Answer:
correct answer is c
v = -0.8 m / s
Explanation:
This is a problem of quantity of movement, for this we must define a system formed by the two cars, so that the forces during the collision are internal and therefore the quantity of movement is conserved
initial
p₀ = m₁ v₁ - m₂ v₂
final
= (m₁ + m₂) v
We have taken the direction to the right as positive
p₀ =p_{f}
m₁ v₁ - m₂ v₂ = (m₁ + m₂) v
v = (m₁ v₁ - m₂ v₂) / (m₁ + m₂)
we calculate
v = (2 4 - 8 2) / (2 + 8)
v = (8 -16) / 10
v = -0.8 m / s
the negative sign indicates that the set is moving to the left
correct answer is c
A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Apply both the Heisenberg uncertainty principle and the equivalence of energy and mass to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.
Answer:
Δm Δt> h ’/ 2c²
Explanation:
Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions
ΔE Δt> h ’/ 2
h’= h / 2π
to relate this to the masses let's use Einstein's relationship
E = m c²
let's replace
Δ (mc²) Δt> h '/ 2
the speed of light is a constant that we can condense exact, so
Δm Δt> h ’/ 2c²
Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.
Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?
Answer:
λ1 = 0.0129m = 1.29cm
λ2 = 0.00923m = 0.92 cm
Explanation:
To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:
[tex]y_m=\frac{m\lambda D}{d}[/tex] (1)
m: order of the bright fringe = 1
λ: wavelength of the light = 660 nm, 470 nm
D: distance from the screen = 5.50 m
d: distance between slits = 0.280mm = 0.280 *10^⁻3 m
ym: height of the m-th fringe
You replace the values of the variables in the equation (1) for each wavelength:
For λ = 660 nm = 660*10^-9 m
[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]
For λ = 470 nm = 470*10^-9 m
[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]
Using the equation for the distance between fringes, Δy = xλ d , complete the following. (a) Calculate the distance (in cm) between fringes for 694 nm light falling on double slits separated by 0.0850 mm, located 4.00 m from a screen. cm (b) What would be the distance between fringes (in cm) if the entire apparatus were submersed in water, whose index of refraction is 1.333? cm
Answer:
Explanation:
Distance between fringe or fringe width = xλ / d
where x is location of screen and d is slit separation
Given x = 4 m
λ = 694 nm
d = .085 x 10⁻³ m
distance between fringes
= 4 x 694 x 10⁻⁹ / .085 x 10⁻³
= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶
= 32.66 x 10⁻³ m
= 32.66 mm .
3.267 cm
b )
when submerged in water , wavelength in water becomes as follows
wavelength in water = wave length / refractive index
= 694 / 1.333 nm
= 520.63 nm
new distance between fringes
3.267 / 1.333
= 2.45 cm .
How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?
Answer:
Explanation:
There's a formula for this:
[tex]F = k*displacement[/tex]
F being force, k being the spring constant, and displacement being the change in x
We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters
Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the spectator A is 0.55 s, and for the spectator B it is 0.45 s. Sight lines from the two spectators to the player kicking the ball meet at an angle of 90°. The speed of sound in the air is 343 m/s.
How far are (a) spectator A and (b) spectator B from the player?
(c) How far are the spectators from each other?
Answer:
a)188.65m
b)154.35m
c)243.7m
Explanation:
Given data:
[tex]t_A=0.55s[/tex]
[tex]t_B=0.45s[/tex]
(a) The distance from the kicker to each of the 2 spectators is given by:
[tex]d_A=v \times t_A[/tex]
where,
v= speed of sound
[tex]t_A[/tex]=time taken for the sound waves to reach the ears
[tex]d_A=343\times 0.55=188.65[/tex]m
(b)[tex]d_B=v \times t_B[/tex]
where,
v= speed of sound
[tex]t_B[/tex]=time taken for the sound waves to reach the ears
[tex]d_B=343\times 0.45=154.35m[/tex]
(c)As the angle b/w slight lines from the two spectators to the player is right angle,
hypotenuse=the distance b/w 2 spectators
and, the slight lines are the other 2 lines
[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]
02
Blue light has a frequency of about 7.5 x 1014 Hz. Calculate the energy, in Joules, of a single photon associated with this frequency
Answer:
49.725× 10^-24J
Explanation:
The Energy associated with a Photon us defined as;
E = hf
Where h is Planck's constant = 6.63× 10^-34m2kg/s
f is the frequency= 7.5 x 10^14 Hz
Hence
E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J
If a cart of 8 kg mass has a force of 16 newtons exerted on it, what is its acceleration?
