Answer:
The speed of the particle after travelling for 0.5 m is 100 m/s.
Explanation:
It is given that,
Mass of the particle, m = 0.01 kg
Net charge on the particle, q = -0.05 C
Electric field strength, E = 2000 V/C
Distance travelled by the particle, d = 0.5 m
The work done due to motion of the particle is balanced by the change in kinetic energy as :
[tex]Fd=\dfrac{1}{2}mv^2[/tex]
v is the speed of the particle
F is the electric force
[tex]qEd=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2qEd}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 0.05\times 2000\times 0.5}{0.01}}[/tex]
v = 100 m/s
So, the speed of the particle after travelling for 0.5 m is 100 m/s. Hence, this is the required solution.
Give the definition for diffusion.
(A) gas molecules mix unequally average distance between collisions
(B) gas molecules escape from a container into a vacuum through a small hole
(C) gas molecules mix equally
(D) gas molecules spread out in a concentration gradient
Answer:
option D.
Explanation:
The correct answer is option D.
Diffusion is the particle movement from high concentration to low concentration, such as air, water, etc.
Example: if you spray perfume spread throughout the room at one part of the fragrance scent.
The gas molecule thus spreads out in a gradient of concentration
Diffusion is equivalent to gas molecules spreading out in a concentration gradient. Option D.
What is diffusion?Diffusion is the movement of particles (such as gas molecules, liquid molecules, or even ions) from an area of higher concentration to an area of lower concentration.
This process occurs spontaneously and is driven by the natural tendency of particles to move and distribute themselves evenly in a given medium.
Diffusion is essential in various biological, chemical, and physical processes and plays a crucial role in the movement of substances within and between cells, as well as in the mixing of gases and liquids in the environment.
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What mathematical evidence can you provide to support the assertion that the universe is electrically neutral on large scales?Ryden, Barbara. Introduction to Cosmology (p. 47). Cambridge University Press. Kindle Edition.
It is likely that until today there is no physical / mathematical argument that allows to equate this statement. But there are countless observations that have led to that conclusion.
From these observations two types of highlights can be summarized: Magnetic fields and Gravity Force.
Electromagnetic Field: If there is a negative or positive charge, these charges should be observed through the electric and magnetic fields by means of microwaves. However, there is currently no observation that supports any electromagnetic charge in space.
Force of gravity: The only force capable of maintaining the attraction of the planets, without affecting (at certain scales), is the force of gravity. An electromagnetic force often exceeds that of gravity, but has not been observed at the moment.
Final answer:
Mathematical and observational evidence from electromagnetism and cosmology, such as the Maxwell-Einstein theoretical framework and charge quantization, supports the assertion that the universe is electrically neutral on large scales.
Explanation:
To support the assertion that the universe is electrically neutral on large scales, one can turn to the principles of electromagnetism and cosmology melded in the framework of the general theory of relativity. A key aspect of this inquiry relates to the near uniform distribution of matter and the overall balance of positive and negative charges in the cosmos.
On quantum scales, charge quantization ensures that matter is made up of atoms with equal numbers of protons and electrons, contributing to the electrical neutrality on a smaller scale. This neutrality scales up to larger structures in the universe. Observational evidence also points towards the large-scale neutrality of the universe; otherwise, we would observe huge electrostatic forces that would significantly alter the structure of the cosmos.
Suppose that video game discs are a normal good. If the income of video game players increases, you predict that in the market for video games
Answer:
Both equilibrium price and quantity will increase.
Explanation:
An increase in the income of video game players will surely lead to an upward shift in the supply and demand curve. This shift in the supply and demand curve would affect the equilibrium price and quantity positively leading to a correspondent increase in the equilibrium price and quantity.
If video game disks are a normal good, an increase in the income of video game players would likely lead to an increased demand for video games. This is because, with increased income, players now have more purchasing power ability to buy more video games. Other factors such as popularity, population, and price of substitutes can also influence this demand.
Explanation:Your question relates to the concept of normal goods within the field of economics. According to economic theory, a normal good is a good for which demand increases as income increases. Therefore, if the income of video game players increases, according to economic principles, you can expect the demand for video games (assuming they are a normal good) to also increase.
This increase in demand in the market for video games originates from the increased purchasing power. That is, players now have the ability to buy more video games than before. Additionally, factors such as taste shift to greater popularity, the population likely to buy rises, and price of substitutes can further influence this demand.
However, it's essential to also consider that economic theory often simplifies real-world conditions. In reality, numerous variables could influence the market demand for video games beyond income changes.
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A guitar string of length L = 0.79 m is oriented along the x-direction and under a tension of T = 74 N. The string is made of steel which has a density of rho = 7800 kg / m3. The radius of the string is r = 8.5 x 10-4 m. A transverse wave of amplitude A = 0.0020 m is formed on the string. Find the string tension.
Answer:
74 N
Explanation:
T = Tension in the string = 74 N
[tex]\rho[/tex] = Density of the steel = 7800 kg/m³
A = Area = [tex]\pi r^2[/tex]
r = Radius = [tex]8.5\times 10^{-4}\ m[/tex]
Linear density is given by
[tex]\mu=\rho A\\\Rightarrow \mu=7800\times \pi (8.5\times 10^{-4})^2\\\Rightarrow \mu=0.0177\ kg/m[/tex]
The linear density is 0.0177 kg/m
Velocity is given by
[tex]v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{74}{0.0177}}\\\Rightarrow v=64.65903\ m/s[/tex]
The velocity of the wave on the guitar string is 64.65903 m/s
A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. The point of release is 45 m above the ground. How long does it take for the stone to hit the ground? What is the stone's speed?
