of force over time, necessitates calculating the definite integral of F(t) from 0 to 0.500 s. Applying the specific values given for A and B, and after the integration process, the impulse imparted to the 12.25-kg object is found to be 12.9 N·s, horizontally.
The half-life of Actinium 227 decays in 20 years. Calculate the mass of the element left when a 2kg sample was left for 160 years.
The mass of Actinium 227 left is 0.0078 kg
Explanation:
The amount of mass left of a radioactive isotope after time t is given by the equation:
[tex]m(t) = m_0 (\frac{1}{2})^{-\frac{t}{\tau_{1/2}}}[/tex]
where
[tex]m_0[/tex] is the initial amount of the sample
t is the time
[tex]\tau_{\frac{1}{2}}[/tex] is the half-life of the isotope
For the sample of Actinium 227 in this problem,
[tex]m_0 = 2 kg[/tex]
[tex]\tau_{1/2}=20 years[/tex]
t = 160 years
Substituting into the equation,
[tex]m(160) = (2 kg) (\frac{1}{2})^{-\frac{160}{20}}=0.0078 kg[/tex]
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The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 18 m wide and averages 22 m deep.(a) What is the average speed (in m/s) of the river in the gorge?_______m/s.(b) What is the average speed (in m/s) of the water in the river downstream of the falls when it widens to 63 m and its depth increases to an average of 42 m________m/s.
Answer
Given,
Flow rate of river is equal to 300,000 L/s.
Width of river = 18 m
and depth of river = 22 m
a) Average speed of river
Q = 300,000 L/s
= 300 m³/s
Q = Av
[tex]v = \dfrac{Q}{A}[/tex]
[tex]v = \dfrac{300}{18 \times 22}[/tex]
[tex]v = \dfrac{300}{396}[/tex]
[tex]v = 0.757\ m/s[/tex]
b) Average speed when river is widen to 63 m and depth is increased to
[tex]v = \dfrac{Q}{A}[/tex]
[tex]v = \dfrac{300}{63 \times 42}[/tex]
[tex]v = \dfrac{300}{2646}[/tex]
[tex]v = 0.113\ m/s[/tex]
Final answer:
The average speed of the river in the gorge is 0.75 m/s, and downstream of the falls, when the river widens and its depth increases, the average speed decreases to 0.125 m/s.
Explanation:
To calculate the average speed of the river at Huka Falls, we can use the formula for flow rate, which is the volume of fluid passing a point in the river per unit of time:
Flow rate (Q) = Area (A) imes Velocity (V)
(a) Average speed in the gorge: Given that the flow rate (Q) is 300,000 liters per second (which is equal to 300 cubic meters per second, since 1,000 liters is equal to 1 cubic meter), and the cross-sectional area of the river in the gorge (A) is 20 meters wide imes 20 meters deep (400 square meters), we can solve for velocity (V) using the formula:
Q = A imes V
300 m³/s = 400 m² imes V
V = 300 m³/s \/ 400 m²
V = 0.75 meters per second
(b) Average speed downstream of the falls: Downstream, the river widens to 60 meters and deepens to an average of 40 meters, so the cross-sectional area is 2,400 square meters. Using the same flow rate, we can find the new velocity:
Q = A imes V
300 m³/s = 2,400 m² imes V
V = 300 m³/s \/ 2,400 m²
V = 0.125 meters per second
(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface is 3.00 m/s2 ? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1000 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?
Final answer:
The escape velocity of a spherical asteroid with a given radius and given gravitational acceleration can be calculated using a special formula. The distance a particle travels from the surface when it leaves it at a given radial velocity can be determined using the projectile height equation. The speed at which an object hits an asteroid after falling from a given height can be calculated using the falling object terminal velocity equation.
Explanation:
(a) Escape velocity can be defined as the minimum speed required for an object to escape the gravitational pull of a celestial body. To calculate the escape velocity on a spherical asteroid we can use the formula:
escape velocity = sqrt(2 * acceleration due to gravity * radius).
