The location of the centre of the mass of the CO molecule relative to the carbon atom is 6.46 ×10⁻¹¹ m.
What is the centre of mass of the molecule?The Centre of Mass of that object is a position that is relative to an object or a system of objects. The average position of all parts of the system that are weighted to their masses is known as the Centre of Mass.
Given, the distance between the carbon and oxygen atoms, r = 1.13 ×10⁻¹⁰m.
The ratio of the mass of the carbon and oxygen:
Mc/Mo = 12/16 = 0.750
Consider that Xc is the position of carbon and Xo is the position of the oxygen atom in CO molecule.
The centre of the mass of the system is equal to the:
[tex]X_{co}=\frac{M_CX_C+M_OX_O}{M_C+M_O}[/tex]
[tex]X_{co}=\frac{(M_C/M_O)X_C+X_O}{M_C/M_O+1}[/tex]
As we are calculating the centre of the mass relative to the carbon atom.
Therefore Xc = 0 and Xo= 1.13 ×10⁻¹⁰m.
[tex]X_{CO} =\frac{1.13\times 10^{-10}}{0.750 +1}[/tex]
Xco = 6.46 ×10⁻¹¹ m
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The figure shows a standing wave oscillating at 100 Hz on a string. What is the wave speed ?
Answer:
The wavelength is 60 cm.
Explanation:
Speed = frequency x wavelength = 100 x 60 = 6000 cm/s
or if you wanted the answer in m/s
Speed = frequency x wavelength= 100 x 0.60 = 60 m/s
If your 1000 kg car ran out of gas and you had to push it into the gas station with a
force of 50 N, what would the acceleration of the car be?
Answer:
a=0.05m/s²
Explanation:
force=mass × acceleration
f=ma
f=50N
m=1000kg
a=?
50=1000×a
50=1000a
a=50/1000
a=0.05m/s²
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It takes 225 kJ of work to accelerate a car from 20.1 m/s to 28.1 m/s. What is the car's mass?
Answer:
THE REMAINIG WILL BE 75
Explanation:
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The tortoise and the hare A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s.
The time taken for the hare to catch up with the tortoise is 10 seconds.
What is the time taken for the hare to catch the tortoise?
The time taken for the hare to catch up with the tortoise is calculated by applying the principle of relative velocity as shown below.
Distance travelled by hare + Distance travelled by tortoise = Total distance
V₁t + V₂t = d
where;
V₁ is the velocity of hareV₂ is the velocity of tortoiset is the time taken for them to meetd is the distance between themSince they are moving in the same direction, the relative velocity becomes
V₁t - V₂t = d
(V₁ - V₂)t = d
(4 - 1)t = 30
3t = 30
t = 30/3
t = 10 seconds
Thus, the time taken for the hare to catch up with the tortoise is 10 seconds.
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The complete question is below:
A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s. At what time does the hare catch up with the tortoise?
For the following questions, refer to the position versus time graph above.a. What is the object's velocity att = 5 s? (2 sig figs)b. What is the object's velocity at t=40s? (2 sig figs)
We are given a graph of position vs time. We are asked to determine the speed at 5 seconds. To do that we need to determine the slope of the line above the time point of 5 seconds. To determine the slope of the line we will use the following formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]This means that we need to choose two points in the line. We use the graph to determine these points as follows:
The points we have chosen are:
[tex]\begin{gathered} (x_1,y_1)=(0,0)_{} \\ (x_2,y_2)=(10,5) \end{gathered}[/tex]Replacing in the formula for the slope we get:
[tex]m=\frac{5-0}{10-0}[/tex]Solving the operations:
[tex]m=\frac{5}{10}=\frac{1}{2}[/tex]Therefore, the speed at 5 seconds is 0.5 meters per second.
