Answer:
The water and peanut butter have different chemical properties.
(Lipids are not soluble in water)
Answer:
a. The water and peanut butter have different chemical properties.
Explanation:
please help me fast please help
Answer:
the net energy Gained per hour equals 30Kcal/h
A 65 kg cart travels at a constant speed of 4.6 m/s. What is its kinetic energy?
Answer:
Explanation:
mass (m) = 65 kg
velocity (v) = 4.6 m/s
Kinetic energy (KE)
= 1/2 * m * v²
= 1/2 * 65 * 4.6²
= 687.7 J
hope it helps :)
Three 5 Ohm resistors are connected in parallel to a 9 Volt power supply. What is the current through each resistor?
Answer:
I₁ = I₂ = I₃ = 12 A
Explanation:
Since the resistors are connected in parallel across a battery. Hence, the voltage across each of them will be the same, which will be equal to the voltage of the power supply. Since each resistor has the same resistance of 5 Ω. Therefore. the current across each resistor will also be the same. We can easily calculate these current values by the use of Ohm's Law, as follows:
[tex]V = IR\\I_{1}=I_{2}=I_{3}=\frac{V}{R} \\\\I_{1}=I_{2}=I_{3}=\frac{9\ Volt}{5 \Omega}\\\\[/tex]
I₁ = I₂ = I₃ = 12 A
A kicker punts a football from the very center of the field to the sideline 41 yards downfield. (A football field is 53 yards wide.)What is the angle between the direction of the net displacement of the ball and the 50-yard line of the field
Answer:
Ur answer is 12 sir!
Explanation:
HAve a good day sir
A uniform 140 g rod with length 57 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 30 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 11 cm on each sides of the center, at which time the system rotates at an angular speed of 23 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.
Answer:
The correct answer is "12 rad/s"
Explanation:
The given values are,
Mass of rod,
M = 140 g
i.e.,
= 0.14 kg
Length,
L = 57 cm
i.e.,
= 0.57 m
Mass of beads,
M = 30 g
i.e.,
= 0.03 kg
Angular speed,
r = 11 cm
i.e.,
= 0.11 m
Now,
The inertia of rods will be:
= [tex]\frac{1}{12}ML ^2[/tex]
On substituting the values, we get
= [tex]\frac{1}{12}\times 0.14\times (0.57)^2[/tex]
= [tex]0.0037905 \ kg-m^2[/tex]
The inertia of beads will be:
= [tex]mr^2[/tex]
On substituting the values, we get
= [tex]0.03\times (0.11)^2[/tex]
= [tex]0.000726 \ kg-m^2[/tex]
The total inertia will be:
= [tex]Inertia \ of \ rods+Inertia \ of \ beads[/tex]
= [tex]0.0037905 + 0.000726[/tex]
= [tex]0.0045165 \ kg-m^2[/tex]
now,
The angular speed of the system will be:
⇒ [tex]L_1w_1=L_2w_2[/tex]
On substituting the values in the above equation, we get
⇒ [tex]0.0045165\times 23 = (0.0037905 + (0.03\times 0.285^2)\times 2 )\times w_2[/tex]
⇒ [tex]0.1038795 = 0.0037905 + (0.00243675\times 2 )\times w_2[/tex]
⇒ [tex]w_2 = 12 \ rad/s[/tex]
Gaseous sulfur dioxide is a compound that combines with water in the atmosphere to form acid rain. What is the primary source of sulfur dioxide?
O volcanic emissions
O combustion of fossil fuels
O destruction of tropical forests
O mining and mineral extraction
Please help
A 90 kg football player is running at 0.9 m/s. Assuming a head on collision, at what speed would a 110 kg football player have to run into the 90 kg
player in order to stop him
Answer:
The two moments must be the same:
p1=p2
m1v1=m2v2
v2=(m1v1)/m2
v2=(90 kg x 0.9 m/s)/110kg=0.7 m/s
compare and contrast series and parallel circuits?
In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.
Hope It Helps!
Please please help I'm stuck!!!
Answer: iDc im in 3grade
Explanation:sorry
A capacitor with air between its plates ischarged to 60 V and then disconnected fromthe battery. When a piece of glass is placedbetween the plates, the voltage across thecapacitor drops to 46 V.What is the dielectric constant of this glass?Assume the glass completely fills the spacebetween the plates.
