The moment of inertia of the wheel is 4.27 kg.m²
The kinematics equation explains the variables associated and related of motion.
From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:
[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]
Making acceleration (a) the subject, we have:
[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]
where;
y = 1.5 mt = 2.0 s[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]
a = 0.75 m/s²
The angular acceleration of the wheel can be estimated by the formula:
[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]
[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]
[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]
Finally, the torque acting on the wheel is:
[tex]\mathbf{\tau = I \alpha}[/tex]
[tex]\mathbf{Tr = I \alpha}[/tex]
where;
T = tensionr = radiusI = moment of inertia∝ = angular acceleration∴
[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]
[tex]\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }[/tex]
I = 4.27 kg.m²
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what is the meaning of word thermodynamics
Answer:
physics that deals with the mechanical action or relations of heat.
After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure.The horizontal steel beam had a mass of 88.50 kg
per meter of length, and the tension in the cable was =11650 N
. The crane was rated for a maximum load of 500 kg
. If =5.580 m
, =0.522 m
, =1.350 m
, and ℎ=2.070 m
, what was the magnitude of L
(the load on the crane) before the collapse? The acceleration due to gravity is =9.810 m/s2
The magnitude of the load L on the crane before the collapse is 3211.81 N
To determine the magnitude of the load on the crane (L), we will need to make use of the equilibrium conditions of the torque.
It is always an ideal process to list out all the parameters given as this will let you understand how you can determine the answer to the question from the given parameters.
From the given information;
The tension in the cable = 11650 NThe length (d) = 5.580 mThe mass of the horizontal steel beam (M) = 88.50 kg/m (d)= 88.50 kg/m × 5.580 m= 493.83 kgDistance (s) = 0.522 mx = 1.350 m and h = 2.070 mAcceleration due to gravity = 9.81 m/s²From the question;
the angle at which the crane is positioned can be determined by taking the tangent of the angle θ. i.e.
[tex]\mathbf{tan \ \theta = \dfrac{h}{d-s}}[/tex]
[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{h}{d-s} \Big )}[/tex]
[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{2.070 }{5.580 - 0.522} \Big )}[/tex]
[tex]\mathbf{\theta =22.26^0}[/tex]
Consider the equilibrium conditions of the torques with respect to the magnitude of the load at point P.
∴
[tex]\mathbf{Tsin \theta (d-s) - W_L (d-x) -(Mg) (\dfrac{d}{2}) = 0}[/tex]
By making the magnitude of the load [tex]\mathbf{W_L}[/tex] the subject of the formula, we have:
[tex]\mathbf{W_L = \dfrac{Tsin \theta (d-x) -(Mg) (\dfrac{d}{2})}{ (d-s) } }[/tex]
[tex]\mathbf{W_L = \dfrac{(11650 )sin (22.26) (5.580-1.350) -(88.50\times 9.81) (\dfrac{5.580}{2})}{ (5.580-0.522) } }[/tex]
[tex]\mathbf{W_L = 3211.81 \ N }[/tex]
Therefore, we can conclude that the magnitude of the load is 3211.81 N
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A small, free-to-rotate magnet is placed in a strong magnetic field. In what orientation will it come to rest
Answer:
South-North
Explanation:
All circuits include
a battery, wires, and a switch.
an energy source, a resistor, and a battery.
a battery, a light bulb, and a switch.
an energy source, a load, and wires.
Answer:
a battery, wires, and a switch.
Explanation:
All circuits include?
science thanks sa points
Answer: Are these free point?
Explanation:
An object moving at a constant velocity of 5.4 m/s travels for 12 s. How far will it move during that time?
Free-fall Acceleration is -10 m/s^2
Answer:
we know that
s=vt
given
v=5.4 m/s
t=12 s
s=5.4 m/s*12 s=64.8m
Explanation:
Hope this helps:)
Read the scenario below and answer the question that follows. Randall is hiring cooks for his restaurant. The first applicant is a handsome man with an average resumé and average job experience. The second applicant is a far less attractive man with a slightly above average resumé and above average job experience. Randall decides to hire the first applicant. Based on this information and on Randall’s decision, what might a psychologist conclude about Randall’s social perception? Randall has an unconscious assumption that attractive people are more competent. Randall has a unconscious assumption that unattractive people are bad cooks. Randall has a conscious assumption that attractive people make better cooks. Randall has a conscious assumption that unattractive people are more competent.