Answer:
Explanation:
From Newton's 2nd Law,
F = m×a
Where F is Force
m is mass
a is acceleration
Hence a= F/m
a= 16/8= 2m/s2
determine the smallest mass of lead that when tied using a string to a wooden boat on a pond will be enough to sink the toy boat. assuming specific gravity of wood is 0.5 and density of water is 1000kg per cubic metre?
The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm1cm and the IMA of the machine is 66 , what is the radius of the handle
Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?
Answer:
Radius of the handle = 3 cm = 0.03 m
Explanation:
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)
Diameter of the rod = 1 cm
Radius of the rod = Diameter/2
Radius of the rod = 1/2
Radius of the rod = 0.5 cm
Mechanical Accuracy of the machine, MA = 6
Substitute the values into equation (1)
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
The Radius of the handle is = 3 cm = 0.03 m
Calculation of the radius of the handle:Since
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)
Here,
Diameter of the rod = 1 cm
We know that
The radius of the rod = Diameter/2
So,
Radius of the rod = 1/2
So,
Radius of the rod = 0.5 cm
Now
Mechanical Accuracy of the machine, MA = 6
Now
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
Learn more about radius here: https://brainly.com/question/18648019
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section), and the power fit of your Trendline equation,calculate the drag coefficient. Solve for it first (see video) and then plug in the values.
Answer:
The drag coefficient is [tex]D_z = 1.30512[/tex]
Explanation:
From the question we are told that
The density of air is [tex]\rho_a = 1.21 \ kg/m^3[/tex]
The diameter of bottom part is [tex]d = 0.15 \ m[/tex]
The power trend-line equation is mathematically represented as
[tex]F_{\alpha } = 0.9226 * v^{0.5737}[/tex]
let assume that the velocity is 20 m/s
Then
[tex]F_{\alpha } = 0.9226 * 20^{0.5737}[/tex]
[tex]F_{\alpha } = 5.1453 \ N[/tex]
The drag coefficient is mathematically represented as
[tex]D_z = \frac{2 F_{\alpha } }{A \rho v^2 }[/tex]
Where
[tex]F_{\alpha }[/tex] is the drag force
[tex]\rho[/tex] is the density of the fluid
[tex]v[/tex] is the flow velocity
A is the area which mathematically evaluated as
[tex]A = \pi r^2 = \pi \frac{d^2}{4}[/tex]
substituting values
[tex]A = 3.142 * \frac{(0.15)^2}{4}[/tex]
[tex]A = 0.0176 \ m^2[/tex]
Then
[tex]D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }[/tex]
[tex]D_z = 1.30512[/tex]
An aluminum wing on a passenger jet is 30 m long when its temperature is 27 C. At what temperature would the wing be 0.03 shorter?
Answer:2000
Explanation:
A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 0.9 m. What is the net force acting on a 65 kg driver who is driving at 18 m/sec and comes to rest in this distance
Answer:
11,700Newton
Explanation:
According to Newton's second law, Force = mass × acceleration
Given mass = 65kg.
Acceleration if the car can be gotten using one of the equation of motion as shown.
v² = u²+2as
v is the final velocity = 18m/s
u is the initial velocity = 0m/s
a is the acceleration
s is the distance travelled = 0.9m
On substitution;
18² = 0²+2a(0.9)
18² = 1.8a
a = 324/1.8
a = 180m/²
Net force acting on the body = 65×180
Net force acting on the body = 11,700Newton
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work
Answer:
Explanation:
The work required to push will be equal to work done by friction . Let d be the displacement required .
force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction
work done by force of friction
mg x μ x d = 660
80 x 9.8 x .272 x d = 660
d = 3 .1 m .
A silver rod having a length of 83.0 cm and a cross-sectional diameter of 2.40 cm is used to conduct heat from a reservoir at a temperature of 540 oC into an otherwise completely thermally insulated chamber that contains 1.43 kg of ice at 0 oC. How much time is required for the ice to melt completely
Answer:
3985 s or 66.42 mins
Explanation:
Given:-
- The length of the rod, L = 83.0 cm
- The cross sectional diameter of rod , d = 2.4 cm
- The temperature of reservoir, Tr = 540°C
- The amount of ice in chamber, m = 1.43 kg
- The temperature of ice, Ti = 0°C
- Thermal conductivity of silver, k = 406 W / m.K
- The latent heat of fusion of water, Lf = 3.33 * 10^5 J / kg
Find:-
How much time is required for the ice to melt completely
Solution:-
- We will first determine the amount of heat ( Q ) required to melt 1.43 kg of ice.