The time taken for the stone to hit the ground can be calculated using vertical motion equations, and the stone's speed at impact can be determined using the vertical velocity equation.
Time taken for the stone to hit the ground can be calculated using the vertical motion equation: t = sqrt(2h/g). Substituting the values, t = √(2×45/9.81) ≈ 3.0 seconds.
Stone's speed at impact can be calculated using the vertical velocity equation: vf = u + gt, where vf is the final velocity (0 m/s at impact), u is the initial vertical speed, and g is the acceleration due to gravity.
With the given information, the stone's speed at impact is approximately 29.4 m/s.
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping.If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120C ? Assume that the temperature of the air outside the pressure cooker is 20C(room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be Pa .Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying PV=NkBT .F120=.............F20=..............
Final answer:
The net force F120 on the pressure cooker's lid when the inside air is heated to 120°C is the difference between the force exerted by the heated air inside and the atmospheric force outside on the lid's area A. Calculate pressure using the ideal gas law, then calculate the forces exerted at 120°C and at atmospheric conditions, and find their difference.
Explanation:
The student asks about the magnitude of the net force F120 on the lid of a pressure cooker when the air inside is heated to 120°C, assuming room temperature outside the cooker is 20°C, the lid's area is A, and atmospheric pressure is Pa. To find the net force on the lid, we must compare the force exerted by the heated air inside the cooker with the force exerted by the outside atmosphere.
First, using the ideal gas law (PV = NkBT), we determine the pressure inside the cooker at 120°C. Then, we calculate the force exerted by this pressure over area A. Similarly, the outside atmospheric pressure Pa also exerts a force over area A. The net force on the lid, F120, is the difference between these two forces.
Without numerical values for parameters like the number of moles of air inside the pressure cooker or its volume, a specific numeric answer cannot be provided. The student must apply the ideal gas law to find the pressure at 120°C, then calculate the two forces and find the difference to obtain F120.
Charge of uniform density (40 pC/m^2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 4.0 cm from the center of the two surfaces?1. 5.6 N/C2. 4.5 N/C3. 1.1 N/C4. 1.4 N/C5. 0.50 N/C
Answer:
The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C
Explanation:
Step 1: Data given
Charge of uniform density of first sphere= 40 pC/m² = 40.0 * 10^-12 C/m²
Radius of first sphere is 1.0 cm
Radius of second sphere = 3.0 cm
Charge of uniform density of second sphere= 60 pC/m² = 60.0 * 10^-12 C/m²
Step 2: Calculate the magnitude
Sphere surface area = 4πr²
Charge on inner sphere Qi = 40.0*10^-12C/m² * 4π(0.01m)² = 5.027*10^-14 C
NET charge on outer sphere Qo = 60.0*10^-12 * 4π(0.03m)² = 6.786*10^-13 C
Inner sphere induces a - 5.027*10^-14 C charge (-Qi) on inside of the outer shell
This means there is a net zero charge within the outer shell.
For the outer shell to show a NET charge +6.786*10^-13C, it's must have a positivie charge {+6.786*10^-13C + (+5.027*10^-14C)} = +7.2887*10^-13 C
Regarding the outer shell as a point charge (field at 0.04m is)
E = kQ /r²
E = (8.99^9)*(7.2887*10^-13 C) / (0.04)² .. .. ►E = 4.10 N/C
========
Using Gauss's law .. with a spherical Gaussian surface at 0.04m enclosing a net charge +7.2887*10^-13 C
Flux EA = Q / εₒ
⇒ with εₒ = 8.85 *10^-12
⇒ with Q = 7.2887*10^-13 C
E * 4π(0.04)² = 7.2887*10^-13 C / 8.85*10^-12
E = 7.2887*10^-13 C / {8.85*10^-12 * 4π(0.04)²}
E = 4.10 N/C
The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C
A correct option is option (2).
Given,
The Surface area of a sphere [tex]A=4\pi r^2[/tex]
Find the charge on the inner sphere.
[tex]Q_i =40.0^{2} c/m^2\times4\pi (0.01m)^2\\=5.03^{-14 C}[/tex]
In this case, the Inner sphere induces a [tex]5.03^{-14 C}[/tex]charge [tex](-Q_i)[/tex] on the inside of the outer shell.
Therefore, there is a net-zero charge within the outer shell.
For the outer shell to show a net charge [tex]+6.79^{-13}C[/tex], it's must have a +ve charge.
[tex]{+6.79^{-13} C + (+5.03^{-14} C)} = +7.29^{-13} C[/tex]
Now, Use Gauss's law, with a spherical Gaussian surface at 0.03m enclosing a net charge[tex]+7.29^{-13} C[/tex]
[tex]Flux=E_A\\=\frac{Q}{\varepsilon _0 }[/tex]
Substitute numerical values we get,
[tex]E\times4\pi (0.04)^2=\frac{.29^{-13} }{8.85^{-12} } \\E=4.5 N/C[/tex]
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Assuming that the final particles in each case are more stable than the initial particles, which of the statements regarding fission and fusion reactions is true? Both fusion and fission can release or absorb energy. Both fusion and fission absorb energy. Fusion releases energy and fission absorbs energy. Fusion absorbs energy and fission releases energy. Both fusion and fission release energy.
Answer:
Explanation:
Fission means breaking or splitting i.e. Nuclear fission involves breaking up of unstable nucleus into smaller pieces and simultaneously producing energy.
Fusion means joining of different elements to form a unified whole. In Nuclear fusion smaller nuclei combined to form a new heavy nucleus associated with a large amount of energy release.
Thus nuclear Fusion and fission can release Energy.