Using the given values, the escape velocity of a spherical asteroid is:
exhaust rate = sqrt(2 * 3.00 m/s2 * 500 000 m) = 6928 m/s.
(b) To calculate the distance from the surface that a particle will travel when it leaves the surface of the asteroid with a radial velocity of 1000 m/s, we can use the formula for the height of the projectile:
Height = (radial velocity)2 / (2 * acceleration due to gravity).
After entering the specified values, the particle travels the distance:
Altezza = (1.000 m/s)2 / (2 * 3,00 m/s2) = 166.666,67 m.
(c) To calculate the speed at which an object hits an asteroid as it falls 1000 km above the surface, we can use the equation for the final velocity of the falling object:
Final velocity = sqrt (initial velocity2 + 2 * acceleration due to gravity * height).
Substituting the given values, the object hits the asteroid with a speed of:
Final velocity = square(0 + 2 * 3.00 m/s2 * 1,000,000 m) = square(6,000,000 m2/s2) = 2,449 m/s.
As electrons are passed through the system of electron carriers associated with photosystem ii
A ledge on a building is 20 m above the ground. A taut rope attached to a 4.0 kg can of paint sitting on the ledge passes up over a pulley and straight down to a 3.0 kg can of nails on the ground. If the can of paint is accidently knocked off the ledge, what time interval does a carpenter have to catch the can before it smashes on the floor?
Answer:
t = 5.4 s
Explanation:
from the question we are given :
height (s) = 20 m
mass of paint (Mp) = 4 kg
mass of nails (Mn) = 3 kg
acceleration due to gravity (g) = 9.8 m/s^{2}
The net force accelerating the can of paint should be equal to the difference in weight of the can of paint and the can of nails.weight of nails = mass of nails x g = 3 x 9.8 = 29.4 N
weight of paint = mass nails x g = 4 x 9.8 = 39.2 N
net force = 39.2 - 29.4 = 9.8 N
net force = total mass x acceleration9.8 = (3 +4) x a
a = 1.4 m/s^{2}
from S = Ut + 0.5at^{2} we can get the time the carpenter has to catch the nailswhere U is the initial velocity and is 0 since the can was initially at
rest
20 = (0 x t) + (0.5 x 1.4 x t^{2})
20 = 0.7 x t^{2}
t^{2} = 28.6
t = 5.4 s
What role do earth’s layers play in the formation of metamorphic rock?
A helicopter lifts a 60 kg astronaut 17 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. (a) How much work is done on the astronaut by the force from the helicopter? J (b) How much work is done on the astronaut by her weight? J (c) What is the kinetic energy? J (d) What is the speed of the astronaut just before she reaches the helicopter?
Answer:
a) W = 10995.6 J
b) W = - 9996 J
c) Kf = 999.6 J
d) v = 5.77 m/s
Explanation:
Given
m = 60 Kg
h = 17 m
a = g/10
g = 9.8 m/s²
a) We can apply Newton's 2nd Law as follows
∑Fy = m*a ⇒ T - m*g = m*a ⇒ T = (g + a)*m
where T is the force exerted by the cable
⇒ T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)
⇒ T = 646.8 N
then we use the equation
W = F*d = T*h = (646.8 N)*(17 m)
W = 10995.6 J
b) We use the formula
W = m*g*h ⇒ W = (60 Kg)(9.8 m/s²)(-17 m)
⇒ W = - 9996 J
c) We have to obtain Wnet as follows
Wnet = W₁ + W₂ = 10995.6 J - 9996 J
⇒ Wnet = 999.6 J
then we apply the equation
Wnet = ΔK = Kf - Ki = Kf - 0 = Kf
⇒ Kf = 999.6 J
d) Knowing that
K = 0.5*m*v² ⇒ v = √(2*Kf / m)
⇒ v = √(2*999.6 J / 60 Kg)
⇒ v = 5.77 m/s
Answer:
Explanation:
mass =60kg d = 17m a=g/10
(a) work done on the astronaut by the force from the helicopter = fd
but f =m(g+a)
w= m( g+g/10)d
wt = 11/10 mgd
w =11/10 * 60 *9.8 * 17 = 10995.6J = 1IKJ
(b) workdone by her weight = -mgh
= 60*9.8* 17 = -9996J
(C) Kinetic energy = wt + w
= (10995.6 - 9996)J = 999.6J
(d) Kinetic energy =1/2m[tex]v^{2}[/tex]
hence velocity = [tex]\sqrt{2ke/m}[/tex] = 5.777m/s
Confidence intervals for the population mean μ and population proportion p _____ as the size of the sample increases.
