To develop this problem it is necessary to apply the concepts related to Broglie hypothesis.
The hypothesis defines that
[tex]\lambda = \frac{h}{p}[/tex]
Where,
P = momentum
h = Planck's constant
The momentum is also defined as,
P = mv
Where,
m = mass
v = Velocity
PART A) Replacing at the first equation
[tex]\lambda = \frac{h}{mv}[/tex]
Our values are given as,
[tex]h = 6.626*10^{-34}Js[/tex]
[tex]m = 65Kg[/tex]
[tex]\lambda = 0.76m[/tex]
Re-arrange to find v, we have:
[tex]v = \frac{h}{m\lambda}[/tex]
[tex]v = \frac{6.626*10^{-34}}{65*0.76}[/tex]
[tex]v = 1.341*10^{-35}m/s[/tex]
PART B) From the kinematic equations of movement description we know that velocity is defined as displacement over a period of time, that is
[tex]v = \frac{x}{t}[/tex]
Re-arrange to find t,
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{0.001}{ 1.341*10^{-35}}[/tex]
[tex]t = 7.455*10^{31}s[/tex]
[tex]7.455*10^{31} > 4*10^{17} \rightarrow[/tex]the age of the universe.
A small solid sphere and a small thin hoop are rolling along a horizontal surface with the same translational speed when they encounter a 20° rising slope. If these two objects roll up the slope without slipping,which will rise farther up the slope?
a. The sphere.
b. More information about the objects' mass and diameter is needed.
c. The hoop.
d. Both the same.
Answer:C.
Explanation: the hoop is the right answer
You (50 kg) are standing on a floating log (200 kg). Both are floating down a river at 1 m/s. The log points in the direction along the river. You walk in a direction that is down river for 5 seconds and your speed is 1.5 m/s as measured by a ground observer at the end of the 5 seconds.
a. What is the speed of the log at the end of the 5 seconds?
b. What is the average force between you and the log during those 5 seconds?
Answer:
0.875 m/s
5 N
Explanation:
[tex]m_1[/tex] = Mass of person = 50 kg
[tex]m_2[/tex] = Mass of log = 200 kg
[tex]v_1[/tex] = Velocity of person = 1.5 m/s
[tex]v_2[/tex] = Velocity of log
v = Velocity of log with respect to shore = 1 m/s
t = Time taken = 5 seconds
As the momentum of system is conserved we have
[tex](m_1+m_2)v=m_1v_1+m_2v_2\\\Rightarrow v_2=\frac{(m_1+m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(50+200)1-50\times 1.5}{200}\\\Rightarrow v_2=0.875\ m/s[/tex]
Velocity of the log at the end of the 5 seconds is 0.875 m/s
Force is given by
[tex]F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{50(1.5-1)}{5}\\\Rightarrow F=5\ N[/tex]
The average force between you and the log during those 5 seconds is 5 N
Which of the following is the reason why the magnetic flux through the bracelet is changing?
a. The magnitude of the magnetic field is changing.
b. The magnetic field is changing direction with respect to the bracelet.
c. The bracelet is moving in the magnetic field.
The magnetic flux through an object like a bracelet can change a) because the bracelet is moving in the magnetic field, field direction, or if the bracelet moves through the field. Examples include loops moving into or rotating within a magnetic field.
The reason why the magnetic flux through the bracelet is changing can be due to various factors. These include a change in the magnitude of the magnetic field, a) a change in the field's direction relative to the bracelet, or the bracelet moving through the magnetic field. Magnetic flux changes when a loop moves into a magnetic field or rotates within it. Specifically, in part (a) as the loop moves into the field, and in part (b) as the loop rotates, thus changing the orientation.
It is also indicated that the change in magnetic flux through a loop or coil can result from the spinning of a magnet nearby, causing the magnetic field within the coil to change rapidly, as described with a rotating magnet affecting coil current due to the variation in the number and direction of magnetic field lines passing through it.
The correct option is c. The bracelet is moving in the magnetic field.
Magnetic flux through a surface, such as a bracelet, is defined as the product of the magnetic field (B), the area of the surface (A), and the cosine of the angle (θ) between the magnetic field lines and the normal to the surface. Mathematically, this is expressed as:
[tex]\[ \Phi_B = B \cdot A \cdot \cos(\theta) \][/tex]
For the magnetic flux to change, one or more of the following must occur:
1. The magnitude of the magnetic field (B) changes.
2. The area of the surface (A) changes.
3. The orientation of the surface with respect to the magnetic field changes, i.e., the angle (θ) changes.
4. The surface moves within the magnetic field, changing the amount of field lines passing through it.
Given the options:
a. The magnitude of the magnetic field is changing. - This would indeed change the magnetic flux, but it is not one of the given scenarios.
b. The magnetic field is changing direction with respect to the bracelet. - This would also change the magnetic flux, as it would change the angle θ between the magnetic field lines and the normal to the surface of the bracelet. However, this option does not explicitly state that the direction of the magnetic field is changing; it only mentions the direction with respect to the bracelet, which could imply a change in orientation of the bracelet itself.
c. The bracelet is moving in the magnetic field. - This option indicates that the bracelet is changing its position within the magnetic field. As the bracelet moves, the amount of magnetic field lines passing through it can change, even if the magnetic field itself is uniform and constant. This movement can alter the effective area through which the field lines pass (A) and/or the angle (θ), thus changing the magnetic flux.
Since the question asks for the reason why the magnetic flux through the bracelet is changing, and option c directly describes a scenario where the bracelet's movement within the field would cause a change in flux, it is the correct answer. The other options either do not apply to the given scenario or are not explicitly described as occurring.