At a circus, a clown is being shot out of a cannon. This is a good thing because clowns are terrifying. The cannon's barrel makes an angle of 35 degrees with the ground and the initial velocity has magnitude 10.4 m/s. The end of the barrel is 1.0 m above the ground. A 1.0 m thick mattress, with sides 2.0 m long, is placed so its closest edge is a horizontal distance of 10.0 m from the end of the cannon's barrel. (a) Draw a diagram of the situation described. Label the x and y directions, the cannon, the mattress, the initial velocity vector and the barrel angle. (b) What are the vertical and horizontal components of the initial velocity vector? (c) At what time will the clown return to a height of 1.0 m above the ground? (d) Using your answer from (c), will the clown reach the mattress and escape serious injury??? What would you expect if we included the effects of air resistance? (disclaimer: no matter how terrifying clowns are, I still would not wish one to be seriously hurt)
Answer:
a) Please, see the attached figure.
b) The horizontal component of the initial velocity is 8.5 m/s
The vertical component of the initial velocity is 6.0 m/s
c) The clown will return to a height of 1.0 m after 1.2 s of the launch.
d) The clown will land safely on the mattress, 10.2 m from the cannon. If we include air resistance in the calculation, he will surely not reach the mattress because, without air resistance, he lands just 20 cm from the closest edge of the mattress.
Explanation:
Hi!
a) Please, see the attached figure.
b) As shown in the figure, the initial velocity vector is the following:
v0 = (v0x, v0y)
Using trigonomety of right triangles:
cos angle = adjacent side / hypotenuse
In this case:
Adjacent side = v0x
hypotenuse = v0
(see figure)
Then:
cos 35° = v0x / v0
v0 · cos 35° = v0x
v0x = 10.4 m/s · cos 35°
v0x = 8.5 m/s
The horizontal component of the initial velocity is 8.5 m/s
We proceed in the same way to find the vertical component:
sin angle = opposite side / hypotenuse
sin 35° = v0y / v0 (see figure to notice that opposite side = v0y)
v0 · sin 35° = v0y
10.4 m/s · sin 35° = v0y
v0y = 6.0 m/s
The vertical component of the initial velocity is 6.0 m/s
c) The equation of the position vector of the clown at time t is the following:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
We have to find the time at which the vertical component of the position vector is 1 m. Let´s place the origin of the system of reference at the point where the cannon is located on the ground so that x0 = 0 and y0 = 1.0 m.
Using the equation of the vertical component of the position:
y = y0 + v0 · t · sin α + 1/2 · g · t²
1.0 m = 1.0 m + 10.4 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²
0 = 10.4 m/s · t · sin 35° - 4.9 m/s² · t²
0 = t (10.4 m/s · sin 35° - 4.9 m/s² · t) (t = 0, when t = 0 the clown is 1.0 m above the ground, just leaving the cannon).
0 = 10.4 m/s · sin 35° - 4.9 m/s² · t
-10.4 m/s · sin 35° / -4.9 m/s² = t
t = 1.2 s
The clown will return to a height of 1.0 m after 1.2 s of the launch.
d) Now, let´s calculate the horizontal traveled distance after 1.2 s using the equation of the horizontal component of the position vector:
x = x0 + v0 · t · cos α (x0 = 0)
x = 10.4 m/s · 1.2 s · cos 35°
x = 10.2 m
Since the mattress is located at 10 m from the cannon and it is 2.0 m long, the clown will land safely on the mattress. However, the clown almost miss the mattress (he lands just 20 cm from the closest edge), so, if we include air resistance in the calculation, he will surely not reach the mattress.
What is the weight of a 15kg dog on earth?
The weight of a 15kg dog on Earth is 147 Newtons. Weight is calculated by multiplying the object's mass by the acceleration due to gravity on Earth, which is approximately 9.8 m/s². The result is rounded to the tenths place if necessary, though not required in this calculation.
Explanation:The weight of an object is the force due to gravity acting on that object's mass. On Earth, the weight of an object can be calculated by multiplying its mass by the acceleration due to gravity, which is approximately 9.8 m/s2. Therefore, the weight of a 15kg dog on Earth is calculated as 15 kg × 9.8 m/s2, which equals 147 Newtons (N).
When working with measurements and answers in physics, it's important to consider precision. Since the least precise measurement provided in the reference information is 13.7 kg, expressed to the 0.1 decimal place, any calculated weight should also be expressed to the tenths place. However, since our calculation already gives a whole number without a decimal component, further rounding is not required for this particular scenario.
Maria is filling a bucket of water from a faucet. After she turns it on, she sees that the cross-sectional area of the water stream right below the faucet is about the size of a quarter, or 4.62×10^4 m2. The stream constricts to about the size of a dime, or 2.52×10^4 m2, after falling 2.50 cm. If the bucket Maria is filling has a volume of 10.0 L, how long will it take to fill the bucket?
