Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
V is the common velocity after the collisions
[tex]V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s[/tex]
So, the two car train will move with a common velocity of 6.79 m/s.
1. What is the longest-wavelength EM radiation that can eject a photoelectron from silver, given that the binding energy is 4.73 eV? Is this in the visible range?
2. Find the longest-wavelength photon that can eject an electron from potassium, given that the binding energy is 2.24 eV. Is this visible EM radiation?
Answer:
1. λ = 2.626 x 10⁻⁷ m = 262.6 nm
This wavelength is not in visible range. It lies in Ultra Violet Region.
2. λ = 5.546 x 10⁻⁷ m = 554.6 nm
This wavelength is in visible range.
Explanation:
The binding energy of in photoelectric effect is also known as work function. The formula for this is given as follows:
Binding Energy = Work Function = hc/λ
λ = hc/Work Function --------------- equation (1)
where,
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = longest wavelength capable of ejecting electron = ?
(1)
Here,
Work Function = (4.73 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.568 x 10⁻¹⁹ J
Therefore,
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.568 x 10⁻¹⁹ J)
λ = 2.626 x 10⁻⁷ m = 262.6 nm
This wavelength is not in visible range. It lies in Ultra Violet Region.
(2)
Here,
Work Function = (2.24 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 3.584 x 10⁻¹⁹ J
Therefore,
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(3.584 x 10⁻¹⁹ J)
λ = 5.546 x 10⁻⁷ m = 554.6 nm
This wavelength is in visible range.
What should be done following health violations from inspection
Answer:
Correct identifies issues
Explanation:
Bridget is riding her bicycle up a hill. Which statements are correct? Select five options.
Bridget is transferring energy to the bicycle.
The bicycle is using energy to do work.
Bridget has kinetic energy.
The bicycle has potential energy.
Bridget has nuclear energy.
The bicycle has mechanical energy.
Answer:
A,B,C,D, and F are correct
Explanation:
Answer:
Explanation:
A,B,C,D, and F
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0 m/s, and the distance between them is 52.0 m. After t1 = 5.00 s, the motorcycle starts to accelerate at a rate of 5.00 m/s2. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car?
Answer:
t=2.14s
Explanation:
Since it asks us for the time it takes from the moment it starts accelerating, the first 5 seconds of movement mentioned is irrelevant here.
Next it'd be simple just to use the kinematic equation involving distance, time, and velocity:
∆d=V₀t+at²/2
From here we can move the change in distance over and use the quadratic formula to solve for time.
My work is in the attachment. Comment with any questions.
how to protect the ozone layer?
Answer:
[tex]\huge{\purple{\underline{\underline{\bf{\pink{Answer}}}}}}[/tex]
we have to protect ozone layer because it is like blanket for earth . It protects the earth form uv rays that are coming from the sunwe have to take some preventive. measures like
Use ecofriendly things that Doesnot contain (cfc) . It is the major reason for ozone DeplitionEffects of uv rays
Uv rays are very harmful . If it directly comes in contact with human it cause skin cancer , eyes irritation etc.Hope it helps.
How many kg are equal to 325 g? Please explain
Answer:
.0325
Explanation:
you take 325 g times 0.001/ 1g and that equals to 0.325
Answer:
there is 0.325 kilograms in a gram
Explanation:
divide the mass by 1000
Whenever you push an object across a frictional surface, it starts out taking a lot of force to push to get the object moving. However, once the object starts moving, it is much easier to keep it moving. What does this tell you about one of the main differences between static and kinetic friction?
a.Static friction only depends on the surface material, while kinetic friction only depends on the motion
b Kinetic and Static friction are the exact same thing
c Static friction and kinetic friction are combined in one at all times, but static friction is just not seen during the motion
d Static Friction is much stronger than kinetic friction, so it takes more force to get an object moving than it does to keep it moving
Answer:
The answer is A
Explanation:
Here's an example. A child is in school taking a test. They have made a mistake on a question, and want to erase it. The eraser is made out of a type of rubber, the rubber has friction, which means the eraser has something that's going to resist movement. Now the child has exerted enough force to get it moving, and it's moving, it won't stop unless the child stops exerting force to keep it moving. Both Newton's 1st and 3rd law explain the action of moving something on a surface with friction.