- The heat required would be used as latent heat for which we require the latent heat of fusion of ice ( Lf ). We will employ the first law of thermodynamics assuming no heat is lost from the chamber ( perfectly insulated ):
[tex]Q = m*L_f\\\\Q = ( 1.43 ) * ( 3.33 * 10 ^5 )\\\\Q = 476190 J[/tex]
- The heat is supplied from the hot reservoir at the temperature of 540°C by conduction through the silver rod.
- We will assume that the heat transfer through the silver rod is one dimensional i.e along the length ( L ) of the rod.
- We will employ the ( heat equation ) to determine the rate of heat transfer through the rod as follows:
[tex]\frac{dQ}{dt} = \frac{k.A.dT}{dx}[/tex]
Where,
A: the cross sectional area of the rod
dT: The temperature difference at the two ends of the rod
dx: The differential element along the length of rod ( 1 - D )
t: Time ( s )
- The integrated form of the heat equation is expressed as:
[tex]Q = \frac{k*A*( T_r - T_i)}{L}*t[/tex]
- Plug in the respective parameters in the equation above and solve for time ( t ):
[tex]476190 = \frac{406*\pi*0.024^2 * ( 540 - 0 ) }{0.83*4}*t \\\\t = \frac{476190}{119.49619} \\\\t = 3985 s = 66.42 mins[/tex]
Answer: It would take 66.42 minutes to completely melt the ice
A solid exerts a force of 500 N. Calculate the pressure exerted to the surface where area
of contact is 2000 cm2.
Answer:
2500 N/m²
Explanation:
Pressure: This can be defined as the force acting normally on a surface per unit area.
The expression for pressure is give as
P = F/A...................... Equation 1
Where P = pressure (N/m²), F = force (N), A = Contact area (m²)
Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.
Substitute into equation 1
P = 500/0.2
P = 2500 N/m²
Hence the pressure exerted to the surface is 2500 N/m²
Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is initially at rest. After the collision, it is observed that the ratio of the final kinetic energy system to its initial kinetic energy is KfK0= 1/6. Determine the ratio of the carts' masses, mBmA. (Assume the track is frictionless.)
Answer:
Explanation:
Initial kinetic energy of the system = 1/2 mA v0²
If Vf be the final velocity of both the carts
applying conservation of momentum
final velocity
Vf = mAvo / ( mA +mB)
kinetic energy ( final ) = 1/2 (mA +mB)mA²vo² / ( mA +mB)²
= mA²vo² / 2( mA +mB)
Given 1/2 mA v0² / mA²vo² / 2( mA +mB) = 6
mA v0² x ( mA +mB) / mA²vo² = 6
( mA +mB) / mA = 6
mA + mB = 6 mA
5 mA = mB
mB / mA = 5 .
You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 85 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
Force = 60.08 N
Explanation:
Given that
Diameter d = 30 mm
Holding pressure = 85 % of Atmospherics pressure
Solution
As we know that here 1 atm = 10⁵ N/m²
and pressure is known as force per unit area
pressure = [tex]\frac{F}{A}[/tex] ................1
put here value and we will get
F = [tex]0.85\times 10^5\times \frac{\pi}{4}\times 0.03^2\ N[/tex]
solve it we get
Force = 60.08 N
The Great Lakes are all part of what? The Mississippi River The St. Lawrence Seaway A large body of salt lakes The Missouri River
Answer:
St Lawrence Sea way
Explanation:
The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.
pls what is the difference between Ac power and dc power
Answer:
The difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction, or "forward." In AC, electrons keep switching directions, sometimes going "forward" and then going "backward."
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens would receive our signals at speed of:_______.
a. 0.99c
b. 1.10c
c. 1.00c
d. 0.90c
e. 0.10c
Answer:
3×10^7 m/s or 0.10c (e)
Explanation: If the actual value of the speed of light were to be put into consideration.
Given that the speed of light is c = 3.0×10^8m/s
The alien spaceship is approaching at the rate of 10% of the speed of light.
10% of 3.0×10^8m/s
10/100 × 3.0×10^8m/s
0.1 ×3.0×10^8m/s
3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c
Answer: 1.00c
Explanation: I got it correct on the homework
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). If a positron is accelerated by a constant electric field of magnitude 286 N/C, find the following.
(a) Find the acceleration of the positron. m/s2
(b) Find the positron's speed after 8.70 × 10-9 s. Assume that the positron started from rest. m/s
Answer:
a) a = 5.03x10¹³ m/s²
b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]
Explanation:
a) The acceleration of the positron can be found as follows:
[tex] F = q*E [/tex] (1)
Also,
[tex] F = ma [/tex] (2)
By entering equation (1) into (2), we have:
[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]
Where:
F: is the electric force
m: is the particle's mass = 9.1x10⁻³¹ kg
q: is the charge of the positron = 1.6x10⁻¹⁹ C
E: is the electric field = 286 N/C
[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]
b) The positron's speed can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed =?