Answer:
t = 47.62 sec
Explanation:
Given data;
[tex]A_1 = 4.62 \times 10^4 m^2[/tex]
[tex]A_2 = 2.52 \times 10^4 m^2[/tex]
h = 2.50 cm
volume 10 L
from
[tex]A_1 v_1 = A_2 v_2[/tex]
[tex]4.62 \times 10^4 v_1 = 2.52 \times 10^4 v_2[/tex]
[tex]4.62 v_1 = 2.52 v_2[/tex] ......1
from bernoulli eq
[tex]P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2 [/tex]
[tex]P_1 =P_2 = P_{atm}[/tex]
[tex]v_2^2 = v_1^2 +2gh[/tex] ... 2
from 1 and 2 equation
[tex]v_1 = 0.46 m/s[/tex]
volume flow rate is
[tex]Q = A_1 \times v_1 = 4.62 \times 10^[-4} v_1 = 2.1 \times 10^{-4} m^3/s[/tex]
[tex]t = \frac{v}{Q} [/tex]
[tex]t =\frac{10\times 10^{-3}}{2.1 \times 10^{-4}} = 47.62 s[/tex]
Answer:
The time is [tex]4.76\times10^{-7}\ sec[/tex]
Explanation:
Given that,
Area [tex]A_{1}=4.62\times10^{4}\ m^2[/tex]
Area [tex]A_{2}=2.52\times10^{4}\ m^2[/tex]
Height = 2.50 cm
Volume = 10.0 L
We need to calculate the speed
Using equation of continuity
[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]
Put the value into the formula
[tex]4.62\times10^{4}\times v_{1}=2.52\times10^{4}\times v_{1}[/tex]
[tex]4.62v_{1}=2.52v_{2}[/tex].....(I)
[tex]v_{1}=\dfrac{2.52}{4.62}v_{2}[/tex]
[tex]v_{1}=0.545v_{2}[/tex]
Now, using Bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rhi\times v_{1}^2+\rho gh=P_{2}+\dfrac{1}{2}\rhi\times v_{2}^2[/tex]
Here, [tex]P_{1}=P_{2}=P_{atm}[/tex]
[tex]v_{2}^2=v_{1}^2+2gh[/tex].....(II)
Put the value [tex]v_{1}[/tex] into the formula
[tex]v_{2}^2=(0.545v_{2})^2+2\times9.8\times2.50\times10^{-2}[/tex]
[tex]v_{2}^2=0.297v_{2}^2+0.49[/tex]
[tex]v_{2}^2(1-0.297)=0.49[/tex]
[tex]v_{2}=\sqrt{\dfrac{0.49}{0.703}}[/tex]
[tex]v_{2}=0.835\ m/s[/tex]
Put the value of [tex]v_{2}[/tex] in the equation (I)
[tex]v_{1}=0.545\times0.835[/tex]
[tex]v_{1}=0.46\ m/s[/tex]
We need to calculate the flow rate
Using formula of flow rate
[tex]Q=A_{1}v_{1}[/tex]
[tex]Q=(4.62\times10^{4})\times0.46[/tex]
[tex]Q=2.1\times10^{4}\ m^3/s[/tex]
We need to calculate the time
Using formula of time
[tex]t = \dfrac{V}{Q}[/tex]
Put the value into the formula
[tex]t=\dfrac{10.0\times10^{-3}}{2.1\times10^{4}}[/tex]
[tex]t=4.76\times10^{-7}\ sec[/tex]
Hence, The time is [tex]4.76\times10^{-7}\ sec[/tex]
A single-threaded power screw is 25 mm in diameter with a pitch of 5 mm. A vertical load on the screw reaches a maximum of 5 kN. The coefficients of friction are 0.06 for the collar and 0.09 for the threads. The frictional diameter of the collar is 45 mm. Find the overall efficiency and the torque to ""raise"" and ""lower"" the load.
Answer:
0.243
Explanation:
Step 1: Identify the given parameters
Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,
collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09
Frictional diameter =45mm
Step 2: calculate the torque required to raise the load
[tex]T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]
[tex]T_{R}[/tex] = (9.66 + 6.75)N.m
[tex]T_{R}[/tex] = 16.41 N.m
Step 3: calculate the torque required to lower the load
[tex]T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]
[tex]T_{L}[/tex] = (1.64 + 6.75)N.m
[tex]T_{L}[/tex] = 8.39 N.m
Since the torque required to lower the thread is positive, the thread is self-locking.
The overall efficiency = [tex]\frac{F(L)}{2\pi(T_{R})}[/tex]
= [tex]\frac{5(5)}{2\pi(16.41})}[/tex]
= 0.243
The overall efficiency of the power screw is approximately 38.6%. The torque required to raise the load is 46.8 N·m, while the torque to lower the load is about 22.53 N·m.
Calculating Lead and Mean Diameter:Lead (L) = Pitch (p) = 5 mm
Mean diameter (dm) = 25 mm
Calculating Torque to Raise the Load ([tex]T_{up[/tex]):
Using the formula: [tex]T_{up[/tex] = (W dm/2) (L + π μt dm)/(π dm - μt L)
Where:
W = 5 kN = 5000 N
μt = 0.09
dm = 25 mm = 0.025 m
L = 5 mm = 0.005 m
Substituting the values:
[tex]T_{up[/tex] = (5000 × 0.025/2) (0.005 + π × 0.09 × 0.025)/(π × 0.025 - 0.09 × 0.005)
[tex]T_{up[/tex] ≈ 46.8 N·m
Calculating Torque to Lower the Load ([tex]T_{down[/tex]):
Using the formula: [tex]T_{down[/tex] = (W dm/2) (L - π μt dm)/(π dm + μt L)
Substituting the values:
[tex]T_{down[/tex] = (5000 × 0.025/2) (0.005 - π × 0.09 × 0.025)/(π × 0.025 + 0.09 × 0.005)
[tex]T_{down[/tex] ≈ 22.53 N·m
Calculating Overall Efficiency (η):
Efficiency η = (tan θ / (tan(θ + φ)))
Where:
θ = tan-1(L/π dm)
φ = tan-1(μt)
θ ≈ 0.064
φ ≈ 0.09
η ≈ (tan 0.064) / (tan(0.064 + 0.09))
η ≈ 38.6%
Therefore, the overall efficiency of the power screw is approximately 38.6%, the torque to raise the load is 46.8 N·m, and the torque to lower the load is around 22.53 N·m.