For part B the following points can be taken:
I just started a new lesson and have this study guide, but the material is unfamiliar to me and
2.79 eV
Explanation:Given
The wavelength, λ = 445 nm
The planck's constant, h = 6.626 x 10^-34 m² kg/s
The speed of light, c = 3 x 10^8 m/s
The energy of the photon is calculated as:
[tex]\begin{gathered} E=\frac{hc}{\lambda} \\ \\ E=\frac{6.626\times10^{-34}\times3\times10^8}{445\times10^{-9}} \\ \\ E=\frac{6.626\times3}{445}\times10^{-34+8+9} \\ \\ E=0.0447\times10^{-17} \\ \\ E=4.47\times10^{-2}\times10^{-17} \\ \\ E=4.47\times10^{-19}J \end{gathered}[/tex]But, 1 eV = 1.6 x 10^-19J
[tex]\begin{gathered} E=\frac{4.47\times10^{-19}}{1.6\times10^{-19}} \\ \\ E=2.79eV \end{gathered}[/tex]Students were experimenting with objects in a collision. Ball A moves with a constant acceleration by sliding down a frictionless incline plane before colliding with Ball B. Students used motion detectors to measure the velocity of Ball A at various points. Ball A (mass of 1.00 kg) began from rest, at position 1, on a ramp at a height of 1.25 m. At position 2, ball A was moving at 5.00 m/s. Ball A continues to roll at a constant 5.00 m/s when it collides with Ball B (mass of 1.00 kg) at position 3. Ball A comes to a complete stop. Ball B moves at a constant velocity after the collision.FORMULAS: PE = m•g•h (g=9.8m/s²) KE = 1/2•m•v² momentum = m•v
Given data:
* The mass of ball A is m_1 = 1 kg.
* The mass of ball B is m_2 = 1 kg.
* The initial velocity of ball A is u_1 = 5 m/s.
* The final velocity of ball A is v_1 = 0 m/s.
* The initial velocity of the ball B is u_2 = 0 m/s.
Solution:
According to the law of conservation of momentum, the net momentum of the system before the collision is equal to the net momentum of the system after the collision.
Thus,
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Substituting the known values,
[tex]\begin{gathered} 1\times5+1\times0=1\times0+p_2 \\ 5+0=0+p_2 \\ p_2=5\text{ kgm/s} \end{gathered}[/tex]where p_2 is the momentum of the ball B,
Thus, the momentum of ball B after the collision is 5 kgm/s.
Hence, the third option is the correct answer.
Given two 2.00μC charges on the horizontal axis are positioned at x=0.8m, and the
other at x=-0.8m, and a test charge q = 1.28x10-18C at the origin.
(a) What is the net force exerted on q by the two 2.00μC charges? [5]
(b) What is the electric field at the origin due to 2.00μC charges? [5]
(c) What is the electric potential at the origin due to the two 2.00μC charges
Answer:Question 1
Given q1=2µC
q2=2µC
q= 1.2×10^-18C at origin
Net force exerted by two charges on q
F_1= force on q due to q1
F_2= force on q due to q2
F_net= F_(1-) F_2
= (Kqq_1)/r^2 - (kqq_2)/r^2 Then q_1=q_2=〖2×10〗^(-6)
F_net=0N
b) The electric field at charge q
E_net= E_1- E_2
= (kq_1)/r^2 - (kq_2)/r^2
Then q_1=q_2
E)_net= 0 N/C
c) The electric potential at origin due to two charge
V_net= V_(1 )- V_2
= (kq_1)/r - (kq_2)/r
Then q1= q2
V_net= 0 V
Explanation:
Calculate the depth in the ocean at which the pressure is three times the atmospheric pressure
ANSWER:
20.17 meters
STEP-BY-STEP EXPLANATION:
Given:
Pressure = Po = 1.013x10^5 Pa
Pressure at depth = P = 3Po
Density of sea water = 1025 kg/m^3
We can calculate the depth as follows:
[tex]\begin{gathered} P=P_o+d\cdot g\cdot h \\ \text{ we solve for h} \\ dgh=P-P_o \\ h=\frac{P-P_o}{d\cdot g} \end{gathered}[/tex]We replacing and calculate the depth:
[tex]\begin{gathered} h=\frac{3\cdot P_o-P_o}{1025\cdot9.8}=\frac{2P_o}{10045}=\frac{2\cdot1.013\cdot10^5}{10045} \\ h=20.17\text{ m} \end{gathered}[/tex]Therefore, the depth is equal to 20.17 meters
A tensile load of 190 kN is applied to a round metal bar with a diameter of 16mm and a gage length Of 50mm. Under this load the bar elastically deforms so that the gage length increases to 50.1349 mm and the diameter decreases to 15.99 mm. Determine the modulus of elasticity and Poisson s ratio for this metal.