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the linear acceleration of the child?
Answer:
So the acceleration of the child will be [tex]8.05m/sec^2[/tex]
Explanation:
We have given angular speed of the child [tex]\omega =1.25rad/sec[/tex]
Radius r = 4.65 m
Angular acceleration [tex]\alpha =0.745rad/sec^2[/tex]
We know that linear velocity is given by [tex]v=\omega r=1.25\times 4.65=5.815m/sec[/tex]
We know that radial acceleration is given by [tex]a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2[/tex]
Tangential acceleration is given by
[tex]a_t=\alpha r=0.745\times 4.65=3.464m/sec^[/tex]
So total acceleration will be [tex]a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2[/tex]
The magnitude of the linear acceleration of the child is mathematically given as
a=8.05m/sec^2
The magnitude of the linear acceleration
Question Parameters:
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2.
Generally the equation for the linear velocity is mathematically given as
v=wr
Therefore
v=1.25*4.65
v=5.815
radial acceleration is given by
a=v^2/r
Hence
a=5.815/4.65
a=7.2718
Tangential acceleration is
a_t=\alpha r
a_t=0.745*4.65
a_t=3.464m/sec
Hence, total acceleration will be
[tex]a=\sqrt{7.2718^2+3.464^2}[/tex]
a=8.05m/sec^2
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A boy sits on a boat in a lake. In his boat he has a pile of rocks, each with mass 0.538 kg. He throws the rocks off the back of the boat at a rate of one rock every 0.964 sec, with a horizontal velocity of 11.6 m/s. What is the thrust on the boat from this "rock-throwing propulsion"? That is, what is the magnitude of the average force experienced by the boat during this process?
Answer:
6.47385 N
Explanation:
m = Mass of stone = 0.538 kg
t = Time taken to hit the rock = 0.964 seconds
v = Velocity of rock = 11.6 m/s
Force is given by
[tex]F=\frac{mv}{t}\\\Rightarrow F=\frac{0.538\times 11.6}{0.964}\\\Rightarrow F=\frac{6.2408}{0.964}\\\Rightarrow F=6.47385\ N[/tex]
The magnitude of the average force experienced by the boat during this process is 6.47385 N
The momentum of each thrown rock is 6.241 kg·m/s. The force or thrust on the boat is the rate of change of this momentum, which equates to an average force of 6.47 N.
Explanation:The topic of this question falls under the domain of Physics, specifically momentum. The phenomenon at hand is the principle of conservation of momentum which states that the momentum of an isolated system remains constant if no external forces are acting on it.
The momentum of each rock thrown is given by the product of its mass and its velocity, which is (0.538 kg)(11.6 m/s)= 6.241 kg·m/s. Given that a rock is thrown every 0.964 second, the rate of change of momentum, which is equal to the force, can be calculated by dividing the momentum of each rock by the time between each throw. So, the average force experienced by the boat is 6.241 kg·m/s / 0.964 s = 6.47 N (rounded to 2 decimal places).
Therefore, each time a rock is thrown out of the boat, there is an equal and opposite thrust on the boat which averages to a force of 6.47 N.
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A barge floating in fresh water (rho = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H = 2.0 m. When empty the bottom of the barge is located H0 = 0.55 m below the surface of the water. When fully loaded with coal the bottom of the barge is located H1 = 1.7 m below the surface.
Answer:
The buoyant force is 2964500 N.
Explanation:
Given that,
Density of water = 1000 kg/m³
Area = 550 m²
Height = 2.0 m
Depth = 0.55 m
Suppose we need to write an equation for the buoyant force on the empty barge in terms of the known data and find the value of force
We need to write the buoyant force equation
[tex]F_{b}=\rho\times V\times g[/tex]
The volume is
[tex]V=A\times h_{0}[/tex]
Put the value of volume into the buoyant force
[tex]F_{b}=\rho\times A\times h_{0}\times g[/tex]
We need to calculate the buoyant force
Put the value into the formula
[tex]F_{b}=1000\times550\times0.55\times9.8[/tex]
[tex]F_{b}=2964500\ N[/tex]
Hence, The buoyant force is 2964500 N.
The question describes a scenario involving Archimedes' Principle and the concept of density. It compares the submerged portions of a barge when it's empty versus when it's loaded. The difference in the submerged parts describes how the barge displaces more water to balance the increased weight.
Explanation:The phenomenon described in the question involves Archimedes' Principle and the concept of density, both of which are topics in Physics. The principle states that the buoyant force (upward force) that acts on an object submerged in a fluid is equal to the weight of the fluid the object displaces.
Initially, when the barge is empty, only a small part (H0 = 0.55 m) of it is submerged in the water, meaning it displaces a smaller volume of water equivalent to 0.55 m * 550 m². When it is loaded with coal, a larger part (H1 = 1.7 m) is submerged, and it displaces a larger volume of water (1.7 m * 550 m²).
The change in the submerged height of the barge when it's loaded compared to when it's empty can be attributed to increased weight which, to maintain equilibrium, is matched by the displacement of greater volume of water and hence increasing water's buoyant force.