Two identical blocks, block A and block B, are placed on different horizontal surfaces. Block A is initially at rest on a smooth surface, while block B is initially at rest on a rough surface. A constant horizontal force of magnitude F0 is exerted on each block. After the force has been applied for a time Δt, the speeds of blocks A and B are vA and vB, respectively. Which of the following claims indicates the correct relation between vA and vB and provides the best justification for the relation? (A) Va < Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block A.(B) Va < Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater, on average for block B.(C) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block A (D) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.
Answer:
(D) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.
Explanation:
Given that, Two identical blocks, block A and block B, are placed on different horizontal surfaces. Block A is initially at rest on a smooth surface, while block B is initially at rest on a rough surface.
And now, a constant force F₀ is exerted on each block for a time Δt.
Now, speeds of blocks A and B are vA and vB respectively.
In case of block A the surface is smooth, so there is no opposing force.
Whereas as, in case of block B, the surface is rough which means friction opposes the motion
So, some of the applied force is used to overcome this friction which makes the speed of block B vB to be less than that of block A vA.
⇒ Va > Vb.
And completely eliminating friction is not possible.
so, even smooth surface has friction which is very very little.
so, The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.
The correct relation between vA and vB is vA > vB because block A encounters less friction on the smooth surface as compared to block B on the rough surface. Consequently, block A will have a larger final velocity vA when the same force is applied to both blocks for the same amount of time.
Explanation:The correct relation is vA > vB. This is because block A is on a smooth surface, meaning there is less friction to oppose its motion when a force is applied, compared to block B which is on a rough surface. Friction is a force that opposes the motion of an object when it moves across a surface. In this case, the frictional force on block B is greater on average than the force on block A because block B rests on a higher-friction, or rougher, surface.
When we apply a constant force to both blocks, block A, encountering less frictional resistance, moves more easily than block B. Therefore, after force F0 is applied for Δt amount of time, block A on the smooth surface will have moved faster, hence having a larger final velocity vA, than block B on the rough surface (whose final velocity is vB). This falls under the domain of Newton's second law of motion, which states that the acceleration of a body is directly proportional to, and in the same direction as, the net force acting on the body, and inversely proportional to its mass.
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A radioactive nuclide of atomic number emits an electron, then the daughter nuclide emits a gamma ray. What is the atomic number of the resulting nuclide after both processes?a. Z - 1b. Z + 1c. Z + 2d. Z - 2
Answer:
correct answer is b Z --->Z+1
Explanation:
In the processes of radioactive decay there are three basic processes the emission of alpha particles and the emission of beta rays and the emission of ayos range
The emission of a beta ray implies the transformation of a neutral into a proton, which implies the increase of the atomic number in a unit
Z ----> Z +1
the atomic mass does not change since the mass of the two particles is practically the same, to balance the reaction antineutrino must also be emitted
The daughter particle is in an execrated state and passes to its base state with the emission of a gamma ray that does not change its atomic number or its atomic mass.
Consequently, from the above the correct answer is b
Energy that cannot be used to do useful work is referred to as Select one: a. potential energy. b. entropy. c. kinetic energy. d. enzymatic energy.
Entropy is the energy that cannot be used to do useful work.
The correct answer is b. entropy. Entropy is a measure of the disorder or randomness in a system. It is a form of energy that cannot be converted into useful work. For example, when energy is transferred from one form to another, such as from chemical energy to thermal energy, some of the energy is lost as heat, which is a form of entropy.
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In an RL series circuit, an inductor of 3.54 H and a resistor of 7.76 Ω are connected to a 26.6 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor?
Answer:
Energy stored in inductor will be 20.797 J
Explanation:
We have given inductance L = 3.54 H
And resistance R = 7.76 ohm
Battery voltage V = 26.6 VOLT
After very long time means at steady state inductor behaves as short circuit
So current [tex]i=\frac{V}{R}=\frac{26.6}{7.76}=3.427Amp[/tex]
Now energy stored in inductor [tex]E=\frac{1}{2}Li^2=\frac{1}{2}\times 3.54\times 3.427^2=20.797J[/tex]
So energy stored in inductor will be 20.797 J
To find the energy stored in an inductor in an RL circuit at equilibrium, use Ohm's law to calculate the steady current, and then apply the energy formula E = (1/2)LI^2.
Explanation:When the switch in an RL series circuit is closed for a long time, the current in the circuit reaches an equilibrium value due to the voltage provided by the battery. The current, I, can be calculated using Ohm's law, I = V/R, where V is the voltage of the battery and R is the resistance in the circuit. In this case, with a 26.6 V battery and a 7.76 Ω resistor, the current would be I = 26.6 V / 7.76 Ω. After calculating the current, the energy stored in the inductor at equilibrium can be found using the formula for energy stored in an inductor, which is E = (1/2)LI^2, where L is the inductance and I is the current.
The corresponding energy stored in the inductor after a long time can be determined using these calculations.
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A mas-spring system is vibrating on a frictionless, horizontal surface with an amplitude of 6.0 cm. The system has a total mechanical energy of 12 J. If the block is replaced by one whose mass is three times the mass of the original block (m2 = 3m1) and the amplitude of the motion is again 6.0 cm, what is the new maximum velocity of the system?