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1515.
Complete Question
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1/5.
Answer:
The angle is
Explanation:
From the question we are told that
The light emerging from second Polaroid is 1/5 the unpolarized
Generally the intensity of light emerging from the first Polaroid is mathematically represented as
[tex]I_1 = \frac{I_o}{ 2 }[/tex]
Generally from the Malus law the intensity of light emerging from the second Polaroid is mathematically represented
[tex]I_2 = I_1 cos^2 (\theta )[/tex]
=> [tex]cos^2 (\theta ) = \frac{I_2}{I_1 }[/tex]
=> [tex]cos (\theta) = \sqrt{ \frac{I_2}{I_1} }[/tex]
From the question [tex]I_2 = \frac{I_o}{5}[/tex]
[tex]cos (\theta) = \sqrt{ \frac{ \frac{ I_o}{5} }{\frac{I_o}{2} } }[/tex]
[tex]cos (\theta) = \sqrt{ \frac{2}{5} }[/tex]
=> [tex]\theta = cos ^{-1} [\sqrt{\frac{2}{5}} ][/tex]
=> [tex]\theta = 50.77^o[/tex]
Water condenses on the side of a glass of ice water because the glass's temperature is below the dew point temperature.A. TrueB. False
Answer:
A. True
Explanation:
Water exists in the air in its gaseous form. If this gaseous water is cooled sufficiently enough below its dew point, it will condense into liquid water. This is what happens to the body of a glass filled with cold water. The glass's temperature drops the temperature of the air immediately around it, cooling it below dew point temperature. This causes water to condense on the body of the glass.
What is the maximum speed with which a 1200-kg car can round a turn of radius 90.0 m on a flat road if the coefficient of static friction between tires and road is 0.70?
2. Is this result independent of the mass of the car?
Answer:
78.6m/s
Explanation:
We know that frictional force also contributes to the centripetal force that keeps the car in circular motion in the turn
And is given as
F= mv²/r
But the frictional force is
F= ugm
= = 0.7*1200*9.8= 8232N
To find maximum velocity v we say
V= √F x r/m
= √ 8232* 90 /1200
= 78.6m/s
2. Yes it is independent of mass of car
Answer:
The value is [tex]v = 24.85 \ m/s[/tex]
Yes
Explanation:
From the question we are told that
The mass of the car is [tex]m= 1200 \ kg[/tex]
The radius is [tex]r = 90 \ m[/tex]
The coefficient of static friction is [tex]\mu_s = 0.70[/tex]
Generally at maximum speed the centripetal force acting on the car is equal to the friction force on the car
So
[tex]F_c = F_f[/tex]
[tex]\frac{m v^2}{r} = \mu_s * m * g[/tex]
=> [tex]v = \sqrt{\mu_s * g * r }[/tex]
=> [tex]v = \sqrt{ 0.70 * 9.8 * 90 }[/tex]
=> [tex]v = 24.85 \ m/s[/tex]
Yes the value is independent of the mass because from the equation above we see that v is independent of mass
A well with vertical sides and water at the bottom resonates at 5.63 Hz and at no lower frequency. (The air-filled portion of the well acts as a tube with one closed end and one open end.) The air in the well has a density of 1.10 kg/m3 and a bulk modulus of 1.33 × 10^5 Pa. How far down in the well is the water surface?
Answer:
The value is s = 15.4 \ m
Explanation:
From the question we are told that
The frequency is [tex]f = 5.63 \ Hz[/tex]
The density of air in the well is [tex]\sigma = 1.33 *10^{5} \ Pa[/tex]
The Bulk modulus is [tex]B = 1.33*10^{ 5} \ Pa[/tex]
Generally the distance of the water surface from the opening is mathematically represented as
[tex]s = \frac{1}{4 f } * \sqrt{\frac{B}{\rho } }[/tex]
=> [tex]s = \frac{1}{ 4 * (5.63 )} \sqrt{\frac{1.33*10^{5}}{ 1.10} }[/tex]
=> [tex]s = 15.4 \ m[/tex]
a car accelerates from 10 m/s to 30 m/s as the traffic break up on the highway. the acceleration occurs over 3.5 seconds of time. determine the acceleration of the car.