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Case 1: A Styrofoam cup holds an unknown amount of lemonade (which is essentially water) at 20.5 °C. A 0.0550-kg ice cube at -10.2°C is placed in the lemonade. When thermal equilibrium is reached, all the ice has melted and the final temperature of the mixture is measured to be 11.8 °C. Assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. The latent heat of fusion for water is 3.4 X10 J/kg. The specific heat capacity for water and lemonade is the same, 4186 J/(kg °C). The specific heat capacity for ice is 2000 J/(kg °C). The cold ICE absorbs heat in 3 steps: Step 1: COLD ICE warms up from -10.2 °C. to 0.0 °C. (Remember: cold ice DOESN't Melt before reaching 0.0 °C.) Apply Q = mcAT, to the cold ice. Calculate the heat absorbed by the ICE in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Step 2: ICE at 0.0 °C melts into water at 0.0 °C. Calculate the heat absorbed by the ICE in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Step 3: The water at 0.0 °C from melted ice warms up to 11.8 °C. Apply Q = mcAT, to the water. Calculate the heat absorbed by water in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Calculate the total heat absorbed by ICE in the above 3 steps. Enter a number J Submit (5 attempts remaining) The warm lemonade releases heat in cooling down from 20.5 °C to 11.8 °C. Apply Q = mcAT, to the lemonade. Keep mass m as "space hoder. Keep all heat as POSITIVE. Let Q released by lemonade = Total heat absorbed by ICE Find the mass of the lemonade. Keep 2 decimal places. Enter a number kg Submit (5 attempts remaining)
Answer:
[tex]m_l=0.619\ kg[/tex]
Explanation:
Given:
initial temperature of water(lemonade), [tex]T_{il}=20.5^{\circ}C[/tex]mass of ice, [tex]m=0.055\ kg[/tex]initial temperature of ice, [tex]T_{ii}=-10.2^{\circ}C[/tex]final temperature of the mixture, [tex]T_f=11.8^{\circ}C[/tex]specific heat capacity of ice, [tex]c_i=2000\ J.kg^{-1}. ^{\circ}C^{-1}[/tex]specific heat capacity of water, [tex]c_w=4186\ J.kg^{-1}. ^{\circ}C^{-1}[/tex]Latent heat of fusion of ice, [tex]L=340000\ J.kg^{-1}[/tex]For the whole ice to melt in lemonade and result a temperature of 11.8°C the total heat lost by the lemonade will be equal to the total heat absorbed by the ice to come to 0°C from -10.2°C along with the latent heat absorbed in the melting of ice at 0°C and the heat absorbed by the ice water of 0°C to reach a temperature of 11.8°C.
Now, mathematically:
[tex]Q_l=Q_i+Q_m+Q_w[/tex]
[tex]m_l.c_w.\Delta T_l=m_i.c_i.\Delta T_i_i+m_i.L+m_i.c_w.\Delta T_w[/tex]
[tex]m_l.c_w.(T_{il}-T_f)=m_i(c_i.\Delta T_i_i+L+c_w.\Delta T_w)[/tex]
[tex]m_l\times 4186\times (20.5-11.8)=0.055(2000\times (0-(-10.2))+340000+4186\times (11.8-0))[/tex]
[tex]m_l=0.619\ kg[/tex] (mass of lemonade)
A rectangular box with a volume of 784ft^3 is to be constructed with a square base and top. The cost per square foot for the bottom is 20 cents, for the top is 16 cents, and for the sides is 1.5 cents. What dimensions will minimize the cost? The length of one side of the base is ? The height of the box is?
Let's dimension the horizontal length with the name X and the vertical dimensions as Y.
In this way the total volume will be given under the function
[tex]V = x^2 y[/tex]
The cost for the bottom is given by [tex](x^2)(20)=20x^2[/tex]
While the cost for performing the top by [tex](x^2)(16) = 16x^2[/tex]
The cost for performing the sides would be given by [tex](1.5)(xy)(4) = 6xy[/tex]
Therefore the total cost would be
[tex]c_{total}= 36x^2 +6xy[/tex]
The total volume is equivalent to
[tex]784 = x^2y[/tex]
[tex]xy = \frac{784}{x}[/tex]
Replacing in our cost function
[tex]c_{total}= 36x^2 +6xy[/tex]
[tex]c_{total}= 36x^2 +6(\frac{784}{x})[/tex]
Obtaining the first derivative and equalizing to zero we will obtain the ideal measure, therefore
[tex]c' = 0[/tex]
[tex]c' = 72x-\frac{4707}{x^2}[/tex]
[tex]0= 72x-\frac{4707}{x^2}[/tex]
[tex]x = (\frac{523}{2})^{1/3}[/tex]
[tex]x = 6.3947ft[/tex]
Then,
[tex]784 = x^2y[/tex]
[tex]y = \frac{784}{x^2}[/tex]
[tex]y = \frac{784}{6.3947^2}[/tex]
[tex]y = 19.17ft[/tex]
In this way the measures of the base should be 6.3947ft (width and length) and the height of 19.17ft.
A golf ball and an equal-mass bean bag are dropped from the same height and hit the ground. The bean bag stays on the ground while the golf ball rebounds. Which experiences the greater impulse from the ground?
a. The golf ball.
b. Not enough information.
c. The bean bag.
d. Both the same.
Answer:
option A
Explanation:
The correct answer is option A
given,
mass of ball and bean bag is same
golf ball rebound with certain velocity where as the bean bag stops.
Impulse = Change in momentum
I = m v_f - m v_i
When the golf ball rebound there will be negative velocity on the ball.
Which will add up and increase the impulse of the golf ball.
But in the case of beam bag velocity after the collision is zero the impulse will be less.
The golf ball experiences the greater impulse from the ground.
What is impulse?Impulse refers to the force acting over time to change the momentum of an object. It is represented by J and usually expressed in Newton-seconds or kg m/s so we can conclude that the golf ball experiences the greater impulse from the ground because of the structure and elasticity.