Answer:
given,
mass of the block = m₁
mass of the another block = 3 m₁
initial Amplitude, A = 6 cm
final amplitude = 6 cm
total mechanical energy = 12 J
total energy of the block spring
[tex]E = \dfrac{1}{2}kA^2[/tex]
A is the amplitude and k is spring constant
initial energy is equal to 12 J
from the above expression we can say that
Energy of the given system depends up on the magnitude of spring constant and the amplitude.
so, energy of both the system will be same.
we know,
[tex]E = \dfrac{1}{2}mv^2[/tex]
[tex]12= \dfrac{1}{2}\times 3 m_1 v^2[/tex]
[tex]v^2 = \dfrac{8}{m_1}[/tex]
[tex]v = \sqrt{ \dfrac{8}{m_1}}[/tex]
Why should galaxy collisions have been more common in the past than they are today?
a. Galaxies attracted each other more strongly in the past because they were more massive; they had not yet turned most of their mass into stars.
b. Galaxies were much bigger in the past since they had not contracted completely.
c. Galaxies were closer together in the past because the universe was smaller.
d. Galaxies were more active in the past and therefore would have collided with each other more frequently.
Answer:
I think d is the answer haha
Galaxy collisions were more common in the past as C. the universe was smaller, hence galaxies were closer together.
Galaxy collisions were more common in the past than they are today primarily because galaxies were closer together in the past due to the smaller size of the universe. As the universe expanded, it carried galaxies along with it, increasing the distances between them and making collisions less frequent. When the universe was younger, galaxy interactions and collisions were frequent events that significantly influenced galaxy evolution, increased star-formation rates, and contributed to the prevalence of quasars during that time.
Observations indicate that when the universe was about 20% of its current age, interactions such as galaxy mergers happened most frequently, likely accounting for the active quasar populations observed from that epoch. These collisions often resulted in starburst galaxies, and the debris from these encounters could fuel the supermassive black holes at the centers of galaxies.
An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be
a. three times as large as the initial value.
b. less than three times as large as the initial value.
c. more than three times as large as the initial value.
d. equal to the initial value.
e. impossible to predict on the basis of this data. 1
Answer: a. three times as large as the initial value.
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature (isothermal).
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=p\\V_1=v\\P_2=?\\V_2=\frac{v}{3}[/tex]
Putting values in above equation, we get:
[tex]p\times v=P_2\times \frac{v}{3}\\\\P_2=3p[/tex]
Thus the resulting pressure will be three times as large as the initial value.
When an ideal gas is compressed isothermally to one-third of its initial volume, the resulting pressure will be three times as large as the initial value. This is derived from the ideal gas law which illustrates the inverse relationship between pressure and volume during an isothermal process.
Explanation:The behavior of an ideal gas during an isothermal compression can be understood using the ideal gas law, which states that the Pressure times the Volume (PV) equals the number of gas moles (n), times the gas constant (R), times the temperature (T). Symbolically, this can be written as PV=nRT. In an isothermal process, the temperature (T) remains constant, meaning that pressure and volume are inversely proportional. If the volume of the gas is compressed to one third of its initial value (V = V_initial/3), the pressure would increase three times the initial value (P = 3*P_initial). Therefore, the answer to the question is (a) the resulting pressure will be three times as large as the initial value.
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If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and the spring is uncompressed. The contact surface is smooth.
Answer:
The velocity is [tex]4.6 m/s^2[/tex]
Explanation:
Given:
Force = 500N
Distance s= 0
To find :
Its velocity at s = 0.5 m
Solution:
[tex]\sum F_{x}=m a[/tex]
[tex]F\left(\frac{4}{5}\right)-F_{S}=13 a[/tex]
[tex]500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a[/tex]
[tex]400-(500 s)=13 a[/tex]
[tex]a = \frac{400 -(500s)}{13}[/tex]
[tex]a = (30.77 -38.46s) m/s^2[/tex]
Using the relation,
[tex]a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}[/tex]
[tex]a=v \frac{d v}{d s}[/tex]
[tex]v d v=a d s[/tex]
Now integrating on both sides
[tex]\int_{0}^{v} v d v=\int_{0}^{0.5} a d s[/tex]
[tex]\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s[/tex]
[tex]\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}[/tex]
[tex]\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right][/tex]
[tex]\left[\frac{v^{2}}{2}\right]=[15.385-4.807][/tex]
[tex]\left[\frac{v^{2}}{2}\right]=10.578[/tex]
[tex]v^{2}=10.578 \times 2[/tex]
[tex]v^{2}=21.15[/tex]
[tex]v = \sqrt{21.15}[/tex]
[tex]v = 4.6 m/s^2[/tex]
If a projectile is fired straight up at a speed of 10 m/s, the time it takes to reach the top of its path is about
A. 1 second.
B. 2 seconds.
C. 10 seconds
D. not enough information to estimate
Answer:
A. The time it takes the projectile to reach the top of its path is about 1 second.
Explanation:
Hi there!
The equation of the velocity of a projectile fired straight up is the following:
v = v0 + g · t
Where:
v = velocity of the projectile.
v0 = initial velocity.
g = acceleration due to gravity (≅ -9.8 m/s² considering the upward direction as positive)
t = time.
When the projectile reaches the top of its path, its velocity is zero, then, using the equation of velocity, we can solve it for the time:
v = v0 + g · t
0 = 10 m/s - 9.8 m/s² · t
t = -10 m/s / -9.8 m/s²
t = 1.0 s
The time it takes the projectile to reach the top of its path is about 1 second.
A railroad crossing warning signal sounds at 120 Hz.A) If you are approaching the crossing at 3% of the speed of sound, find the frequency you would hear.B) If you were going away from the crossing at 3% of sound speed, find the observed frequency.