The elasticity and Poisson's ratio for this metal is 0.232.
What is ratio?
The realation between two numbers which shows how much bigger one quantity is than another.
Sol-
As per the given question
P=190KN
d=16 mm
Lo=50mm
X=50.1349-50=0.1349mm
Y=15.99-16=-0.01mm
The formula-
E=ó/€
Ó=P/A
A=r/4 d^2 =π/4(16)^2=201.062 mm
ó={190(1000)}201.062=944.982 Mpa
E=944.982/0.002698=350.253 GPa
€y=-0.000625
v=0.232(answer)
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What is the relationship of the force on the spring and stretch of the spring.(This is the spring we are looking at weights can be attached to it)
We will have the following:
We know that for a simple spring the following is true:
[tex]\begin{cases}F=-kx \\ \\ F=ma\end{cases}[/tex]So:
[tex]-kx=ma[/tex]So, the stretch of the spring is directly proportional to the force.
Diffraction is:A.the difference in density of the compression and rarefaction parts of a sound wave.B.the change of frequency heard by an observer when sound waves come from a moving source.C.when waves suddenly appear in a medium without a source.D.the apparent bending of sound waves around obstacles.
Diffraction is a phenomenon that occurs when a wave hits an object or it passes through a small gap.
The propagation of the wave will have a circular pattern after the diffraction effect:
Therefore the correct option is D.
A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?
The true statement is the vertical component of the velocity is half of the magnitude of the velocity.
What is vertical component of velocity?
The vertical component of a velocity is the velocity of the object acting along vertical direction.
The vertical component of the velocity acting along the vertical direction is calculated as follows;
Vy = V sinθ
where;
Vy is the vertical component of the velocityθ is the angle of projectionV is the magnitude of the velocitySubstitute the value of the angle of projection and evaluate the vertical component of the velocity.
Vy = V sin(30)
Vy = V(0.5)
Vy = ¹/₂V.
Thus, the vertical component of the velocity is half of the magnitude of the velocity.
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The complete question is below
A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?
the vertical component of the velocity is half of the magnitude of the velocitythe vertical component of the velocity is one-third of the velocitythe vertical component of the velocity is the same as the magnitude of the velocity.5. Was you hypothesis supported by the data collected for question 2? Why or why not ?
Answer:
Your hypothesis was not supported by the data collected
Explanation:
On question 2 you said that the point with the greatest kinetic energy would be point 5. However, when you analyze the data, the point with the greatest velocity was point 2 which means that this is the point with the greatest kinetic energy.
Therefore, your hypothesis was not supported by the data collected because based on the picture, it is hard to say where is the lowest point of the roller coaster.
mick took his friend out to dinner the bill was $40 but he applied a coupon then the total price was $33 what was the % off?
ANSWER
17.5 %
EXPLANATION
We have to find the percent change, given that the initial price was $40 and the final price was $33.
[tex]\text{ \% }change=\frac{final.price-initial.price}{initial.price}\times100[/tex][tex]\text{\% }change=\frac{33-40}{40}\times100=\frac{-7}{40}\times100=-0.175\times100=-17.5\text{\%}[/tex]The coupon was for 17.5% off
Information givenknown: mass of Christine=60 kgmass of cart= 22 kg mass of hailey=69The two girls on the cart to the left pushing off of Conner take .3833 s to reach a distance of 0.3m. Conner reaches the same distance in 0.2333s. What is the mass of Conner?
We will have the following:
We use conservation of momentum to solve, that is:
[tex]\begin{gathered} (60kg+22kg+69kg)(0.3m/0.3833s)=(m+22kg)(0.3m/0.233s) \\ \\ \Rightarrow118.1841899kg\ast m/s=(m+22kg)(\frac{300}{233}m/s) \\ \\ \Rightarrow m+22kg=91.78968976kg\Rightarrow m=69.78968976... \\ \\ \Rightarrow m\approx69.8 \end{gathered}[/tex]So, Conner's mass is approximately 69.8 kg.