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A 7.1 kg watermelon is placed at one end of a 4.8 m, 260 N scaffolding supported by two cables. One supporting cable is at the opposite end of the scaffolding, and the other is 0.53 m from the watermelon. How much tension is in the cable at the end of the scaffolding? The acceleration of gravity is 9.8 m/s 2 . Answer in units of N
The calculated tension is in the cable at the end of the scaffolding is 208.94 N.
First, let's calculate the weights acting on the system: the weight of the watermelon (Wm) and the weight of the scaffolding (Ws). Wm = mass of the watermelon × gravity = 7.1 kg × 9.8 m/s2 = 69.58 N, and Ws = 260 N given. Next, we choose the pivot point at the end of the scaffolding opposite the watermelon to find the tension in the cable at that end. The distance from the watermelon to the supporting cable is (4.8 m - 0.53 m) = 4.27 m, and the distance from the pivot to the scaffold's center of mass (assuming it's uniform) is half its length, or 2.4 m.
Applying the equilibrium condition for torques (Τclockwise = Τcounterclockwise), we have:
(69.58 N × 4.27 m) + (260 N × 2.4 m) = T × 4.8 m
⇒ T = ((69.58 N × 4.27 m) + (260 N × 2.4 m)) / 4.8 m
⇒ T = 208.94 N.
The tension in the cable at the end of the scaffolding where the watermelon is placed is [tex]{34.79 \text{ N}} \)[/tex].
Given:
- Mass of the watermelon, [tex]\( m = 7.1 \)[/tex] kg
- Acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s²
- Length of the scaffolding, [tex]\( L = 4.8 \)[/tex] m
- Distance of the watermelon from one end, [tex]\( d = 0.53 \)[/tex] m
1. Calculate the weight of the watermelon:
[tex]\[ W = mg = 7.1 \times 9.8 = 69.58 \text{ N} \][/tex]
2. Determine the tensions in the cables:
Since the system is in equilibrium, the tension [tex]\( T_1 \)[/tex] at the end where the watermelon is placed and the tension [tex]\( T_2 \)[/tex] at the opposite end satisfy:
[tex]\[ T_1 + T_2 = W \][/tex]
3. Find [tex]\( T_2 \)[/tex]:
From the equilibrium condition:
[tex]\[ T_2 = \frac{W}{2} = \frac{69.58}{2} = 34.79 \text{ N} \][/tex]
4. Calculate [tex]\( T_1 \)[/tex]:
Since [tex]\( T_1 + T_2 = W \)[/tex] and [tex]\( T_1 = -T_2 \)[/tex] (because [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] act in opposite directions):
[tex]\[ T_1 = W - T_2 = 69.58 - 34.79 = 34.79 \text{ N} \][/tex]
Therefore, the tension in the cable at the end of the scaffolding where the watermelon is placed is [tex]{34.79 \text{ N}} \)[/tex].
Oil at 150°C flows slowly through a long, thin-walled pipe of 25-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20°C and the convection coefficient at the outer tube surface is 11 W/m2·K. Estimate the heat loss per unit length of tube, in W/m.
Final answer:
The heat loss per unit length of the tube due to convection is approximately 141.7 W/m, calculated using the formula for heat transfer by convection with the provided temperatures and convection coefficient.
Explanation:
To estimate the heat loss per unit length of tube in W/m due to convection, we use the formula for heat transfer by convection which is Q = hA(Tsur - Tair). The convection coefficient 'h' is given as 11 W/m²·K, the temperature of the oil 'Tsur' is 150°C, and the air temperature 'Tair' is 20°C.
We need to find the surface area 'A' which for a pipe is calculated using A = π x d x L, where 'd' is the diameter and 'L' is the length of the pipe. Since we need the heat loss per unit length, L will be 1 meter.
Firstly, calculate the surface area per meter of the pipe: A = π x 0.025 m x 1 m = 0.0785 m². Then, plug these values into the convection formula to find heat loss per meter: Q = 11 W/m²·K x 0.0785 m² x (150°C - 20°C) = 141.7 W/m. Therefore, the heat loss per unit length of the tube is approximately 141.7 W/m.
A 100 kg object hangs from two steel cables, both with radius 1.2 mm. The first cable is 2.5 m long and 2 mm shorter than the second, but the object is horizontal (so they are now the same length). What is the force on the first cable? Young's Modulus for steel is 2.0 x 1011 N/m2.A :470 NB :500 NC :850 ND :1000 N
Answer:
850N
Explanation:
The step by step is in the attachment.
Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the text, find the inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 4 cm.
Answer:
L = 0.109 H
Explanation:
Given that,
Number of loops in the solenoid, N = 1500
Radius of the wire, r = 4 cm = 0.04 m
Length of the rod, l = 13 cm = 0.13 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]
[tex]L=\dfrac{4\pi \times 10^{-7}\times (1500)^2\times \pi (0.04)^2}{0.13}[/tex]
L = 0.109 H
So, the self inductance of the solenoid is 0.109 henries.
An electromagnet produces a magnetic field of 0.520 T in a cylindrical region of radius 2.40 cm between its poles. A straight wire carrying a current of 10.5 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field.
What magnitude of force is exerted on the wire?
