Answer:
The energy stored in the spring would be 1 joule.
Explanation:
hope that helps?
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
A 2890-lb car is traveling with a speed of 58 mi/hr as it approaches point A. Beginning at A, it decelerates uniformly to a speed of 18 mi/hr as it passes point C of the horizontal and unbanked ramp. Determine the total horizontal force F exerted by the road on the car just after it passes point B.
Answer:
4592.57 lb
Explanation:
The missing diagram for this question is attached in the image below.
Given that:
the weight of the car = 2890 lb
At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr
At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr
To ft/s:
[tex](V_A)[/tex] = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_A)[/tex] = 85.07 ft/s
[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_C)[/tex] = 26.4 ft/s
Between A to C, the total distance is;
[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]
Now, we need to determine the deceleration of the car using the formula:
[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]
[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]
[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]
[tex]696.96-7236.9049 = 2 a (654.154)[/tex]
[tex]-6539.9449 = 2 a (654.154)[/tex]
[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]
a = -4.99 ft/s²
The velocity of the car as it passes via B
[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]
[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]
[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]
[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]
[tex]v_B =\sqrt{ 10540.2849}[/tex]
[tex]v_B =102.67 \ ft/s[/tex]
Along B, the car's acceleration is:
[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]
[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]
[tex]a_B = 51.17 \ ft/s^2[/tex]
Finally, the total horizontal force F exerted = m[tex]a_B[/tex]
[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]
= 4592.57 lb
An airplane, starting from rest, moves down the runway at constant acceleration for 23 s and then takes off at a speed of 66 m/s. What is the average acceleration of the plane (in m/s2)?
Answer:
46
Explanation:
On topographic maps, contour lines that are farther apart indicate what ?
Answer:
if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.
Answer:
gentle slopes
Explanation:
The time-average power carried by a UPEMW propagating in vacuum is 0.05 W/m2. i) What is the amplitude value of the electric field and the amplitude value of the magnetic field in the wave
Answer:
The correct solution is "11.51 mA".
Explanation:
Given:
Time average power,
[tex]P_{avg}=0.05 \ W/m^2[/tex]
n = 377
As we now,
⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]
or,
⇒ [tex]E_0^2=0.05\times 377[/tex]
⇒ [tex]=4.341 \ V[/tex]
hence,
⇒ [tex]H_0=\frac{E_0}{n}[/tex]
By putting the values, we get
[tex]=\frac{4.341}{377}[/tex]
[tex]=11.51 \ mA[/tex]
12. What type of circuit is the diagram below?
series circuit
parallel circuit
Answer:
parallel circuit
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;
I. Series circuit
II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.
Hence, the type of circuit that the above diagram above represents is a parallel circuit.
Answer:
parallel circuit
Explanation:
I got it right on my exam
When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.
Answer: When it comes to the movement of air, friction is greater near the ground surface.
Explanation:
A resistance in motion observed by an object while on another object is called friction.
For example, a vehicle moving on road will have friction between its tires and the road.
Friction is more near the ground surface rather than away from the ground surface.
Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.
A 1540-kg truck has a wheel base of 3.13 m (this is the distance between the front and rear axles). The center of mass of the truck is 1.09 m behind the front axle.a) What is the force exerted by the ground on each of the front wheels?b) What is the force exerted by the ground on each of the back wheels? Hint: Remember that the truck has four wheels.
a. The force exerted by the ground on each of the front wheels is 4918.16 Newton.
b. The force exerted by the ground on each of the back wheels is 2627.84 Newton.
Given the following data:
Mass of truck = 1540 kgDistance between the front and rear axles = 3.13 meters.Center of mass of the truck = 1.09 meters.a. To determine the force exerted by the ground on each of the front wheels:
First of all, we would take moment about the rear wheels.
[tex]F(3.13) - 1540(9.8) \times (3.13 - 1.09) = 0\\\\3.13F - 15092 \times 2.04 =0\\\\3.13F -30787.68=0\\\\F=\frac{30787.68}{3.13}[/tex]
Force, F = 9836.32 Newton
For each front wheel:
[tex]Force = \frac{9836.32}{2}[/tex]
Force = 4918.16 Newton.
b. To determine the force exerted by the ground on each of the back wheels:
We would determine the sum of the vertical forces acting on the wheels.
[tex]9836.32 + B - 1540(9.8) = 0\\\\9836.32 + B - 15092 = 0\\\\B=15092-9836.32[/tex]
B = 5255.68 Newton.
For each back wheel:
[tex]Force = \frac{5255.68}{2}[/tex]
Force = 2627.84 Newton.
Read more: https://brainly.com/question/22210180
A jogger travels a route that has two parts. The first is a displacement ->A of 2.20 km due south, and the second involves a displacement ->B that points due east.