Show that entropy change due to heat transfer by conduction is given by
∆S=mC (4 marks)
ln
2
T
T
2
The change in the reservoir, the system or device, and the surroundings are added to determine the overall entropy change. The reservoir's entropy change is. Since entropy is a function of state and we are contemplating a complete cycle (return to initial state), the device's entropy change is zero.
For finite variations at constant T, entropy changes (S) are calculated using the relation G=ΔH - TΔS.
The generation, consumption, conversion, and exchange of thermal energy across physical systems is the focus of the thermal engineering field of study known as heat transfer. Different heat transmission techniques, including thermal conduction, thermal convection, thermal radiation, and energy transfer by phase changes, are categorized.
Conduction is the process through which heat is transported from an object's hotter end to its colder end. Heat naturally transfers from a hotter body to a colder body.
For instance, heat is transferred from the hotplate of an electric stove to the bottom of a saucepan that comes into touch with it.
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make G the subject of the formula
F = GMM²/1²
HENCE WRITE THE DIMENSION FOR G
The value of G is FR²/M₁M₂. and the dimension of G is [M⁻¹L³T⁻²]..
The provided formula is of gravitational force F between two objects,
F = GM₁M₂/R²
Where M₁ is the mass of first object and M₂ is the mas of the other object while R is the distance between there centers and G is the universal gravitation constant.
To find the dimension of G, making G the subject of formula,
G = FR²/M₁M₂.
As we know, unit of mass is Kilogram (Kg), unit of force is Newton (N) and unit of distance is Meter (M).
Putting all the values, Units in the place of quantities,
G = N.R²/Kg.Kg
Now, using Dimensional analysis, and writing the dimensions of all the other units,
G = [MLT⁻²][L²]/[M][M]
G = [ML³T⁻²]/[M²]
G = [M⁻¹L³T⁻²]
The dimensions of G are [M⁻¹L³T⁻²].
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When is a secondary source more helpful than a primary source?
A. When you want to confirm the conclusion with your own experiment
B. When you are not an expert in the field being studied in the experiment
C. When you do not need to know the conclusion of the experiment
D. When you want to know the exact data values
Option A is the correct answer: A secondary source is more helpful than a primary source when you want to confirm the conclusion with your own experiment.
Primary and secondary sources work well together to support the argument you are trying to make. Although secondary sources demonstrate how your work connects to earlier research, primary sources are more reliable as evidence. When you want to confirm the conclusion of your experiment, secondary sources prove to be more useful.
Original research is built on primary sources. You are able to discover novel information, offer solid justification for your claims, and provide reliable facts on your subject.
Secondary sources are useful for getting a comprehensive perspective of your subject and learning other researchers' methods. They frequently combine numerous primary materials that would take a lot of time and effort to obtain independently.
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A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°. How far is the ball from the football player when it lands? How much farther would the ball go if he kicked it with the same speed, but at a 45° angle? Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta) \\ v_y=v\sin (\theta) \end{gathered}[/tex]Where v = 16 m/s and θ = 63°
[tex]\begin{gathered} v_x=16\cos (63\degree)=7.26\; \frac{m}{s} \\ v_y=16\sin (63\degree)=14.26\; \frac{m}{s} \end{gathered}[/tex]How far is the ball from the football player when it lands?
The range of the ball is given by
[tex]x=v_x\times t[/tex]Where t is the time the ball remains in the air.
The time (t) can be found as
[tex]y=v_yt+\frac{1}{2}at^2[/tex]y = 0 when the ball is in the air.
The acceleration is due to gravity (-9.8 m/s²)
[tex]\begin{gathered} 0=14.26t+\frac{1}{2}(-9.8)t^2 \\ 0=14.26t-4.9t^2 \\ 0=t(14.26-4.9t) \\ 0=14.26-4.9t \\ 4.9t=14.26 \\ t=\frac{14.26}{4.9} \\ t=2.91\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=7.26\times2.91=21.13\; m[/tex]Therefore, the ball will land 21.13 meters far from the football player.