Answer:
becomes narrower
Explanation:
Confidence intervals for the population mean μ and population proportion p becomes narrower as the size of the sample increases.
As,the sample size increases,standard error decreases,so margin of error decreases and hence width of CI decreases.
A car with a mass of 1.50 X 10^3 kg starts from rest and accelerates to a speed of 18.0 m/s in 12.0 s. Assume that the force of resistance remains constant at 400.0 N during this time. what is the average power developed by the car's engine?
The average power is 20.3 kW
Explanation:
First of all, we calculate the work done on the car: the work-energy theorem states that the work done on the car is equal to the change in kinetic energy of the car, so we have
[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
W is the work done
[tex]K_i, K_f[/tex] are the initial and final kinetic energy of the car
[tex]m=1.50\cdot 10^3 kg = 1500 kg[/tex] is the mass of the car
u = 0 is the initial velocity
v = 18.0 m/s is the final velocity
Substituting,
[tex]W=\frac{1}{2}(1500)(18)^2=2.43\cdot 10^5 J[/tex]
Now we can find the average power developed by the car's engine, which is given by
[tex]P=\frac{W}{t}[/tex]
where
[tex]W=2.43\cdot 10^5 J[/tex] is the work done
t = 12.0 s is the time taken
Substituting,
[tex]P=\frac{2.43\cdot 10^5 J}{12.0}=20,250 W = 20.3 kW[/tex]
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The blood plays an important role in removing heat from th ebody by bringing the heat directly to the surface where it can radiate away. nevertheless, this heat must still travel through the skin before it can radiate away. we shall assume that the blood is brought to the bottom layer of skin at a temperature of 37.0 degrees C and that its outer surface of the skin is at 30.0 degrees C. Skin varies in thickness from 0.500mm to a few millimeters on the palms and the soles so we shall assume an average thickness off 0.740mm. a 165lb, 6 ft person has a surface area of about 2.00 m^2 and loses heat at a net rate of 75.0 w while resting. On the basis of our assumptions, what is the thermal conductivity of this persons skin?
Answer: Thermal comductivity (K) is 3.964x 10 ^-3 W/m.k
Explanation:
Thermal comductivity K = QL/A∆T
Q= Amount of heat transferred through the material in watts = 75W
L= Distance between two isothermal planes = 0.740mm
A= Area of the surface in square metres = 2m^2
∆T= Temperature change = (37-30) °C.
Solving this : K =( 75 x 0.740 x 10^-3)/ 2 x (37-30)
K = 3.964x 10 ^-3 W/m.k
The question pertains to calculating the thermal conductivity of human skin, a Physics concept linked to heat transfer. By using Fourier's Law of heat conduction, and rearranging the formula to solve for thermal conductivity using given data, an approximate thermal conductivity can be obtained.
Explanation:The query is related to the determination of thermal conductivity of human skin based on the known parameters. This phenonmenon belongs to the domain of Physics, specifically heat transfer. Here, thermal conductivity is the measure of a material's ability to conduct heat. In this scenario, you have to consider the heat conduction through the skin, which relies on Fourier's Law of heat conduction. It can be represented as:
Q = (k*A*(T1 - T2))/d
Where, Q is the heat transfer rate, k is the thermal conductivity, A is the area of heat transfer, T1 and T2 are the initial and final temperatures, and d is the thickness of the material, in this case, the skin.