(a) The resultant displacement ->A + ->B has a magnitude of 3.81 km.
What is the magnitude of B?
______ km
What is the direction of A + B relative to due south?
_____° west of south or east of south?
(b) Suppose that A - B had a magnitude of 3.81 km. What then would be the magnitude of B and what is the direction of A - B relative to due south?
Answer:
a) B = 3.11 km. θ= 54.7º E of S
b) B = 3.11 km θ= 54.7º W of S
Explanation:
a)
Since we know the value of the total displacement, and the value of the displacement A and its direction, we can find the magnitude of B just applying the Pythagorean Theorem, as follows:[tex]C=\sqrt{(2.2km)^{2} + B^{2} } = 3.81 Km (1)[/tex]
Solving for B, the only unknown, we get:[tex]B=\sqrt{(3.81km)^{2} -(2.2km)^{2} } = 3.11 Km (2)[/tex]
Now, applying some simple trig, we can find the angle that (A+B) makes with the S axis, as follows:[tex]\theta = arc tg (\frac{B}{A} )= arc tg ( \frac{3.11}{2.2} )= arctg (1.414) = 54.7 deg (3)[/tex]
Since it's a positive number, applying the convention that the positive angles are measured counterclockwise, this means that this angle is measured East of South.b)
If the magnitude of A-B is the same than the one for A+B, this means that the magnitude of B remains the same, i.e. 3.11 Km.But if we do graphically A-B, as it is the same as adding A + (-B), we find that the angle of A-B is different to the one in A+B, even the magnitudes of both displacements are the same.In this case, B is a negative number, because it's a displacement due west.So, applying the same trig that for a) we can find the angle that (A-B) makes with the S Axis, as follows:[tex]\theta = arc tg (\frac{-B}{A} )= arc tg ( \frac{-3.11}{2.2} )= arctg (-1.414) = -54.7 deg (4)[/tex]
So, since it is negative, it's measured clockwise from the S axis, so it's 54.7º W of S.Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
Ahmed is pushing a 4 Kg box to the right and Rashid is Pushing it to the right as well with a force of 12 N , the box accelerates by 5 m/s^2. What is the Force that is applied by Ahmed
Answer:
8 N
Explanation:
Applying,
(F'+F) = ma............... Equation 1
Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.
From the question,
Given: F = 12 N, m = 4 kg, a = 5 m/s²
Substitute these values into equation 1
(F'+12) = 4×5
(F'+12) = 20
F' = 20-12
F' = 8 N.
Hence Ahmed's force is 8 N
what is measured by the ammeter
Answer:
amperes
Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.
Explanation:
hope it helps
Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
In an experiment to measure the temperature of a Bunsen burner flame, a 250 g piece of iron is held in the flame for several minutes until it reaches the same temperature as the flame . The hot metal is then quickly transferred to 285 g of water contained in a 40.0 g copper calorimeter at 15.0 oC. The final temperature of the copper and water is 80.0 oC.
Using your answer from determine the temperature of the Bunsen flame.
Answer:
wait
Explanation:
Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2 < r1) and whose turns per unit length are n1 and n2.
Answer:
M' = μ₀n₁n₂πr₂²
Explanation:
Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.
So, M = N₂Ф₂₁/i₁
substituting the values of the variables into the equation, we have
M = N₂Ф₂₁/i₁
M = N₂B₁A₂/i₁
M = n₂lμ₀n₁i₁πr₂²/i₁
M = lμ₀n₁n₂πr₂²
So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²
M' = μ₀n₁n₂πr₂²
A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 14.5 m away
Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,
[tex](97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s[/tex]
negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:
[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]
Now, the distance covered by the player in this time will be:
[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]
s₁ = 0.022 m
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
The question is incomplete, the complete question is;
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
A) The atom moves to a state of lower energy
B) The atom is ionized
C) One of the electrons leaves the atom
D) The atom can be excited to a higher energy state
Answer:
The atom can be excited to a higher energy state
Explanation:
According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.
If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.
Rather, the atom is excited from ground state to a higher energy level.
Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿ a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
g the total mechanical energy of the satellite-Earth system when the satellite is in its current orbit is E. In order for the satellite to orbit Earth in a new stable circular orbit at an altitude of 12RE, the energy of the satellite-Earth system must be
Answer:
The correct answer is "[tex]\frac{4E}{3}[/tex]".
Explanation:
According to the question,
Energy of satellite,
⇒ [tex]E_s=-\frac{GM_sM_E}{2r}[/tex]
For the very 1st case:
[tex]r = R_E+R_E[/tex]
[tex]=2R_E[/tex]
or,
⇒ [tex]E=-\frac{GM_sM_E}{4R_E}[/tex]...(1)
For the new case:
[tex]r = R_E+\frac{R_E}{2}[/tex]
[tex]=\frac{3R_E}{2}[/tex]
then,
⇒ [tex]E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }[/tex]
[tex]=-\frac{GM_sM_E}{3R_E}[/tex]...(2)
From equation (1) and (2), we get
⇒ [tex]E'=\frac{1}{3}(4E)[/tex]
[tex]=\frac{4E}{3}[/tex]
you are stowing items and come across an aerosol bottle of hairspray.what should you do?