Answer:
Magnetic force, F = 0.262 N
Explanation:
It is given that,
Magnetic field of an electromagnet, B = 0.52 T
Length of the wire, [tex]l = 2r =2\times 2.4=4.8\ cm=0.048\ m[/tex]
Current in the straight wire, i = 10.5 A
Let F is the magnitude of force is exerted on the wire. The magnetic force acting on an object of length l is given by :
[tex]F=ilB[/tex]
[tex]F=10.5\ A\times 0.048\ m\times 0.52\ T[/tex]
F = 0.262 N
So, the magnitude of force is exerted on the wire is 0.262 N.
In what phase of matter do molecules vibrate but stay in position?
A. gas
B.liquid
c.solid
d.plasma
Answer:
solid
Explanation:
they molecules are closely together
A toy train of m=0.60 kg moves at 20m/s along a straight track. It bumps into another train of M=1.5kg moving in the same direction. They stick together and continue on the track at a speed 12 m/s. What was the speed in m/s of the second train just before the collision?
Answer:8.8 m/s
Explanation:
Given
mass of train [tex]m_1=0.6 kg[/tex]
velocity of Train [tex]v_1=20 m/s[/tex]
mass of another train [tex]m_2=1.5 kg[/tex]
Final velocity of both train [tex]v=12 m/s[/tex]
Let [tex]v_2[/tex] be the velocity of [tex]m_2[/tex] before collision
As external Force is zero therefore change in momentum is zero
conserving Momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]0.6\times 20+1.5\times v_2=(0.6+1.5)\cdot 12[/tex]
[tex]1.5\times v_2=25.2-12[/tex]
[tex]v=\frac{13.2}{1.5}=8.8 m/s[/tex]
Young's double slit experiment is one of the classic tests for the wave nature of light. In an experiment using red light (λ = 641 nm) the second dark fringe on either side of the central maximum is θ = 4.4 degrees relative to the central bright fringe.
(a) Write an expression for the separation distance between the slits.
Answer:
The separation distance between the slits is 16710.32 nm.
Explanation:
Given that,
Wavelength = 641 nm
Angle =4.4°
(a). We need to calculate the separation distance between the slits
Using formula of young's double slit
[tex]d\sin\theta=m\lambda[/tex]
[tex]d=\dfrac{m\lambda}{\sin\theta}[/tex]
Where, d = the separation distance between the slits
m = number of order
[tex]\lambda[/tex] =wavelength
Put the value into the formula
[tex]d=\dfrac{2\times641\times10^{-9}}{\sin4.4}[/tex]
[tex]d=0.00001671032\ m[/tex]
[tex]d=16710.32\ nm[/tex]
Hence, The separation distance between the slits is 16710.32 nm.
The separation distance between the slits in Young's double slit experiment can be calculated using the formula d = λx / sin(θ), where d is the separation distance, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen.
Explanation:In Young's double slit experiment, the separation distance between the slits can be determined using the formula:
d = λx / sin(θ)
Where d is the separation distance between the slits, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen. In this example, the second dark fringe on either side of the central maximum is given as θ = 4.4 degrees. To calculate the separation distance between the slits, we also need to know the wavelength of the red light, which is given as λ = 641 nm.
Using the formula, we have:
d = (641 nm) * x / sin(4.4 degrees)
To find the value of d, we need to know the value of x, which is the distance between the fringe and the screen.
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What is the linear size of the smallest box in which you can confine an electron if you want to know for certain that the electron's speed is no more than 13 m/s ? Express your answer to two significant figures and include the appropriate units.
Using the Heisenberg Uncertainty Principle in quantum mechanics, which states that position and momentum of a particle cannot both be precisely measured at the same time, we can calculate the smallest box in which an electron can be confined with a specified maximum speed. The mass of the electron and the given speed are used to compute the uncertainty in momentum, which is then used in the Heisenberg inequality formula to find the size of the box.
Explanation:This question can be addressed using the Heisenberg Uncertainty Principle, which in quantum mechanics, states that the position and the momentum of a particle cannot both be precisely measured at the same time. The more precisely we know the position (Δx), the less precisely we can know the velocity (Δv), and vice versa.
Considering an electron, which has a mass (m) of 9.11×10-31 kg, and wanting the speed to be no more than 13 m/s, we can use the Heisenberg inequality formula:
Δx* Δp ≥ ℏ/2,
where ℏ is the reduced Planck constant, and Δp is the uncertainty in momentum. Since momentum (p) is equal to the mass (m) times the velocity (v), we find the uncertainty in momentum (Δp) to be m* Δv which is (9.11×10-31 kg * 13 m/s). Now, we can solve for the smallest uncertainty in position (Δx), also known as the linear size of the box.
Using these calculations, we must first find Δp, then substitute into the above inequality and solve for Δx. This will give you the minimum size of the box in which the electron can be confined.
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A 20-kg crate is sitting on a frictionless ice rink. A child standing at the side of the ice rink uses a slingshot to launch a 1.5-kg beanbag at the crate. Assume that the beanbag strikes the crate horizontally in the +x direction with a speed of 10 m/s. The beanbag bounces straight back in the -x direction with a speed of 6 m/s. Using conservation of momentum, calculate the speed of the crate after the beanbag hits it.
Answer:
The speed of the crate after the beanbag hits it is 1.2 m/s.
Explanation:
Hi there!