Answer:
k = 1.30
Explanation:
For this exercise let's write the capacitance in air and with dielectric
air C₀ = Q / DV
dielectric C = k Q / DV
They tell us that the capacitor is charged and then the battery is disconnected, therefore the charge stored on the plate remains constant.
therefore the capacitance a changes to the value
C = k C₀
The voltage in the presence of dielectric must meet the relationship
ΔV = ΔV₀ / k
k = ΔV₀ /ΔV
let's calculate
k = 60/46
k = 1.30
If you lean against a wall, the wall pushes back on you with a weaker force
true
false
false - (.do not ) if you lean against a wall the wall pushes back on you with a( n) weaker force
A uniform 2.00-kg circular disk of radius 20.0 cm is rotating clockwise about an axis through its center with an angular speed 30.0 revolutions per second. A second uniform 1.50-kg circular disk of radius 15.0 cm that is not rotating is dropped onto the first disk so that the axis of rotation of the first disk passes through the center of the second disk. What is the final angular speed of the two disks when they are rotating together
Answer:
w = 132.57 rad / s
Explanation:
To solve this exercise we must define a system formed by the two discs, therefore the angular momentum is conserved
initial
L₀ = I₀ w₀
final
L_f = I w
how the moment is preserved
L₀ = L_f
I₀ w₀ = I w
the moment of inertia of disk 1 is
I₀ = ½ m₁ r₁²
The moment of inertia of the set is
I = I₀ + I₂
I = ½ m₁ r₁² + ½ m₂ r₂²
we substitute
½ m₁ r₁² w₀ = (½ m₁ r₁² + ½ m₂ r₂²) w
w = [tex]\frac{ m_1r_1^2 }{ m_1 r_1^2+ m_2r_2^2} \omega_o[/tex]
let's reduce the magnitudes to the SI system
w₀ = 30.0 rev / s (2π rad / 1 rev) = 188.5 rad / s
let's calculate
w = ( [tex]\frac{ 2 \ 0.20^2}{ 2 \ 0.20^2 + 1.5 \ 0.15^2 }[/tex]) 188.5
w = ( [tex]\frac{0.08}{ 0.11375}[/tex] ) 188.5
w = 132.57 rad / s
27. The traffic officer issued violation tickets to traffic
violators. If 68 of them composed 80% of the total
violators, how many violators are there in all?
A. 83
B. 84
C. 85
D. 86
28. There are 18 roses in a bunch of 24 flowers.
What percent of the flowers are roses?
A. 18%
B. 24% C. 60% D. 75%
Please help me ❤️❤️ thank u
Hope god
Answer:
27: 85
28:75%
Explanation:
27:68=80
?=100 hence (68×100)÷80
=85
28:18/24× 100
=75%
Which scale do geologists use to estimate the total energy released by an earthquake?
Answer:
Richter scale Magnitude
Richter scale. Magnitude is a measure of the amount of energy released during an earthquake, and you've probably heard news reports about earthquake magnitudes measured using the Richter scale. Something like, "A magnitude 7.3 earthquake struck Japan today.
Explanation:
Hope I helped!
A 1,250 kg car is moving due to 6,500 N engine force. If the kinetic friction coefficient between the car and the road is 0.32, what is the car's acceleration?
A) 32m/s²
B) 200m/s²
C) 50m/s²
D) 2m/s²
Answer and I will give you brainiliest
Answer:
b i hope this is correct answee
You have two cylindrical bar magnets with the same mass, length, and dipole moment, but different moments of inertia, I1 and I2. You place each bar magnet so that it is initially oriented perpendicular to a uniform magnetic field. When you let each bar magnet go from rest, you find that magnet 1 takes less time than magnet 2 to rotate so that it is parallel to the magnetic field. Is I1 greater than, less than, or equal to I2
Answer:
[tex]l1 < l2[/tex]; [tex]l1[/tex] is less than [tex]l2[/tex]
Explanation:
Firstly, we take a look at the concept of moment of inertia, This concept comes from the fact that it resists change in motion. A body that has a higher value of moment of inertia requires more force ( torque) to change its position as compared to a body with lesser value of moment of inertia.
Now taking a look at the situation in the question, both the magnets are placed in similar magnetic field, so having the same amount of force, hence torque acts on both magnets.
Since magnet 1 takes less time to rotate compared to magnate 2, hence magnet 1's moment of inertia should be less than that of magnet 2.
Therefore, [tex]l1 < l2[/tex]; [tex]l1[/tex] is less than [tex]l2[/tex]
Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why
1. Glider 2 will be always kicked toward the same direction as glider 1 comes in (v2,f and v1,i have the same sign)
2. Glider 1 will bounce back (v1f and v1,i have opposite sign) if it is lighter than glider 2 (m1= m2).
3. Glider 1 will keep moving forward (v1,f and v1 i have the same sign) if it is heavier than glider 2 (m1 m2)
4. Glider 1 will stop (v1f =0) if it weighs the same as glider 2 (m1= m2)
v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2
v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2
Answer:
Explanation:
1 )
Put v2,i =0, in second equation
v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2
v2,f = 0 + 2m1v1,i/m1+m2
v2,f = 2m1v1,i/m1+m2
In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.