Answer:
[tex] \boxed{\sf Acceleration \ (a) = 5.7 \ m/s^2} [/tex]
Given:
Initial velocity (u) = 10 m/s
Final velocity (v) = 30 m/s
Time taken (t) = 3.5 seconds
To Find:
Acceleration (a) of the car
Explanation:
From equation of motion:
[tex] \boxed{ \bold{ v = u + at}}[/tex]
Substituting value of v, u & t in the equation:
[tex] \sf \implies 30 = 10 + a(3.5) \\ \\ \sf \implies 30 - 10 = 3.5a \\ \\ \sf \implies 20 = 3.5a \\ \\ \sf \implies 3.5a = 20 \\ \\ \sf \implies a = \frac{20}{3.5} \\ \\ \sf \implies a = 5.7 \: m/s^2[/tex]
[tex] \therefore[/tex]
Acceleration (a) of the car = 5.7 m/s²
You are comparing a reaction that produces a chemical change and one that produces a physical change. What evidence could you use to determine which type of change is occurring?
Answer:
''
Explanation:
A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.
Answer:
If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.
in your own words provide two advantages of using meters as a measurement of length rather than old measurements of length such as hands or steps
Answer:
it is accepted all over the world
it is more reliable than other kinds of measuring system
Does the distance between diffraction minima increase or decrease when the slit width is increased? Explain with reference to your data (observation) and theory
Answer:
the width of the slit is dividing therefore the distance to the minimums increases as the width of the slit decreases
Explanation:
In the diffraction phenomenon the intensity minima are described by
a sin θ = (m + ½) λ
without using trigonometry we can find an expression for the sine
tan θ = y / L
for small angles
Tan θ = sin θ/ cos θ = sin θ = y / L
a y / L = (m + ½) lam
y = (m + ½) λ L / a
we see that the width of the slit is dividing therefore the distance to the minimums increases as the width of the slit decreases
In the observation of the experiment the length of the pattern increases with the decrease in the width of the slit
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initially 5.0 km east of the flagpole and is running with a constant velocity of 8.0 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross
Answer:
From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km
So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr
So runner towards the west will be
distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8
So equating east and west time we have
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km
That is the distance covered by runner towards the east and he will meet the runner toward the west at
6-5.92=0.08 km west of the flagpole.
A rock is dropped from a tall building. You can ignore air resistance. Determine its final speed and distance traveled after 4 seconds.
(A) speed = 20 m/s, distance = 80 m
(B) speed = 40 m/s, distance = 20 m
(C) speed = 40 m/s, distance = 80 m
(D) speed = 80 m/s, distance = 40 m
(E) speed = 20 m/s, distance = 40 m
Answer:
(C) speed = 40 m/s, distance = 80 m
Explanation:
According to Newton law of motion for a falling body:
v = u + gt
Where v is the final velocity, u is the initial velocity, t is the time taken to fall and g is the acceleration due to gravity.
Given that: t = 4 s, g = 10 m/s², u = 0 m/s(at top of building). Therefore substituting values:
v = 0 + 10(4)
v = 40 m/s
To find the distance h, we use the formula:
v² = u² + 2gh
Substituting values:
40² = 0² + 2(10)h
20h = 40² - 0²
20h = 1600
h = 1600 / 20
h = 80 m
The speed = 40 m/s, distance = 80 m
Suppose the same car had at one point an acceleration of 12m/s^2 and an initial velocity of 8m/s. After 2 seconds, what would be the car's final velocity
Answer: 32 m/s
Explanation: v = u + at
= 8 + 12*2
= 8 + 24
= 32 m/s
You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through the inductor and increase the current at a steady rate. What is the current through the inductor at the end of 2.60 s
Answer:
The current through the inductor at the end of 2.60s is 9.7 mA.