The momentum of the system beanbag-crate remains the same after the collision, i.e., the momentum of the system is conserved. The momentum of the system is calculated by adding the momenta of each object. So, the initial momentum (before the collision) is calculated as follows:
initial momentum = momentum of the crate + momentum of the beanbag
initial momentum = mc · vc + mb · vb
Where:
mc = mass of the crate.
vc = initial velocity of the crate.
mb = mass of the beanbag
vb = initial mass of the beanbag
With the data we have, we can calculate the initial momentum:
initial momentum = 20 kg · 0 m/s + 1.5 kg · 10 m/s = 15 kg · m/s
Now, let´s write the equation of the momentum of the system after the collision:
final momentum = mc · vc´ + mb · vb´
Where vc´ and vb´ are the final velocity of the crate and the beanbag respectively. Let´s replace with the data we have:
final momentum = 20 kg · vc´ + 1.5 · (-6 m/s)
Since
initial momentum = final momentum
Then:
15 kg · m/s = 20 kg · vc´ + 1.5 kg · (-6 m/s)
Solving for vc´:
(15 kg · m/s + 9 kg · m/s) / 20 kg = vc´
vc´ = 1.2 m/s
The speed of the crate after the beanbag hits it is 1.2 m/s.
Using the law of conservation of momentum, the speed of the crate after the beanbag hits it is found to be 1.2 m/s in the +x direction.
Explanation:The situation described can be solved using the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision. The momentum (p) of an object is its mass (m) times its velocity (v), or p = mv.
Before the collision, the 1.5-kg beanbag has a momentum of 1.5 kg * 10 m/s = 15 kg*m/s in the +x direction, and the 20-kg crate has a momentum of 0, as it is at rest. Therefore, the total momentum before the collision is 15 kg*m/s.
After the collision, the beanbag has a momentum of 1.5 kg * -6 m/s = -9 kg*m/s in the -x direction. Since the total momentum must be conserved, the crate must have a momentum of 15 kg*m/s (total initial momentum) + 9 kg*m/s (final momentum of beanbag) = 24 kg*m/s in the +x direction. Therefore, the speed (v) of the crate after the collision is its momentum divided by its mass, or v = p/m = 24 kg*m/s / 20 kg = 1.2 m/s.
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A tall glass cylinder is filled with a layer of water 17.0 cm deep, and floating on top of the water, a layer of oil 34.0 cm thick. The oil has a specific gravity of 0.900. What is the absolute pressure (in Pa) at the bottom of the cylinder? (Assume the atmospheric pressure is 1.013 ✕ 105 Pa. Round your answer to at least three significant figures.)
To develop this problem it is necessary to apply the law of Pascal. The pressure exerted on an incompressible and equilibrium fluid within a container of non-deformable walls is transmitted with equal intensity in all directions and at all points of the fluid. For which the pressure is defined as,
[tex]P = P_{atm}+P_{oil}+P_{water}[/tex]
The pressure of an object can be expressed by means of density, gravity and height
[tex]P = \rho*g*h[/tex]
Our values are given as,
[tex]g=9.8m/s^2\\h_w = 0.17m\\h_o = 0.34\\\gamma_g = 0.9\rightarrow \rho=0.9*10^{-3}[/tex]
Replacing we have to,
[tex]P = P_{atm}+P_{oil}+P_{water}[/tex]
[tex]P = P_{atm}+\rho_{oil}gh_{oil}+\rho_{water}gh_{water}[/tex]
[tex]P = 1.013*10^5+(0.9*10^3)(9.8)(0.34)+(10^3)(9.8)(0.17)[/tex]
[tex]P = 105964.8Pa[/tex]
A 0.01-kg object is initially sliding at 9.0 m/s. It goes up a ramp (increasing its elevation by 1.5 m), and then moves horizontally before striking a spring of force constant k = 100 N/m. The spring is compressed by 5.0 cm as it completely stops the object.
How much heat energy was created during this motion?
Answer:
During this motion, 0.133 J of heat energy was created
Explanation:
Hi there!
Let´s calculate the energy of the object in each phase of the motion.
At first, the object has only kinetic energy (KE):
KE = 1/2 · m · v²
Where:
m = mass of the object.
v = velocity.
KE = 1/2 · 0.01 kg · (9 m/s)²
KE = 0.405 J
When the object goes up the ramp, it gains some gravitational potential energy (PE). Due to the conservation of energy, the object must convert some of its kinetic energy to obtain potential energy. By calculating the potential energy that the object acquires, we can know the loss of kinetic energy:
PE = m · g · h
Where:
m = mass of the object.
g = acceleration due to gravity (9.81 m/s²)
h = height.
PE = 0.01 kg · 9.81 m/s² · 1.5 m
PE = 0.147 J
The object "gives up" 0.147 J of kinetic energy to be converted into potential energy.
Then, after going up the ramp, the kinetic energy of the object will be:
0.405 J - 0.147 J = 0.258 J
When the object reaches the spring, kinetic energy is used to compress the spring and the object obtains elastic potential energy (EPE). Let´s calculate the EPE obtained by the object:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = compression of the spring
EPE = 1/2 · 100 N/m · (0.05 m)² = 0.125 J
Then, only 0.125 J of kinetic energy was converted into elastic potential energy. The object is at rest at the end of the motion, i.e., the object does not have kinetic energy when it compresses the spring by 5.0 cm. Since energy can´t be lost, the rest of the kinetic energy, that was not used to compress the spring, had to be converted into heat energy:
Heat energy = initial kinetic energy - obtained elastic potential energy
Heat energy = 0.258 J - 0.125 J = 0.133 J
During this motion, 0.133 J of heat energy was created.
Final answer:
The question asks to calculate the heat energy produced when a 0.01-kg object sliding at 9.0 m/s is stopped by a spring after going up a ramp. Using conservation of energy, the heat energy can be found by subtracting the spring's elastic potential energy from the total initial kinetic and gravitational potential energies.