2 )
Put m1 < m2 and v2,i =0 in first equation
v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2
v1,f= (m1-m2)v1,i
As m1-m2 is negative , v1f and v1i will have opposite sign.
3 )
Put m1 > m2 and v2,i =0 in first equation
v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2
v1,f= (m1-m2)v1,i
m1 - m2 is positive so v1f and v1i will have same sign.
4 )
Put m1 = m2 and v2,i =0 in first equation
v1,f= (m1-m2)v1,i
= 0 because m1 = m2
So glider 1 will stop because v1,f = 0 .
When the glider 2 is initially at rest (v2,i =0) and collision is elastic, the different equation are explained for different conditions of glider 1.
What is conservation of momentum?Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.
When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.
The given equations are,
[tex]v_{1f}= \dfrac{(m_1-m_2)v_{1i} + 2m_2v_{2i}}{m1+m2}[/tex]
[tex]v_{2f}= \dfrac{(m_2-m_1)v_{2i} + 2m_1v_{1i}}{m1+m2}[/tex]
Given that the collision is elastic and glider 2 is initially at rest (v2,i =0),
1. Glider 2 will always be kicked toward the same direction as glider 1 comes in (v2,f and v1,i have the same sign)Put [tex]v_{2i}[/tex] equal to zero, in the given equation as,
[tex]v_{2f}= \dfrac{(m_2-m_1)(0) + 2m_1v_{1i}}{m1+m2}\\v_{2f}= \dfrac{ 2m_1v_{1i}}{m1+m2}[/tex]
v2,f and v1,i have the same sign in the above equation.
2. Glider 1 will bounce back (v1f and v1,i have opposite sign) if it is lighter than glider 2 (m1<m2).Put [tex]v_{1i}[/tex] equal to zero, in the given equation as,
[tex]v_{1f}= \dfrac{(m_2-m_1)(0) + 2m_2v_{2i}}{m1+m2}\\v_{1f}= \dfrac{ 2m_2v_{2i}}{m1+m2}[/tex]
Now, if we consider (m1<m2), for the above equation, the result will be negative. Thus, v1f and v1,i have opposite sign.
3. Glider 1 will keep moving forward (v1,f and v1 i have the same sign) if it is heavier than glider 2 (m1 >m2)Put [tex]v_{2i}[/tex] equal to zero, in the first equation as,
[tex]v_{1f}= \dfrac{(m_2-m_1)(v_{1i}) + 2m_2(0)}{m1+m2}\\v_{1f}= \dfrac{(m_2-m_1)(v_{1i})}{m1+m2}[/tex]
Now, if we consider (m1>m2), for the above equation, the result will be positive. v1,f and v1 i have the same sign.
4. Glider 1 will stop (v1f =0) if it weighs the same as glider 2 (m1= m2)Put [tex]v_{2i}[/tex] equal to zero, in the first equation as,
[tex]v_{1f}= \dfrac{(m_2-m_1)(v_{1i}) + 2m_2(0)}{m1+m2}\\v_{1f}= \dfrac{(m_2-m_1)(v_{1i})}{m1+m2}[/tex]
Now, if we consider (m1=m2), for the above equation, the result will be zero. Thus the glider 1 will stop (v1f =0).
Hence, when the glider 2 is initially at rest (v2,i =0) and collision is elastic, the different equation are explained for different conditions of glider 1.
Learn more about the conservation of momentum here;
https://brainly.com/question/7538238
what is the magnitude of electric field at apoint 2 meter from a point charge q= 4nc
Your heart pumps blood into your aorta (diameter 2.5 cm) with a maximum flow rate of about 500 cm^3/s. Assume that blood flow in the aorta is laminar (which is not a very accurate assumption) and that blood is a Newtonian fluid with a viscosity similar to that of water.
a. Find the pressure drop per unit length along the aorta. Compare the pressure drop along a 10 cm length of aorta to atmospheric pressure (105 Pa).
b. Estimate the power required for the heart to push blood along a 10 cm length of aorta, and compare to the basal metabolic rate of 100 W.