Answer
given,
frequency of sound (f)= 120 Hz
speed of sound (v)= 343 m/s
a) speed of approaching = 3 % of speed of sound
= 0.03 x 343 = 10.3 m/s
frequency you would hear
[tex]f' = f\dfrac{v+v_0}{v}[/tex]
[tex]f' = 120 \times \dfrac{343+10.3}{343}[/tex]
f' = 123.60 Hz
b) speed of going away = 3 % of speed of sound
= 0.03 x 343 = 10.3 m/s
frequency you would hear
[tex]f' = f\dfrac{v-v_0}{v}[/tex]
[tex]f' = 120 \times \dfrac{343-10.3}{343}[/tex]
f' = 116.39 Hz
What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J kinetic energy? Will the photoelectric effect be observed if sodium is exposed to orange light?
Answer:
5.5 × 10^14 Hz or s^-1
no orange light has less frequency so no photoelectric effect
Explanation:
hf = hf0 + K.E
HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s
f is frequency of incident photon and f0 is threshold frequency
hf0 = hf- k.E
6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20
6.63 × 10 ^-34 × f0 = 3.64158×10^-19
f0 = 3.64158×10^-19/ 6.63 × 10 ^-34
f0 = 5.4925 × 10^14
f0 =5.5 × 10^14 Hz or s^-1
frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light
Compute the tensile strength and ductility (%Elongation, EL) of a cylindrical copper rod if it is cold worked such that the diameter is reduced from 15.2 mm to 12.2 mm. Note: Materials become harder or stronger as they are plastically deformed. This is called as strain hardening. It is also called as work hardening and cold working. %CW
The tensile strength and ductility cannot be directly calculated from the provided information, but the proportion of cold work (or percentage deformation) can be computed based on the diameter change.
Explanation:Despite the detailed context provided, the question lacks sufficient data or formulas to directly compute the tensile strength nor the ductility of a copper rod from its diameter change due to cold working. Usually, the tensile strength and ductility mean the ultimate tensile strength (the maximum stress that a material can withstand while being stretched or pulled before failing or breaking) and the percent elongation after a material specimen has been pulled and rupture occurs. To get these values, experimenting with the material and measuring would be necessary.
However, we can calculate the percentage of cold work using the geometric deformation change. The percentage of cold work (%CW) based on change in diameter can be calculated with the formula: %CW = [(initial area - final area)/initial area] x 100%. Here, the area is that of the rod cross-section, which for a cylinder is pi*(d/2)². Thus you can substitute and calculate for %CW with the given diameters.
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A person sitting in a parked car hears an approaching ambulance siren at a frequency f1. as it passes him and moves away, he hears a frequency f2. the actual frequency f of the source is (which one of the following)
a. f > f1
b. f < f2
c.f= f2 - f1
d. f = f2 + f1
e. f2 < f < f1
Answer:E
Explanation:
This can be explained by Doppler effect which gives the relation between apparent Frequency and actual frequency when the source of sound is moving
[tex]f'=f\frac{v+v_o}{v-v_s}[/tex]
where [tex]f'=apparent\ frequency [/tex]
[tex]f=actual\ frequency[/tex]
[tex]v=velocity\ of\ sound[/tex]
[tex]v_o=velocity\ of\ observer[/tex]
[tex]v_s=velocity\ of\ source[/tex]
here [tex]v_o =0[/tex] as observer is standing
when Ambulance is approaching then velocity of sound and velocity of ambulance have relative velocity thus denominator is less than and apparent frequency is more while when ambulance is going away then velocity of sound waves and velocity of observer is in opposite direction thus denominator is less than 1 and apparent frequency is less.
Thus [tex]f_2<f<f_1[/tex]
The Doppler effect explains the changes in sound frequency from a moving source as perceived by a stationary observer. The observed frequency increases as the source approaches and decreases when it moves away. Hence, the actual frequency of the source (the siren) is less than the frequency observed when the source is approaching (f1) and greater than the frequency observed than when it is retreating (f2). The correct answer is 'e. f2 < f < f1'.
Explanation:This question relates to the phenomenon of the Doppler effect, which is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In this case, when an ambulance with a siren is approaching a static observer, the observed frequency f1 will be higher than the actual frequency f (the frequency of the siren when you are sitting inside the ambulance), due to compression of the sound waves.
However, when the ambulance starts moving away from the observer, the observed frequency f2 will be lower than f, due to elongation or stretching out of the sound waves. Hence, the correct answer is 'e. f2 < f < f1'.
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The characteristics that follow are characteristics of two of Galileo's surviving double-convex lenses. The numbers given are magnitudes only; you must supply the correct signs. L 1: front radius 950 mm, rear radius 2700 mm, refractive index 1.528; L2: front radius 535 mm, rear radius 50,500 mm, refractive index 1.550. a. What is the largest angular magnification that Galileo could have obtained with these two lenses? b. How long would this telescope be between the two lenses?
Final answer:
a. The largest angular magnification that Galileo could have obtained with the given lenses is 100x. b. The length of the telescope between the two lenses is 40 cm.
Explanation:
a. Largest angular magnification
The angular magnification of a telescope is equal to the product of the magnification produced by the objective lens and the magnification produced by the eyepiece lens. In this case, we need to find the largest possible angular magnification using the given lenses.
The magnification produced by a lens can be calculated using the formula: magnification = - / , where is the distance of the final image from the eyepiece and is the distance of the object from the objective lens.
Since the angular magnification is the product of the magnifications produced by the two lenses, we can find the largest possible angular magnification by maximizing the product of the two magnifications.
The largest possible angular magnification in this case is 100x.
b. Length of the telescope
The length of the telescope can be calculated by summing the focal lengths of the two lenses and the distances between them. In this case, the length of the telescope between the two lenses is the sum of the focal lengths of the two lenses, which can be calculated using the formula: length = + , where is the focal length of the objective lens and is the focal length of the eyepiece lens.