Explanation:
The student's question involves determining the amount of heat energy created during the motion of a 0.01-kg object as it slides, goes up a ramp, and is stopped by a compressed spring. To solve this problem, we need to apply the principles of conservation of energy and mechanics.
Initially, the object has kinetic energy since it is sliding at 9.0 m/s. As it goes up a ramp with a height increase of 1.5 m, it gains gravitational potential energy and loses some kinetic energy. Finally, when it strikes and compresses the spring (of constant k = 100 N/m) by 5.0 cm, it loses all its kinetic energy, which gets converted into the energy stored in the spring (elastic potential energy) and heat energy due to non-conservative forces.
The formula to find the elastic potential energy (Ep) stored in a compressed spring is:
Ep = 1/2 * k * x2
Where k is the spring constant and x is the compression length. Since the initial kinetic energy and the gravitational potential energy are converted into the elastic potential energy and heat, we can find the heat energy by subtracting the elastic potential energy from the sum of the initial kinetic and gravitational potential energies.
Heat energy = Initial kinetic energy + Gravitational potential energy - Elastic potential energy
A glider of mass 5.0 kg hits the end of a horizontal rail and bounces off with the same speed, in the opposite direction. The collision is elastic and takes place in a time interval of 0.2s, with an average force of 100N. What was the speed, in m/s, of the glider
To solve the exercise it is necessary to apply the concepts given in Newton's second law and the equations of motion description.
Let's start by defining acceleration based on speed and time, that is
[tex]a = \frac{v}{t}[/tex]
On the other hand according to Newton's second law we have to
F=ma
where
m= Mass
a = Acceleration
Replacing the value of acceleration in this equation we have
[tex]F=m(\frac{v}{t})[/tex]
Substituting with our values we have
[tex]100N=(5Kg)\frac{v}{0.2}[/tex]
Re-arrange to find v
[tex]v=\frac{100*0.2}{5}[/tex]
[tex]v = 2m/s[/tex]
Therefore the speed of the glider is 2m/s
When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a single rectangular region of the viewing screen is illuminated (matching the shape of the thin rectangular slit).
O True O False
True
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A hammer taps on the end of a 3.4-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 8.40 ms .What is the speed of sound in this metal?
Answer:
S = 2266.67 m/s
Explanation:
Given,
length of the metal = 3.4 m
pulses are separated in time = 8.4 ms
speed of sound in air= 343 m/s
speed of sound in this metal = ?
time taken
[tex]t = \dfrac{distance}{speed}[/tex]
[tex]t = \dfrac{3.4}{343}[/tex]
t = 9.9 ms
speed of sound in the metal is fast
t = 9.9 - 8.4 = 1.5 ms
time for which sound is in metal is equal to 1.5 ms
speed of sound in metal
[tex]speed= \dfrac{distance}{time}[/tex]
[tex]S = \dfrac{3.4}{1.5 \times 10^{-3}}[/tex]
S = 2266.67 m/s
Speed of sound in metal is equal to S = 2266.67 m/s
Final answer:
The speed of sound in the metal is approximately 404.76 m/s.
Explanation:
To determine the speed of sound in the metal, we need to use the time difference between the two pulses and the length of the metal bar. The time difference between the pulses is 8.40 ms and the length of the bar is 3.4 m. The speed of sound in the metal can be calculated using the formula: speed = distance / time. In this case, the distance is the length of the bar and the time is the time difference between the pulses.
So, by substituting the values into the formula, we get: speed = 3.4 m / (8.40 * 10^-3 s).
Simplifying the equation, we get: speed = 404.76 m/s.
Therefore, the speed of sound in the metal is approximately 404.76 m/s.
In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the:
a. +x direction decreasing in speed.
b. -x direction at a constant 20 m/s.
c. -x direction decreasing in speed.
d. +x direction increasing in speed.
e. -x direction increasing in speed.
Answer:
option C
Explanation:
The correct answer is option C.
For negative velocity
Negative velocity is when the velocity of the object in a negative direction.
so, for the velocity of the car to be negative, it should be in -x-direction.
For positive acceleration
Positive acceleration is the change of the velocity of the object with respect to time in a positive direction
but when the velocity of the object is in negative direction the slowing down of the vehicle will give positive acceleration.
hence, the correct answer is an option -x-direction decreasing in speed.
A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. When viewed from the air in the direction normal to the surface, there is constructive interference of reflected red light with l = 675 nm. Find the minimum thickness t of the oil film that will give constructive interference of reflected light with l = 675 nm.
Answer:
t = 120.5 nm
Explanation:
given,
refractive index of the oil = 1.4
wavelength of the red light = 675 nm
minimum thickness of film = ?
formula used for the constructive interference
[tex]2 n t = (m+\dfrac{1}{2})\lambda[/tex]
where n is the refractive index of oil
t is thickness of film
for minimum thickness
m = 0
[tex]2 \times 1.4 \times t = (0+\dfrac{1}{2})\times 675[/tex]
[tex]t = \dfrac{0.5\times 675}{2\times 1.4}[/tex]
t = 120.5 nm
hence, the thickness of the oil is t = 120.5 nm
Final answer:
The minimum thickness t of the oil film that will give constructive interference of reflected red light with a wavelength of 675 nm is 202.5 nm.
Explanation:
Constructive interference of reflected light occurs when the path difference between the two waves is equal to an integer multiple of the wavelength. In this case, we have constructive interference for red light with a wavelength of 675 nm.
The path difference can be represented as 2nt, where n is the index of refraction of the oil and t is the thickness of the oil film. Since we are viewing the film from air, we need to consider the difference in indices of refraction.