How much farther would the ball go if he kicked it with the same speed, but at a 45° angle?
We need to repeat the above calculations
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta)=16\cos (45\degree)=11.31\; \frac{m}{s} \\ v_y=v\sin (\theta)=16\sin (45\degree)=11.31\; \frac{m}{s} \end{gathered}[/tex]The time (t) is given by
[tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=11.31_{}t+\frac{1}{2}(-9.8)t^2 \\ 0=11.31_{}t-4.9t^2 \\ 0=11.31_{}-4.9t \\ 4.9t=11.31_{} \\ t=\frac{11.31_{}}{4.9} \\ t=2.31\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=11.31\times2.31=26.13\; m[/tex]Therefore, the ball will land 26.13 meters far from the football player.
Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
The ball kicked at 16 m/s and at a 63° angle takes 2.91 s to land.
The ball kicked at 9 m/s and at a 45° angle will take
[tex]v_y=9\sin (45\degree)=6.36\; \frac{m}{s}[/tex][tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=6.36t+\frac{1}{2}(-9.8)t^2 \\ 0=6.36t-4.9t^2 \\ 0=6.36-4.9t \\ 4.9t=6.36 \\ t=\frac{6.36}{4.9} \\ t=1.30\; s \end{gathered}[/tex]So, the ball kicked at 9 m/s and at a 45° angle takes 1.30 s to land.
Therefore, the ball kicked at 9 m/s and at a 45° angle will land first.
You throw an object up with an initial velocity of Voy = 11 m/s from a height of y = 25 m. How long, in seconds, does it take for the object to reach the ground? What is the objects final velocity, in meters per second, as it impacts the ground? Find the time, in seconds, if you instead threw the object DOWN with the same velocity Voy
Calculate the ball's greatest height using the vertical motion model, h = -16t2 + vt + s, where v is the beginning velocity in feet/second and s is the height in feet.
What does a ball being thrown upwards accelerate to?A ball is thrown into the air, where it gradually loses speed until it abruptly comes to a rest at the peak of the motion. The body is travelling upward against gravity at the top, hence the acceleration there is 9.8 ms2. For example, g=9.8 ms2 is the formula for the acceleration caused by gravity.
Only at the greatest point of a body being hurled vertically upwards would velocity be zero because of the constant downward acceleration brought on by the gravitational force. As a result, velocity is zero throughout the rest of the motion.
A ball is originally moving upward when it is tossed into the air, for instance.
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A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. What isthe acceleration of the skier?(Unit = m/s?)Enter
A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. The acceleration of the skier is approximately 2.97 m/s².
To find the acceleration of the skier, we need to use Newton's second law of motion and consider the forces acting on the skier.
Identify the forces acting on the skier:
The forces acting on the skier are the force of gravity (mg) and the force exerted by the rope (T), where m is the mass of the skier (72.5 kg) and g is the acceleration due to gravity (9.8 m/s²). The force exerted by the rope is parallel to the slope and can be calculated using the given value of 383 N.
Resolve the forces:
Since the rope is parallel to the slope, we need to resolve the force of gravity into components parallel and perpendicular to the slope. The component parallel to the slope is m * g * sin(21.7°).
Apply Newton's second law of motion:
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, Fnet = ma.
Determine the net force:
The net force acting on the skier is the difference between the force exerted by the rope and the component of the force of gravity parallel to the slope. Fnet = T - m * g * sin(21.7°).
Calculate the acceleration:
Using Newton's second law, we can rearrange the equation Fnet = ma to solve for the acceleration (a). a = Fnet / m.
Substitute the values and solve:
Substitute the known values into the equation to find the acceleration.
Therefore, the acceleration of the skier is approximately 2.97 m/s².
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Answer:
about 0.567 m/s²
Explanation:
You want the acceleration of a 72.5 kg skier up a 21.7° slope with µk = 0.120, towed by a rope exerting a force of 383 N.
ForcesThe force up the slope is 383 N.
The forces down the slope will be the sum of the force due to gravity and the friction force.
GravityThe force down the slope due to gravity is ...