c. Determine and sketch the velocity profile across the aorta (assuming laminar flow). What is the velocity at the center
Answer:
a. i) The pressure drop per unit length is 52,151.89 Pa
ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta
b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts
ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta
c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel
ii) The velocity at the center is approximately 2.04 m/s
Explanation:
The given diameter of the aorta, D = 2.5 cm = 0.025 m
The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s
Assumptions;
The blood flow is laminar
The blood is a Newtonian fluid
The viscosity of water ≈ 0.01 poise = 1 cp
a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;
[tex]Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)[/tex]
Where;
Q = The flow rate = A·v
A = The cross sectional area
R = The radius = D/2
Δp/L = The pressure drop per unit length of the pipe
Therefore, we have;
[tex]\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89[/tex]
The pressure drop per unit length ΔP/L = 52,151.89 Pa
ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;
∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa
Given that the atmospheric pressure, [tex]P_{atm}[/tex] = 10⁵ Pa, we have;
[tex]P_{atm}[/tex]/ΔP = 10⁵/5,215.189 ≈ 19.175
Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta
b. i) The power, P = Q × ΔP
Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts
ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;
P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35
The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length
c. i) The velocity profile across the aorta is given as follows;
[tex]v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2[/tex]
Where;
[tex]v_m[/tex] = The velocity at the center
We get;
[tex]v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04[/tex]
The velocity at the center, [tex]v_m[/tex] ≈ 2.04 m/s
ii) The velocity profile, v(r), is given by the following formula;
[tex]v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right][/tex]
Therefore, we have;
[tex]v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2[/tex]
The velocity profile of the pipe is created with Microsoft Excel
Which bond(s) shown are double bonds?
Answer:
in the diagram shown, Oxygen is the only one that has a double bond.
Explanation:
Chemical bonds are the way that electors are shared between atoms to form molecules whose energy is less than the energy of individual atoms.
A graphical way to show these electrons is to represent them by point in the atoms. Every other point represents a shared bond or pair of electrons.
Therefore for simple bonds two electrons are drawn. For double bonds four electrons and for triple bonds six electrons.
Therefore, in the diagram shown, Oxygen is the only one that has a double bond.
Your physics teachers want to race each other using wagon rockets.
The fastest wagon accelerates at 4 ms-2.
Calculate the distance traveled in 10 seconds, if it starts from rest.
Answer:
s=1/2at^2
s=1/2 x 4 x(10)^2=200m
Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (560 nm) on a double slit with a separation of 0.108 mm. The diffraction pattern shines on the classroom wall 3.0 m away. Calculate the fringe separation between the third order and central fringe.
Answer:
y = 4.666 10⁻² m
Explanation:
The constructive interference experiment for the double slit
d sin sin θ = m λ
Let's use trigonometry to find a sine relationship.
Tan θ = y / L
tan θ = sin θ/ cos θ
in these experiments the angles are very small
tan θ = sin θ
sin θ = y / L
[tex]d \frac{y}{L}[/tex] = m λ
y = [tex]\frac{ m \lambda \ L}{d}[/tex]
we replace the values
y = 3 560 10⁻⁹ 3.0 / 0.108 10⁻³
y = 4.666 10⁻² m
What is optics???????
see my percentage of answering and thanks , it's crazy
Answer:
hi..bro
make pinky sis answer as brinlist ok
have a good day
A skier starts from rest at the top of a 20 degree incline and skis in a straight line to the bottom of the slope, a distance d (measured along the slope) of 400 m. If the coefficient of kinetic friction between the skis and the snow is 0.2, calculate the skier's speed at the bottom of the run.
Answer:
Explanation:
Loss of potential energy = mgh.
h = d sin 20
= 400 sin20 = 136.8 m
Loss of potential energy = m x 9.8 x 136.8
= 1340.64 m
negative work done by friction = μ mg cosθ x d
= .2 x m x 9.8 x cos 20 x 400
= 736.72 m
Net loss of potential energy = 1340.64 m - 736.72 m
= 603.92 m
= gain of kinetic energy = 1/2 m v²
1/2 m v² = 603.92 m
v² = 1207.84
v = 34.75 m /s .
Explain why the sound waves always reach the observer after the light waves
Answer:
I think it's because the light waves travel faster than the sound waves.
The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.
What is the speed of light?Light occurs in the electromagnetic spectrum. Recall that light can be transmitted through vaccuum unlike sound.
The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.
Learn more about speed of light:https://brainly.com/question/8832859
#SPJ2
A 50kg bicyclist on a 10kg bicycle speeds up from 5.0m/s to 10m/s. What was the total kinetic energy after accelerating?