[tex]V_{0}[/tex]: is the initial speed =0
t: is the time = 8.70x10⁻⁹ s
[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]
I hope it helps you!
A student drives 105.0 mi with an average speed of 61.0 mi/h for exactly 1 hour and 30
minutes for the first part of the trip. What is the distance in miles traveled during this
time?
Answer:
91.5 miles
Explanation:
61 miles per hour so 61(x amount of hours)
so 61 x 1.5 hours is 91.5 miles
Suppose the demand for air travel decreases (as illustrated in the graph below). A decrease in demand _____ the equilibrium price for air travel and _____ the equilibrium quantity for air travel. decreases, decreases increases, increases decreases, increases
Answer:
decreases, decreases
Explanation:
A decrease in the demand will create a fall in equilibrium prices and the quantity supplied will also decrease. As the equilibrium prices in the market are the price in which the quantity demanded equals to quantity supplied. If the demand for the air decreases then the quantity of the air travel will also decrease and thus when the supply and demand change so do the changes associated with the equilibrium prices.A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A. her moment of inertia decreases and her angular speed decreases B. her moment of inertia decreases and her angular speed increases C. her moment of inertia increases and her angular speed decreases D. her moment of inertia increases and her angular speed decreases E. her moment of inertia increases and her angular speed remains the same.
Answer:
C. her moment of inertia increases and her angular speed decreases
D. her moment of inertia increases and her angular speed decreases
Explanation:
The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.
The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.
This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.
When the person extends her arms, her moment of inertia increases and her angular speed decreases.
_____
Note to those looking for a letter answer
Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.
Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentum (in kg · m2/s) relative to the point (0, 5.0) meters?
Answer:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
Explanation:
In order to calculate the angular momentum of the particle you use the following formula:
[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex] (1)
r is the position vector respect to the point (0 , 5.0), that is:
r = 0m i + 5.0m j (2)
p is the linear momentum vector and it is given by:
[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex] (3)
the direction of p comes from the fat that the particle is moving along the i + j direction.
Then, you use the results of (2) and (3) in the equation (1) and solve for L:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
The angular momentum is -30 kgm^2/s ^k
Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m
Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]
Calculate the inlet and exit velocity of water jet
[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]
The conservation of mass equation of steady flow
[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]
[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]
[tex]A_e\ \texttt {is the exit area of the jet}[/tex]
since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal
The expression for thickness of the jet
[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]
(b)
[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]
[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]
[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
Two parallel plates having charges of equal magnitude but opposite sign are separated by 21.0 cm. Each plate has a surface charge density of 39.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density.
Answer:
E = 3.45*10^-19 N/C
Explanation:
a) The electric field between two parallel plates id given by the following formula:
[tex]E=\frac{\sigma}{\epsilon_o}[/tex] (1)
where:
σ: surface charge density of the plates = 39.0nC/m^2
εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2
You replace these values in the equation (1):
[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]
The electric field in between the parallel plates is 3.45*10^-19 N/C
g: To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
I =1.8 kgm^2
Explanation:
In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:
[tex]\tau=I\alpha[/tex] (1)
τ: torque applied to the door
I: moment of inertia of the door
α: angular acceleration = 5 rad/s^2
The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).
F = 10 N
d = 0.9 m
You replace the expression for τ, and solve for I:
[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]
The moment of inertia of the door is 1.8 kgm^2
To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
The moment of inertia is [tex]I = 1.8 \ kg m^2[/tex]
Explanation:
From the question we are told that
The force applied is [tex]F = 10 \ N[/tex]
The distance of the knob to the hinge is [tex]d = 0.9 \ m[/tex]
The angular acceleration is [tex]a = 5 \ rad/s[/tex]
The moment is mathematically represented as
[tex]I = \frac{d Fsin(\theta)}{a}[/tex]
Here [tex]\theta = 90^o[/tex] This is because the force direction is perpendicular to the plane of the door
substituting values
[tex]I = \frac{0.9 * 10 * sin (90)}{5}[/tex]
[tex]I = 1.8 \ kg m^2[/tex]
Two radio antennas A and B radiate in phase. Antenna B is a distance of 100 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 50.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
Required:
a. What is the longest wavelength for which there will be destructive interference at point Q?
b. What is the longest wavelength for which there will be constructive interference at point Q?
Answer:
a. 200 m
b. 100 m
Explanation:
Solution:-
- We will first draw three points marked A,B and Q from left most to right most.
- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).
- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:
AQ - BQ = n*λ/2
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ/2
- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ/2
λ = 200 m .... Answer
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:
AQ - BQ = n*λ
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ
- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ
λ = 100 m .... Answer