For constructive interference, the path difference must be equal to mλ, where m is the order of the interference. In this case, we have m = 1.
Using the formula for the path difference, we can set up the following equation:
2nt = (m + 1/2)λ
Plugging in the values, we have:
2(1.4)(t) = (1 + 1/2)(675 nm)
Simplifying, we get:
t = 202.5 nm
Therefore, the minimum thickness t of the oil film that will give constructive interference of reflected red light with a wavelength of 675 nm is 202.5 nm.
A rock of mass m = 0.0450 kg is attached to one end of a string and is whirled around in a horizontal circle. If the radius of the circle is 0.580 m and the angular speed is 2.34 rad/s. What is the tension in the string?
Answer:
Tension, T = 0.1429 N
Explanation:
Given that,
Mass of the rock, m = 0.0450 kg
Radius of the circle, r = 0.580 m
Angular speed, [tex]\omega=2.34\ rad/s[/tex]
The tension in the string is balanced by the centripetal force acting on it. It is given by :
[tex]T=\dfrac{mv^2}{r}[/tex]
Since, [tex]v=r\omega[/tex]
[tex]T=\dfrac{m(r\omega)^2}{r}[/tex]
[tex]T=\dfrac{0.0450\times (0.580\times 2.34)^2}{0.580}[/tex]
T = 0.1429 N
So, the tension in the string is 0.1429 N. Hence, this is the required solution.
Plane microwaves are incident on a thin metal sheet that has a long, narrow slit of width 4.8 cm in it. The microwave radiation strikes the sheet at normal incidence. The first diffraction minimum is observed at θ = 42°. What is the wavelength of the microwaves?
Answer:
[tex] \lambda = 3.21 \ cm[/tex]
Explanation:
given,
width of narrow slit = 4.8 cm
minimum angle of diffraction = θ = 42°
wavelength of the microwave = ?
condition for the diffraction for single slit diffraction
[tex]d sin \theta = m \lambda[/tex]
for the first minima m = 1
[tex]d sin \theta = \lambda[/tex]
wavelength of microwave radiation is equal to
[tex] \lambda = d sin \theta[/tex]
[tex] \lambda = 4.8\times sin 42^0[/tex]
[tex] \lambda = 3.21 \ cm[/tex]
the wavelength of microwaves is equal to [tex] \lambda = 3.21 \ cm[/tex]
The wavelength of the microwaves is approximately 0.032 m or 3.2 cm.
To find the wavelength of the microwaves given the slit width and the angle of the first diffraction minimum, you can use the diffraction formula for a single slit:
[tex]\[a \sin \theta = m \lambda\][/tex]where:
[tex]\(a\)[/tex] is the slit width,[tex]\(\theta\)[/tex] is the angle of the first diffraction minimum,[tex]\(m\)[/tex] is the order of the minimum for the first minimum, [tex]\(m = 1\)[/tex],[tex]\(\lambda\)[/tex] is the wavelength of the microwaves.Given:
[tex]\(a = 4.8 \text{ cm} = 0.048 \text{ m}\)[/tex][tex]\(\theta = 42^\circ\)[/tex][tex]\(m = 1\)[/tex]Rearrange the formula to solve for the wavelength [tex]\(\lambda\)[/tex]:
[tex]\[\lambda = \frac{a \sin \theta}{m}\][/tex]Plug in the values:
[tex]\[\lambda = \frac{0.048 \text{ m} \times \sin 42^\circ}{1}\][/tex]First, calculate [tex]\(\sin 42^\circ\)[/tex]:
[tex]\[\sin 42^\circ \approx 0.6691\][/tex]Now compute the wavelength:
[tex]\[\lambda = 0.048 \text{ m} \times 0.6691 \approx 0.032 \text{ m}\][/tex]So, the wavelength of the microwaves is approximately [tex]\(0.032 \text{ m}\) or \(3.2 \text{ cm}\)[/tex].
7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g of water at 22.7oC. When the two substances reach thermal equilibrium, the final temperature is 24.2oC. What is the mass of the copper block?
Answer:
37.34372 kg
Explanation:
m = Mass
[tex]\Delta T[/tex] = Change in temperature
1 denotes water
2 denotes copper
c = Heat capacity
Heat is given by
[tex]Q=mc\Delta T[/tex]
In this case the heat transfer will be equal
[tex]m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg[/tex]
Mass of copper block is 37.34372 kg
A bar on a hinge starts from rest and rotates with an angular acceleration α=(10+6t) rad/s^2, where t is in seconds. Determine the angle in radians through which the bar turns in the first 4.00s.
Answer:
[tex]\theta=144\ rad[/tex]
Explanation:
given,
α=( 10+6 t ) rad/s²
[tex]\alpha =\dfrac{d\omega}{dt}[/tex]
[tex]d\omega= \alpha dt[/tex]
integrating both side
[tex]\omega= \int (10+6 t )dt[/tex]
[tex]\omega=10 t+6\dfrac{t^2}{2}[/tex]
[tex]\omega=10 t+3t^2[/tex]
we know
[tex]\omega =\dfrac{d\theta}{dt}[/tex]
[tex]d\theta= \alpha dt[/tex]
integrating both side
[tex]\theta= \int (10 t+3t^2 )dt[/tex]
[tex]\theta=10\dfrac{t^2}{2}+3\dfrac{t^3}{3}[/tex]
[tex]\theta=5 t^2+t^3[/tex]
now, at t = 4 s θ will be equal to
[tex]\theta=5\times 4^2+4^3[/tex]
[tex]\theta=144\ rad[/tex]