Answer:
Explanation:
Ek=1/2mv^2
m=50+10=60,v=10
Ek=1/2*60*100=3000J
A frictionless piston-cylinder contains carbon dioxide gas (CO2) initially at 500oC and 2 MPa. The system is o3 cooled in an isobaric process until the final temperature becomes 350 C and the final volume is 1 m . For this process, determine: (a) The reduced pressure and the reduced temperature of the initial state. (b) The initial volume of the piston-cylinder (in m3). (c) The mass of CO2 in the piston-cylinder (in kg). (d) The total boundary work for the process (in kJ). (e) The amount of heat transfer during the cooling process (in kJ).
Answer:
(a) The reduced pressure is 0.2711 MPa
The reduced temperature is 2.54 K
(b) The initial volume of the piston is approximately 0.806 m³
(c) The mass of CO₂ is approximately 16.9884 kg
(d) The work done, W is approximately 388.023 kJ
(e) The heat transfer is approximately -2,650.1904 kJ
Explanation:
The initial temperature of the piston-cylinder, T₁ = 500°C = 773.15 K
The initial pressure of the gas, P₁ = 2 MPa
The final temperature of the gas, T₂ = 350°C
The final volume of the gas = 1 m³
(a) For an isobaric process, we have;
The reduced pressure,
[tex]P_r = \dfrac{P}{P_c}[/tex]
The critical pressure of carbon dioxide, [tex]P_c[/tex] = 7.3773 MPa
[tex]P_r = \dfrac{2 \, MPa}{7.3773 \, MPa} \approx 0.2711 \, MPa[/tex]
The reduced pressure, [tex]P_r[/tex] = 0.2711 MPa
The critical temperature, [tex]T_c[/tex] = 304.13 K
The reduced temperature, [tex]T_r[/tex], is given by the following formula;
[tex]T_r = \dfrac{T}{T_c}[/tex]
Therefore, [tex]T_r[/tex] = (773.15 K)/(304.13 K) = 2.54216947 K
The reduced temperature, [tex]T_r[/tex] ≈ 2.54 K
(b) The initial volume of the piston, V₁ = (V₂/T₂) × T₁
∴ V₁ = (1 m³/773.15) × 623.15 = 0.80598848865 m³ ≈ 0.806 m³
The initial volume of the piston, V₁ ≈ 0.806 m³
(c) The number of moles of CO₂ in the cylinder, 'n', is given according to the following formula;
n = P·V/(T·R)
The universal gas constant, n = (2 × 10⁶Pa × 1 m³)/(623.15 K × 8.3145 J/(mol·K)) ≈ 386.0124 moles
The mass of CO₂ ≈ 386.0124 moles × 44.01 g/mol = 16.9884 kg
(d) The work done, W = P·([tex]V_f - V_i[/tex])
W = 2 × 10⁶ × (1 - 0.80598848865) = 388023.0227
The work done, W ≈ 388.023 kJ
(e) The heat transfer dQ = m·[tex]c_p[/tex] ×(T₂ - T₁)
[tex]c_p[/tex] for CO₂ ≈ 1.04 kJ/(kg·K)
∴ dQ = 16.9884 × 1.04 × (350 - 500) = -2,650.1904 kJ
Therefore, the heat transfer = dQ = -2,650.1904 kJ
The bus lay 40 km at a speed of 72 km / h, and then another 60 km at a speed of 30 m / s. Determine the average speed of the bus along the way.
Answer:
25 m/s
Explanation:
From the question,
Average speed = Total distance /total time.
S' = D/T........................... Equation 1
D = 40+60 = 100 Km = 100000 m.
T = t₁+t₂
t₁ = (40×3600/72) s = 2000 s
t₂ = 60000/30 = 2000 s
T = 2000+2000 = 4000 s.
SUbstitute the values of T and D into equation 1
S' = 100000/4000
S' = 25 m/s
What year did Badminton become a full-medal Olympic sport?
Answer:
1992
Explanation:
Badminton made its debut as a demonstration sport at the 1972 Olympic Games in Munich. It was not until the 1992 Games in Barcelona that it was officially included on the Olympic programme, with men's and women's singles and doubles events.
An automobile tire with a radius of 0.30 m starts at rest and accelerates at a constant angular acceleration of 2.0 rad/s^2 for 7.0 s. What is the angular displacement of the tire?
A. 32 rad
B. 25 rad
C. 28 rad
D. 49 rad
The tire undergoes an angular displacement θ in time t according to
θ = θ₀ + ω₀ t + 1/2 α t ²
where
θ₀ = initial angular displacement
ω₀ = init. ang. velocity
α = ang. acceleration
Here we have θ₀ = 0 and ω₀, so after t = 7.0 s with acceleration α = 2.0 rad/s², the wheel will have been displaced
θ = 1/2 (2.0 rad/s²) (7.0 s)² = 49 rad
making the answer D.