Randall has an unconscious assumption that attractive people are more competent.
What is meant by assumption ?The term assumption can be described as an unspoken premise that underlies the conclusion.
Here,
The capacity to accurately evaluate and draw conclusions about other individuals based on their overall physical appearance, verbal behaviour, and nonverbal attitudes is referred to as social perception.
Given that, for his restaurant, Randall is employing chefs. The first applicant is a dashing man with an average resume and career history. The second candidate is a far less appealing man with an average to slightly above average resume and work experience. Randall chooses to hire the first candidate.
This shows the social perception of Randall and his unconscious assumptions against unattractive people.
Hence,
Randall has an unconscious assumption that attractive people are more competent.
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How long will it take a car, starting from rest, accelerating at 2 meters per second square to travel the same distance that another car traveling at a constant rate of 20m/s will travel?
20 seconds
Explanation:
Let [tex]x_a[/tex] be the distance traveled by the accelerating car and [tex]x_c[/tex] be the distance traveled by the car moving with a constant velocity. When they cover the same distance, we can write
[tex]x_a = x_c \Rightarrow v_{0a}t + \frac{1}{2}at^2 = v_ct[/tex]
where [tex]v_c[/tex] is the velocity of car moving at a constant rate and a is the acceleration of the accelerating car. Since the accelerating car started from rest, then [tex]v_{0a}[/tex] is zero so our equation above simplifies to
[tex]\frac{1}{2}at^2 = v_ct[/tex]
Note that the variable t cancels out so solving for t, we get
[tex]\frac{1}{2}at = v_c \Rightarrow t = \dfrac{2v_c}{a}[/tex]
Plugging in the given values,
[tex]t = \dfrac{2(20\:\text{m/s})}{2\:\text{m/s}^2} = 20\:\text{s}[/tex]
To get the dimmest bulbs with two batteries and two bulbs you would connect the batteries in ____ and the bulbs in ____.
Answer:
batteries in parallel connection and bulbs in serial connection
To get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.
What is Parallel and series circuits?When two-terminal components and electrical networks that can be connected in series or parallel. This will result in two terminals in the electrical network, and may themselves participate in a series or parallel topology. When a two-terminal "object" is an electrical component or electrical network is a matter of perspective.
A circuit is said to be in series when the same current flows through all the components in the circuit where the current has only one path. A circuit is said to be parallel when there are multiple paths for the electric current to flow through it where the components which are part of the parallel circuit will have a constant voltage across all their ends.
Thus, to get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.
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Velocity and Acceleration Quick Check
C
D
E
During which of the labeled time segments is the object moving forward but slowing down?
(1 point)
Ο Α
0 С
OD
ОВ
Answer:
Explanation:
1 Object C has an acceleration that is greater than the acceleration for D.
2 B
3 17M
4 The velocity is zero.
5 a straight line with negative slope
just took it
difference between speed and velocity
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
Explanation:
A car with an initial position of 10.0 m
and an initial velocity of 16.0 m/s accelerates at an average rate of 0.50 m/s2 for 4.0 s. What is the car’s position after 4.0 s?
Answer:
78
Explanation:
x=xi+vi(t)+1/2a(t)^2
x=10+16(4)+1/2(0.50)(4)^2
x=74+4
x=78 m
The figure shown above is the circuit diagram for a simple dc power supply. Identify the type of rectifier circuit represented in the figure and explain the operation of the circuit with reference to the function of each component within the circuit.
Answer:
D1 FG 12 15×AG+5T×G7+3F
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time?