From the given data, you can plug in the values into this formula. However, our primary objective is to find out 'k'. Rearranging the formula to find k gives us:
k = (Q * d) / (A * (T1 - T2))
Now, if we put all the given values into the formula, we get:
k = (75 W * 0.00074 m) / (2 m^2 * (37°C - 30°C))
Solving this would provide us with the estimate of thermal conductivity of the skin.
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Russell drags his suitcase 15.0 M from the door of his house to the car at a constant speed with a horizontal force of 95.0 N. How much work does Russell do to overcome the force of
The work done is 1425 J
Explanation:
The work done by a force to move an object is given by
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
For the suitcase in this problem, we have:
F = 95.0 N is the force applied
d = 15.0 m is the displacement
[tex]\theta=0[/tex], since the force is parallel to the displacement
Substituting, we find
[tex]W=(95.0)(15.0)(cos 0)=1425 J[/tex]
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Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is transverse, travels to the right, and has a displacement of 0.19 m at t = 0 and x = 0.
Answer:
[tex]y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})[/tex]
Explanation:
As we know that the wave equation is given as
[tex]y = A sin(\omega t - k x + \phi_0)[/tex]
now we have
[tex]A = 0.19 m[/tex]
[tex]\lambda = 2.6 m[/tex]
so we have
[tex]k = \frac{2\pi}{\lambda}[/tex]
[tex]k = \frac{2\pi}{2.6}[/tex]
[tex]k = 2.42 per m[/tex]
also we have
[tex]T = 1.2 s[/tex]
so we have
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{1.2}[/tex]
[tex]\omega = 5.23 rad/s[/tex]
now we know that at t = 0 and x = 0 wave is at y = 0.19 m
so we have
[tex]\phi_0 = \frac{\pi}{2}[/tex]
so we have
[tex]y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})[/tex]
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 34.3 m/s2 . The acceleration period lasts for time 6.00 s until the fuel is exhausted. After that, the rocket is in free fall. True or False.
Answer:
True
Explanation:
t = Time taken = 6 s
u = Initial velocity = 0
v = Final velocity
a = Acceleration = 34.3 m/s²
[tex]v=u+at\\\Rightarrow v=0+34.3\times 6\\\Rightarrow v=205.8\ m/s[/tex]
After the six seconds the acceleration from the rocket stops. After the fuel is exhausted the velocity of the rocket will be 205.8 m/s which will reduce by the acceleration due to gravity (i.e., 9.81 m/s²).
Free fall is the state of a body where the only the force of gravity is acting on it other forces are not acting on it.
Hence, the statement here is true.
A sled having a certain initial speed on a horizontal surface comes to rest after traveling 10 m. If the coefficient of kinetic friction between the object and the surface is 0.20, what was the initial speed of the object
Answer:
the initial speed of the object is 6.26 m/s
Explanation:
given information:
distance, s = 10 m
the coefficient of kinetic friction, μ = 0.2
we use the equation where the kinetic energy is equal to the friction force.
kinetic energy, KE = [tex]\frac{1}{2} mv^{2}[/tex]
friction work, W = F(friction) s
KE = W
[tex]\frac{1}{2} mv^{2}[/tex] = F(friction) s
where, F(friction) = μ N, N is normal force (N = m g)
= μ m g
so,
[tex]\frac{1}{2} mv^{2}[/tex] = μ m g s
[tex]\frac{1}{2} v^{2}[/tex] = μ g s
[tex]v^{2}[/tex] = 2 μ g s
= 2 (0.2) (9.8) (10)
= 39.2
hence,
v = [tex]\sqrt{3.92}[/tex]
= 6.26 m/s
The initial speed [tex]\( v_i \)[/tex] of the sled was approximately[tex]\( 6.26 \, \text{m/s} \).[/tex]
The initial speed of the object can be found using the work-energy principle, which states that the work done by friction will be equal to the change in kinetic energy of the sled.