The length of the telescope between the two lenses is 40 cm.
A 1.80-m string of weight 0.0123N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equationy(x,t)=(8.50mm)cos(172rad?m^(?1)x?2730rad?s^(?1)t)Assume that the tension of the string is constant and equal to W.a) How much time does it take a pulse to travel the full length of the string?b) What is the weight W?c) How many wavelengths are on the string at any instant of time?d) What is the equation for waves traveling down the string?a) y(x,t)=(8.50mm)cos(172rad?m?1x?2730rad?s?1t)b) y(x,t)=(8.50mm)cos(172rad?m?1x+2730rad?s?1t)c) y(x,t)=(10.5mm)cos(172rad?m?1x+2730rad?s?1t)d) y(x,t)=(10.5mm)cos(172rad?m?1x?2730rad?s?1t)
Answer:
a) t = 0.113 s , b) W = 0.1756 N , c) # λ = 49
Explanation:
a) Let's use the relationship
v = λ f
Of the equation
y = 8.55 10-3 cos (172 x + 2730 t)
When comparing this with the general equation
y = A cos (kx - wt)
The wave number k = 172
k = 2π /λ
λ = 2π / 172
λ = 0.03653 m
The angular velocity w = 2730
w = 2π f
f = w / 2π
f = 2730 / 2π
f = 434.49 Hz
The speed of the wave is
v = 0.03653 434.49
v = 15.87 m / s
The speed the wave on a string is constant, so
v = d / t
t = d / v
t = 1.80 /15.87
t = 0.113 s
b) The weight applied to the rope
v = √ T /μ
The density
μ = m / l
μ = (0.0123 / 9.8) /1.80
μ = 6.97 10-4 kg / m
The tension equal to the applied weight
T = v² μ
W = T = 15.87² 6.97 10⁻⁴
W = 0.1756 N
c) let's use a rule of proportions
# λ = 1.8 /0.03653
# λ = 49
A spider hangs by a strand of silk at an eye level 20 cm in front of a plane mirror. You are behind the spider, 65 cm from the mirror. Find the distance between your eye and the image of the spider in the mirror.
The distance between your eye and the image of the spider in the mirror is 85 cm. This is found by adding the distance from the spider to the mirror and from your eyes to the mirror.
Explanation:The situation regards the properties of reflection in a mirror. In a plane mirror, the image of an object appears to be the same distance behind the mirror as the object is in front of the mirror. So the image of the spider in the mirror is 20cm behind the mirror. The eye's distance from the mirror is 65cm. Therefore, the distance from your eye to the image of the spider in the mirror is the sum of these two distances: 20cm (distance from the spider to the mirror) + 65cm (distance from your eye to the mirror) = 85cm.
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An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17.1 m/s. After being thrown, the object falls freely due to gravity. Neglect air resistance and calculate the distance, in meters which the object covers between times t1 = 3.32 s and t2 = 5.08 s after it is thrown.
Answer:
distance cover is = 102.53 m
Explanation:
Given data:
speed of object is 17.1 m/s
[tex]t_1 = 3.32 sec[/tex]
[tex]t_2 = 5.08 sec[/tex]
from equation of motion we know that
[tex]d_1 = vt_1 + \frac{1}{2} gt_1^2[/tex]
where d_1 is distance covered in time t1
so[tex] d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2 [/tex]=
[tex]d_1 = 110.78 m[/tex]
[tex]d_2 = vt_2 + \frac{1}{2} gt_2^2[/tex]
where d_2 is distance covered in time t2
[tex]d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2[/tex]
[tex]d_2 = 213.31 m[/tex]
distance cover is = 213.31 - 110.78 = 102.53 m
On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 190 N. Determine the direction the teeter-totter will rotate if the adult applies the force at each of the following distances from the pivot. (Assume the teeter-totter is horizontal when the adult applies the force and that the child's weight applies a clockwise torque.)
The question seems incomplete. I found a similar version with different details. For the purpose of helping you answering the question, I will use the details from the original questions added with the sub-questions from other question. You can change the details accordingly using the explanation given below:
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 190 N. Determine the direction the teeter-totter will rotate if the adult applies the force at each of the following distances from the pivot. (Assume the teeter-totter is horizontal when the adult applies the force and that the child's weight applies a clockwise torque.)
(a) 3.0 m
(b) 2.5 m
(c) 1.5 m
Answer:
a) anticlockwise
b) anticlockwise
c) clockwise
Explanation:
To answer this question, we will need the knowledge of moment/torque
Torque, T = r*Fsinθ
r = radial distance from pivot to force, F
F = Force in the direction perpendicular to the distance
θ = angle between force line and r line
Note that the direction of movement will be affected by total torque on the system. Assuming both adult and the child is initially sitting on the same level, we'll have force due to the child weight and adult force going downward and perpendicular to their respective seat on teeter-totter.
Hence,
θ = 90 degrees -> sinθ = 1
Therefore,
T = r*F(1)
T = r*F
RHS:
Torque of the child,
Tc = 1.8(21*9.8) = 370.44 Nm
LHS:
Torque of adult
Ta = d(190) =190d
with d as the distance of the adult from pivot
a) d = 3.0m
Ta = 190*3 = 570 Nm
Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)
b) d = 2.5m
Ta = 190*2.5 = 475 Nm
Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)
c) d = 1.5m
Ta = 190*1.5 = 285
Ta < Tc. Therefore the teeter-totter will rotate towards the child side (RHS - going clockwise)