Free-fall Acceleration is -10 m/s^2
Answer:
Explanation:
s = s₀ + v₀t + ½at²
s = 0 + 0(15) + ½(6)(15²)
s = 675 m
Not sure what the free fall acceleration is needed for, but if the object is dropped from a high enough point, it will travel in 15 seconds
s = ½10(15²) = 2250 m if air resistance is ignored
A 6.5 N ball is thrown with an initial velocity of 20 m/s at a 35° angle from a height of 1.5 m, what is the velocity if it is caught at 1.5 m?
Answer:
20 m/s at -35°
Explanation:
Ignoring air resistance, the initial vertical velocity will be reversed and the initial horizontal velocity will remain constant.
A metal wire of length 1.2 m and cross-sectional area 2.0 x 10 raised to power-7 m 2 is stretched by a force of 50 N. assuming the force constant of the metal is 6000 Nm-1. Calculate the tensile stress
L=1m
=2mm²=(2/1000²)=2(10^-6)m
y=4x10¹¹N/m²
∆l=2mm=2/10.00=0.002m
=(4x10¹¹x2x10^-6x2x10^-3)÷1m
=16x10¹¹-⁶-³
=16x10¹¹-⁹
=16x10²
=1600N
where a=cross sectional area=2x10^-6m
C=2mm= 2x10^-3m
L=1m
hope it helps
Air is pumped into the tyre to inflate it.
This increases the temperature and the pressure of the air in the tyre.
Use ideas about molecules to explain why the air pressure in the tyre increases. *
A gold doubloon 6.1 cm in diameter and 2.0 mm thick is dropped over the side of a pirate sheep. When it comes to rest on the ocean floor at a depth of 770 m, how much has its volume changed?
The definition of volume modulus and the variation of pressure with depth allows to find the result for the variation of the volume of the coin is:
ΔV = 2.15 10⁻⁸ m³
The pressure with the depth is given by the relation
P = P₀ + ρ g h
Where P is the pressure, ρ is the density anf h depth.
The size of the bodies is determined by the distance of their atomic and molecular bonds, therefore the size of the bodies changes under external interations, in the case of hydrostatic pressure a constant called volumetric modulus is defined.
[tex]B = - \frac{\Delta P}{\frac{\Delta V}{Vo} } \\\Delta V = - \frac{\Delta P }{B} \ V_o[/tex]
Where ΔP is the pressure change, V₀ and V are the volume change and the initial volume of the body, the negative sign is introduced so that the volumetric modulus is a positive quantity.
They indicate the diameter and thickness of the coin (d = 6.1 cm and e =0.20 cm) on the sea surface and the depth to which it is submerged
h = 770 m
Let's look for the volume of the coin.
V₀ = π r² h = [tex]\pi \ \frac{d^2}{4} \ e[/tex]
V₀ = [tex]\pi \ \frac{0.061^2 }{4} \ 0.002[/tex]
V₀ = 5.84 10-6 m³
Let's find the pressure at the depth of y = 770 m, the density of sea water is ρ = 1025 kg / m³, the pressure at the surface is the atmospheric pressure P₀ = 1 10⁵ Pa, the volumetric modulus of water is B = 0.21 10¹⁰ Pa.
P = 1 10⁵ + 1025 9.8 770
P = 1 10⁵ + 7,735 10⁶
P = 7.84 10⁶ Pa
Let's calculate
ΔV =[tex]- \frac{1 \ 10^5 - 7.84 \ 10^6 }{0.21 \ 10^{10}} \ 5.845 \ 10^{-6}[/tex]
ΔV = 2.15 10-8 m³
In conclusion using the definition of volume modulus and the variation of pressure with depth we can find the result for the variation of the volume of the coin is:
ΔV = 2.15 10-8 m³
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a coconut falls from the top of a tree and takes 3.5 seconds to reach the ground. How tall is the tree?
Hello!
To solve, we can begin by using the kinematic equation:
[tex]d = v_it + \frac{1}{2}at^2[/tex]
Where:
vi = initial velocity (m/s)
t = time (s)
a = acceleration (in this case, due to gravity. g = 9.8 m/s²)
Since the object falls from rest, the initial velocity is 0 m/s.