F = m·g·sin(θ) = (72.5 kg)(9.8 m/s²)(sin(21.7°) ≈ 262.7 N
FrictionThe force due to friction will be the product of µk and the force normal to the slope:
F = m·g·cos(θ)·µk = (72.5 kg)(9.8 m/s²)cos(21.7°)·0.120 ≈ 79.22 N
Net ForceThe net force up the slope is ...
383 N -262.7 N -79.22 N ≈ 41.08 N
This will accelerate a mass of 72.5 kg in the amount of ...
A = F/m = 41.08 N/(72.5 kg) ≈ 0.567 m/s²
The acceleration of the skier is about 0.567 m/s² up the slope.
__
Additional comment
We don't have to figure the forces. Rather we can figure the gross acceleration due to the tow rope, then subtract the accelerations due to gravity and friction. This saves a few math operations as we don't have to multiply, then divide, by 72.5 kg.
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n Fig. P9.82, the cylinder Figure P9.82 and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is Pulley wrapped around the cylinder, passes over the pulley, and has a 3.00 kg box Cylinder Box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 2.50 m.
The speed of the box when it has fallen 2.50 m is 4.22 m/s.
What is the speed of the box?
The speed of the box when it has fallen through the given height is calculated as follows;
Apply the principle of conservation of energy to determine the speed of the box.
ΔK.E = ΔP.E
K.Ef - K.Ei = mg(hf - hi)
K.Ef - 0 = mg(hf - 0)
K.Ef = mghf
where;
K.E is the final kinetic energy = rotational + translational kinetic energyhf is the final height of the box¹/₂mv² + ¹/₂I_pω² + ¹/₂I_cω²= mghf
¹/₂mv² + ¹/₂(I_p + I_c)ω² = mghf
where;
I_p is moment of inertia of the pulleyI_c is the moment of inertia of the cylinderω is the angular speed of the boxm is mass of the boxI_p = ¹/₂MR²
where;
M is mass of the pulleyR is the radius of the pulleyI_p = ¹/₂(2)(0.2)² = 0.04 kgm²
I_c = MR²
I_c = (5)(0.4)²
I_c = 0.8 kgm²
¹/₂mv² + ¹/₂(I_p + I_c)(v/r)² = mghf
¹/₂mv² + ¹/₂r²(I_p + I_c)v² = mghf
¹/₂v²[m + 1/r²(I_p + I_c)] = mghf
v²[m + 1/r²(I_p + I_c)] = 2mghf
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
8.25v² = 147
v² = 147/8.25
v² = 17.8
v = √17.8
v = 4.22 m/s
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A 12 N force acts at a 25 degrees and an 8 N force acts at 65 degrees. Determine the magnitude and direction (include angle) of the resultant . Scale is 1cm = 1N
NEEDDD HELLPP ASAPPPPP
A 12 N force acts at a 25-degree angle, while an 8 N force acts at a 65-degree angle. The magnitude and direction (including angle) of the resultant force are 21.43.
Consider the formula for a force, F=M×Cos∝
As M is a Mass of an object and, ∝ Is the angle at which force is acting on an object.
[tex]F_{1}[/tex]=M × Cos ∝
M=12N, and Cos∝= Cos25 = 0.906
[tex]F_{1}[/tex] = 12× 0.906
∴ [tex]F_{1}[/tex]= 10.87N
For [tex]F_{2}[/tex] = 25×Cos65
= 10.56N
According to Superposition resulting force is the Sum of total forces.
[tex]F_{RESULTANT} = F_{1} + F_{2}[/tex]
= 10.87+ 10.56
=21.43N
∴The resultant force is 21.43.N
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A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is.......
A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is....... 400 J.
What is Speed?
Speed is the time rate at which an object is travelling along a path, whereas velocity is the pace and direction of an object's movement. In other words, velocity is a vector, whereas speed is a scalar valu
Work = Change in Kinetic energy
= ½m(v² - u²)
= ½ × 20 kg × [(7 m/s)² - (3 m/s)²]
= 10 kg × 40 m²/s²
= 400 J
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In a physics lab a student discovers that the magnitude of the magnetic field in a specific location near a long wire is 21.919 microTesla. If the wire carries a current of 35.483 A, what is the distance from the wire to that location ?