First, let's calculate the work done by friction. The force of friction [tex]\( F_{\text{friction}} \)[/tex] is given by the product of the normal force \( N \) and the coefficient of kinetic friction [tex]\( \mu_k \):[/tex]
[tex]\[ F_{\text{friction}} = \mu_k N \][/tex]
[tex]\[ W = F_{\text{friction}} d = \mu_k m g d \][/tex]
[tex]\[ \Delta KE = 0 - \frac{1}{2} m v_i^2 = -\frac{1}{2} m v_i^2 \][/tex]
According to the work-energy principle, the work done by friction is equal to the change in kinetic energy:
[tex]\[ \mu_k m g d = -\frac{1}{2} m v_i^2 \][/tex]
Solving for \( v_i \), we get:
[tex]\[ v_i^2 = -2 \mu_k g d \][/tex]
[tex]\[ v_i = \sqrt{-2 \mu_k g d} \][/tex]
Given that [tex]\( \mu_k = 0.20 \), \( g = 9.81 \, \text{m/s}^2 \), and \( d = 10 \, \text{m} \),[/tex] we can plug in these values:
[tex]\[ v_i = \sqrt{-2 \times 0.20 \times 9.81 \times 10} \][/tex]
[tex]\[ v_i = \sqrt{39.24} \][/tex]
[tex]\[ v_i \approx 6.26 \, \text{m/s} \][/tex]
Therefore, the initial speed \( v_i \) of the sled was approximately[tex]\( 6.26 \, \text{m/s} \).[/tex]
The answer is: [tex]6.26 \, \text{m/s}.[/tex]
You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 23.0 cm. Starting from rest at t = 0, the flywheel rotates with constant angular acceleration 3.00 rad/s2 about an axis perpendicular to the flywheel at its center.If the flywheel has a density (mass per unit volume) of 8600 kg/m3, what thickness must it have to store 800 J of kinetic energy at t = 8.00 s?
Answer:
t = 0.0735 m
Explanation:
Angular acceleration of the flywheel is given as
[tex]\alpha = 3 rad/s^2[/tex]
now after t = 8 s the speed of the flywheel is given as
[tex]\omega = \alpha t[/tex]
[tex]\omega = 3 \times 8 [/tex]
[tex]\omega = 24 rad/s[/tex]
now rotational kinetic energy of the wheel is given as
[tex]K = \frac{1}{2}I\omega^2[/tex]
[tex]K = \frac{1}{2}(\frac{1}{2}mR^2)(24^2)[/tex]
[tex]800 = \frac{1}{4}m(0.23)^2(24^2)[/tex]
[tex]m = 105 kg[/tex]
now we have
[tex]m = \rho (\pi R^2) t[/tex]
[tex]105 = 8600(\pi \times 0.23^2) t[/tex]
[tex]t = 0.0735 m[/tex]
We have that for the Question, it can be said that thickness must it have to store 800 J of kinetic energy at t = 8.00 s
h=0.0735m
From the question we are told
You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 23.0 cm. Starting from rest at t = 0, the flywheel rotates with constant angular acceleration 3.00 rad/s2 about an axis perpendicular to the flywheel at its center.If the flywheel has a density (mass per unit volume) of 8600 kg/m3,what thickness must it have to store 800 J of kinetic energy at t = 8.00 s?Generally the equation for the is mathematically given as
[tex]N=\pir^2h*P\\\\N=3.14*(23*10^{-2}^2)*h*8600\\\\N=1428.5h\\\\[/tex]
Therefore
[tex]E=\frac{1}{2}Iw^2\\\\800=\frac{1}{2}*(\frac{NR^2}{2}w^2)\\\\800=\frac{1}{2}*(\frac{(1428.523*10^{-2})^2}{2}(2*8)^2)[/tex]
h=0.0735m
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