[tex]d = \frac{1}{2}at^2[/tex]
Plug in the given values:
[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]
a. How fast should a boy of 40kg mass run so that the kinetic energy becomes 800J?
Answer:
[tex]k.e. = \frac{1}{2}mv {}^{2} \\ 800= \frac{1}{2} \times 40 \times v {}^{2} \\ 800 = 20v {}^{2} \\ v {}^{2} = 400 \\ v {}^{2} = 20ms {}^{ - 1} [/tex]
k.e. = kinetic energy
hope helpful <3
Which is the main gas that makes up the Earth's atmosphere?
Answer:
78 percent nitrogenExplanation:
I hope it's helpful for you
please help me with these four i dont rlly get the question itself tbh. 20 points
Explanation:
These prefixes are very commonly used in naming chemical compounds.
Di- means two.
For example, carbon dioxide's formula is be written as [tex]\text{CO}_2,[/tex] and it has 2 oxygen atoms, hence "di-oxide."
Tetra - means four.
For example, carbon tetrachloride's chemical formula is written as [tex]\text{CCl}_4[/tex], and there are four chlorine atoms
Deca- means ten
For example, lanthanum decahydride's chemical formula is written as [tex]\text{LaH}_{10}.[/tex] In this case there are 10 hydrogen atoms for every lanthanum atom.
Hepta - means seven
For example, iodine heptafluoride is written as [tex]\text{IF}_7[/tex]. Note the seven fluorine atoms attached to the iodine atom, hence the name "hepta-fluoride."
Why do you suppose Km values are so frequently standardized and published, drawing attention to the value of Vmax/2, rather than Vmax itself
Km values are standardized because half the Vmax (Vmax/2) is more informative than Vmax. This value (Km) can be used to calculate the affinity of the enzyme by a given substrate.
The Km (Michaelis constant) of the enzyme refers to the value in which the concentration of substrate is equal to half of its maximum velocity (Vmax/2).
This value (Km) is inversely proportional to the affinity of an enzyme by a given substrate.
An enzyme showing a high Km also exhibits a low affinity for its specific substrate, and thereby this enzyme requires a high concentration of the substrate to reach its maximum velocity (Vmax).
In consequence, the Km value is a more informative value than the maximum velocity (Vmax), which only indicates the concentration of an enzyme catalyzing a reaction under ideal conditions.
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A 115 kg hockey player, Adam, is skating east when he tackles a stationary 133 kg player, Bob. Afterward, they move at 1.35 m/s east. What was Adam's velocity before the collision? (Unit = m/s)
Answer:
Explanation:
Conservation of momentum
115v + 133(0) = (115 + 133)1.35
v = 2.911304...
v= 2.91 m/s east
Answer:
The velocity east is 2.91
Explanation:
Fill in the box
2.91
what is photosynthesis
: [tex] \implies[/tex] The Photosynthesis is the process of capturing light energy and transforming it into chemical energy. Green plants and several other organisms use light energy and convert carbon dioxide and water into glucose. In this process, oxygen is produced as a by-product
→ We also who how it's process occur
In plants and blue-green algae, the photosynthesis process takes place in chloroplasts. The chloroplast is present in all green parts of a plant – the leaves, green stems, sepals, and even in the flowers, in the form of green colour plastids. The chloroplast is found only in plant cells and is essential for photosynthesis reaction.Photosynthesis Equation
Carbon dioxide and water are the two major factors involved in the photosynthesis reaction. It’s an endothermic reaction, and the products resulting from it are oxygen and glucose. The formula is:6CO2 + 6H2O = C6H12O6 + 6O2
However, some bacteria don’t produce oxygen as a by-product of photosynthesis. They are called anoxygenic photosynthetic bacteria, and those who do it are called oxygenic photosynthetic bacteria.Importance of Photosynthesis
The photosynthesis process is very important for the survival of living beings, and to continue the food chain. It also produces oxygen, which is required for breathing.Photosynthetic Pigments
Four types of photosynthetic pigments are present in the leaves of the plants. They are: Chlorophyll a chlorophyll b xanthophylls CarotenoidsThe Factors Affecting Photosynthesis
Various factors influence/affect the photosynthesis process. These are:
Light Intensity: More the light, the more will be the rate of photosynthesis. Similarly, low light will lead to a low rate of photosynthesis.The Concentration of CO2: A higher CO2 concentration rate in a plant also accelerates the photosynthesis process. The required amount of CO2 is 300-400 PPM.Temperature: If the temperature is between the range of 25 to 35 degrees Celsius, the photosynthesis takes place effectively.Water: An essential amount of water is required for stomatal opening, and it’s a key factor in the process of photosynthesis.Pollution: The increasing rate of polluting particles in the atmosphere block the pores of somatic cells, and the intake of carbon dioxide becomes difficult.▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Hope it's helps !!