We will have the following:
First, we have that the permeability of free space is:
[tex]\mu_0=4\pi\ast10^{-7}Tm/A[/tex]Then:
[tex]\begin{gathered} B=\frac{\mu_0I}{2r\pi}\Rightarrow r=\frac{\mu_0I}{2B\pi} \\ \\ \Rightarrow r=\frac{(4\pi\ast10^{-7}Tm/A)(35.483A)}{2\pi(2.1919\ast10^{-5}T)}\Rightarrow r=0.3237647703...m \\ \\ \Rightarrow r\approx0.32m \end{gathered}[/tex]So, the distance is approximately 0.32 m.
Set the cannon to have an initial speed of 20 m/s. For which situation do you think the cannon ball will go father: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?
Question 2 options:
60 degrees
70 degrees
The cannon ball will go farther if the the angle of projection is set at 60 degrees
How to determine which angle will result in farther distance
Case 1:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 60 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×60) / 9.8
R = 346.41 / 9.8
R = 35.35 m
Case 2:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 70 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×70) / 9.8
R = 257.12 / 9.8
R = 26.24 m
From the above calculations, we can conclude that the ball will go farther, if the angle is 60 °
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A resistor uses energy at a rate of 2.50W when there is a current of 4.00A passing through it. What must be the potential difference across the resistor?1.25V1.50V0.625V10.0V
As we know
[tex]P=\text{ V}\times I;[/tex]Where,
P= electric power= 2.50 W
V= potential difference= ?
I= current = 4.00A
Using above formula we get,
[tex]\begin{gathered} P=V\times I; \\ \therefore2.50=\text{ V}\times4; \\ V=\text{ }\frac{2.50}{4}=\text{ 0.625V} \end{gathered}[/tex]Final answer is :- 0.625 V
3. A boulder rolls with speed of 3.5 m/s off a cliff. It hits the ground 2.25 m from the base ofthe ledge. A) How high is the ledge? B) How long did it take the boulder to fall to the bottomof the cliff?DrawingVerticalHorizontal
Given data
*The given distance from the base of the ledge is R = 2.25 m
*The given speed is v = 3.5 m/s
The diagram is given below
(a)
Let (h) be the height of the edge
The formula for the distance from the base of the ledge is given as
[tex]\begin{gathered} R=v\times t \\ R=v\times\sqrt[]{\frac{2h}{g}} \\ h=\frac{R^2\times g}{2v^2} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} h=\frac{(2.25)^2\times9.8}{2\times(3.5)^2} \\ =2.025\text{ m} \end{gathered}[/tex]Hence, the height of the ledge is h = 2.025 m
(b)
The formula for the time taken by the boulder to fall to the bottom of the cliff is given as
[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times2.025}{9.8}} \\ =0.642\text{ s} \end{gathered}[/tex]Hence, the time taken by the boulder to fall to the bottom of the cliff is t = 0.642 s
009 (part 1 of 2) 10.0 points When a water gun is fired while being held horizontally at a height of 1.19 m above ground level, the water travels a horizontal distance of 1.98 m. Find the initial velocity of the water. The acceleration of gravity is 9.81 m/s^2
Answer in units of m/s.
010 (part 2 of 2) 10.0 points A child, holding the same gun in a horizontal
position, slides down a 33.0◦incline at a constant speed of 1.40 m/s. The child fires the gun when it is 4.36 m above the ground and the water takes 0.868 s to reach the ground. How far will the water travel horizontally?
Answer in units of m.
Answer:
Speed = 4 m/s
Explanation:
Given:
009 (part 1 of 2)
H =1 .19 m
L = 1.98 m
g = 9.81 m/c²
____________
V₀ - ?
Equation of motion horizontally:
L = V₀*t (1)
Equation of vertical motion:
H = g*t² / 2 (2)
From equation (2):
Time:
t = √ (2*H / g) = √ ( 2*1.19 / 9.81) ≈ 0,49 s
From equation (1):
Horizontal speed:
V₀ = L / t = 1.98 / 0.49 ≈ 4 m/s