Answer:
below
Explanation:
A bike and rider of total mass 75.0 kg moving at a velocity of 30 m/s to the right
collides with a skater of mass 60,0 kg moking at a velocity of 6.0 m/s to the left.
After they collide, the skater has a velocity of O m/s. What is the final velocity of the
bike and rider?
O A. O m/s
O B. 1.8 m/s to the left
O C. 3.0 m/s to left
O D. 1.8 m/s to the right
Answer:
[tex](75.0 \times 30) + (60.0 \times 6.0) = (75.0 \times V) + (60.0 \times 0) \\ 2250 + 360 = 75V \\ 75V = 2610 \\ V = 34.8 \: m {s}^{ - 1} [/tex]
What is the efficiency of a machine that uses 102 kJ of energy to do 98 kJ of work?
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ( EASY QUESTION.. PLZZ HELPPP MEEE I WILL MARK YOU THE BRAINLIEST PLZZ)
Answer:
144 meters
Explanation:
the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t
S (height cliff)=24m/s×6s=144
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
A 55 kg pole vaulter falls from rest from a height of 5.4 m onto a foam rubber pad. The pole vaulter comes to rest 0.24 s after landing on the pad.
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision.
Answer:
a) 10.3 m/s
b) 566 N
Explanation:
[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]
[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]
The athelete velocity will be 10.3 and constant force 566 N.
What is velocity?The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).
Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.
The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.
Therefore, The athelete velocity will be 10.3 and constant force 566 N.
To learn more about velocity, refer to the link:
https://brainly.com/question/18084516
#SPJ5
A particle of unit mass moves so that displacement after t seconds is given by x = 2 cos (t - 2). Find the acceleration and kinetic energy at the end of 3 seconds. (K.E = (1/2) m v²)
Answer:
a₃ = -1.08 m/s², K = 1.42 J
Explanation:
The particle is in a periodic motion, so the general expression is
x = A cos (wt + Ф)
let's compare the terms with the expression they give us
x = 2 cos (t - 2)
the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.
to find the acceleration we use its definition
v = dx / dt
a = dv / dt
a = [tex]\frac{ d^2x}{dt^2}[/tex]
let's perform the derivative
v = - A w sin (wt + Ф)
a = - A w² cos wt + Ф)
substituting the values
a = - 2 1² cos (t-2)
for t = 3 s
a₃ = 2 cos (3-2)
remember angles are in radians
a₃ = -1.08 m/s²
To calculate kinetic energy, let's find the velocity for t = 3 s
v = - 2 sin (t-2)
v = -2 sin (3-2)
v = - 1.683 m / s
body mass is m = 1 kg
we calculate
K = ½ m v²
K = ½ 1 (-1.683) ²
K = 1.42 J
Assume you are going to race the three objects (hollow sphere, disk and ring used in Experiment 8.2) by releasing them from rest at the top of an inclined plane.Which object do you expect to reach the bottom of the inclined plan first? Why?
Answer:
a. The disk
b. Because it has the smallest rotational inertia
Explanation:
a. Which object do you expect to reach the bottom of the inclined plan first?
I would expect the disk to reach the bottom first.
b. Why?
This is because the disk has the smallest rotational inertia.
The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.
Since the three objects are rolling from the same height, they have the same mechanical energy.
But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy. The disk's smaller rotational kinetic energy will cause to rotate less but translate more than the other objects and thus reach the bottom first.
The object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
What is a moment of inertia?Moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
There are three objects, hollow sphere, disk and ring.
The moment of inertia of the hollow sphere object is given as,[tex]I=\dfrac{2}{3}mr^2[/tex]
The moment of inertia of the ring is,[tex]I=mr^2[/tex]
The moment of inertia of the disk is,[tex]I=\dfrac{1}{2}mr^2[/tex]Here, (m) is the mass and (r) is the radius of the object.
These three objects are going to race by releasing from rest at the top of an inclined plane to the bottom of the plane.
As moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
Thus the less the value of inertia will result in less the time required to reach at the bottom of the inclined plane.
Hence, the object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
Learn more about the force of inertia here;
https://brainly.com/question/10454047
The graph below shows the distance traveled by the skateboarder on each of the different road conditions. Using the graph, determine which of the roads was dry, wet, or muddy. Explain your answer using complete sentences.
Answer:
Road A- dry
Road B- mud
Road C- wet
Explanation:
Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.
The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.
The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.