The element which does not show variable valency a) AI b)Fe c) Cu d) Hg
Answer:
None of these elements.
A student stomps sternly on a super-sized stomp rocket. They notice that the rocket lands 54
m away in a time of 5.5 s. Find the magnitude of the total initial velocity of the rocket.
The magnitude of the total initial velocity of the rocket is determined as 9.82 m/s.
What is the total initial velocity of the rocket?
The magnitude of the total initial velocity of the rocket is calculated as follows;
V = D/T
where;
D is the distanceT is time of motionV = (54)/(5.5)
V = 9.82 m/s
Thus, the magnitude of the total initial velocity of the rocket is determined as 9.82 m/s.
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A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in m, would a 1000-kg mass have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water
A.
The energy of the hot water is 482630400 J
Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,
c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C
Substituting the values of the variables into the equation, we have
Q = mcΔT
Q = ρVcΔT
Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C
Q = 482630400 J
So, the energy of the hot water is 482630400 J
B.
The elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.
Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m
Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J
So, ΔU = mgΔh
ΔU = mg(h - h')
making h subject of the formula, we have
h = h' + ΔU/mg
Substituting the values of the variables into the equation, we have
h = h' + ΔU/mg
h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)
h = 0 m + 482630400 J/(9800 kgm/s²)
h = 0 m + 49248 m
h = 49248 m
So, the elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.
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A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
The figure shows a top view of a merry-go-round of radius capital R rotating counterclockwise. A sandbag is located on the merry-go-round at a distance lowercase r from the center.
Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.
The angular speed of the merry-go-round reduced more as the sandbag is
placed further from the axis than increasing the mass of the sandbag.
The rank from largest to smallest angular speed is presented as follows;
[m = 10 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 20 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 1.0·R]
Reasons:
The given combination in the question as obtained from a similar question online are;
1: m = 20 kg, r = 0.25·R
2: m = 10 kg, r = 1.0·R
3: m = 10 kg, r = 0.25·R
4: m = 15 kg, r = 0.75·R
5: m = 10 kg, r = 0.5·R
6: m = 40 kg, r = 0.25·R
According to the principle of conservation of angular momentum, we have;
[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]
The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]
Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of m·r² increases, the value of [tex]\omega _f[/tex] decreases.
The values of m·r² for each combination are;
Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²
Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²
Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²
Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²
Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²
Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²
Therefore, the rank from largest to smallest angular speed is as follows;
Combination 3 > Combination 1 > Combination 5 = Combination 6 >
Combination 2
Which gives;
[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =
10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].
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5 kg block of iron is heated to 800°C. It is placed in the tub containing 2 L of water at 15°C. Assuming all the water is brought to the boil rapidly, calculate the mass of water which boils off. (The specific heat capacity of iron 800°C is 220 J kg-1 K-1)
Answer:
Heat Loss = 5 kg * 700 deg K * 220 J / (kg*deg K) = 7.70E5 J
Since there are 4.186 J/cal
Heat Loss = 7.70E5J / 4.186 J/cal = 1.84E5 cal
Heat Gain = 2000 g * 85 deg K / cal / (deg K g) + M * 540 cal/g
Heat Gain = 1.70E5 cal + M * 540 cal/g
M = (1.84 - 1.70) E5 g / 540
25.9 g
25.9 g or 25.9 cm^3 or .0259 L of water will boil away