Explanation:
Given;
emf of the inductor, V = 41.0 mV
inductance of the inductor, L = 13 H
initial current in the inductor, I₀ = 1.5 mA
change in time, Δt = 2.6 s
The emf of the inductor is given by;
[tex]V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA[/tex]
Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.
Nina and Jon are practicing an ice skating routine. Nina is standing still. Jon, who is twice as heavy as Nina, skates toward her, pushing Nina away with force f. Assuming the system is closed, which statement is correct about this system? A. Nina experiences a force equal to f. B. Nina experiences a force equal to f/2.
If the system is closed, Nina will experience only a force equal to F exerted by Jon.
According to Newton's third law of motion, action and reaction is equal and opposite. That is the force applied to a stationary object is equal to the reaction exerted by the stationary object.
Applying this law, the force Jon applied on Nina is equal to the reaction Nina exerts on Jon.
If Jon exerts a force of F magnitude on Nina, consequently, Nina will experience the force of equal magnitude and exert equal reaction in opposite direction towards Jon.Also, if the system is closed, no additional force will be experienced by Nina, except the force applied by Jon.Thus, assuming the system is closed, Nina will experience only a force equal to F exerted by Jon.
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From the top of a cliff, a person throws a stone straight downward. The initial speed of the stone just after leaving the person's hand is 9.7 m/s. (a) What is the acceleration (magnitude and direction) of the stone while it moves downward, after leaving the person's hand? magnitude m/s^2direction Is the stone's speed increasing or decreasing? a. increasing b. decreasingAfter 0.51 s, how far beneath the top of the cliff is the stone? (Give just the distance fallen, that is, a magnitude.)________ m.
Answer:
a
[tex]a = 9.8 \ m/s[/tex]
Increasing
b
[tex]s = 6.22 \ m[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 9.7 \ m/s[/tex]
The time taken is [tex]t = 0.51 \ s[/tex]
Generally given that the stone is moving downward the acceleration is equivalent to the generally value of acceleration due to gravity and it would be increasing as the stone approaches the ground(toward the center of the earth )
Thus the acceleration is [tex]a = 9.8 \ m/s[/tex]
Generally from the equation of motion we have that
[tex]s = ut + \frac{1}{2} at^2[/tex]
=> [tex]s = 9.7 * 0.51 + \frac{1}{2} *9.8 * 0.51^2[/tex]
=> [tex]s = 6.22 \ m[/tex]
How could you test the hypothesis that elephants interpreted the ground signal as being farther away than the air signal?
Answer:
One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.
Explanation:
This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.
The speed of sound in solid soil is an average of the speed of its constituent media, giving values between
wood 3900 m / s
concrete 4000 m / s
fabrics 1540 m / s
earth 5000 m / s wave S
ground 7000 m / s P wave
we can see that the speed on solid earth is an order of magnitude greater than in air.
One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.
From the initial information, the wave going through the ground should arrive first.
To test the hypothesis that elephants interpreted the ground signal as being farther away the air signal is to see how fast they react by high and low frequency sounds. This however can be done by creating two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. The reaction of the animal is then observed and via the first sound from one of them
HypothesisA hypothesis is an assumption, an idea that is proposed for the sake of argument so that it can be tested to see if it might be true.
In non-scientific use, hypothesis and theory are often used interchangeably.
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One of the great upcoming sports in the Olympics is the sport of curling. Write a brief essay on the uses of momentum collisions in curling. Be sure to discuss the importance of massive stones and of going last. in at least 150 words
The Olympic sport of curling is one that is practically designed to show Physics in motion. Curling is a sport in which two teams alternate sliding smoothed stone pucks down an ice rink court with the intent to seat their stone closest to the center of the target (called the house). Each team has eight stones, meaning that the team that goes second has the (could be) massive advantage of sending the last stone.
The mass of the stone is important in that the more massive a stone (m) and the speed at which it travels (v) dictates it's momentum (momentum=mxv). As the curling stone slides down the ice (which is relatively frictionless unless acted upon by other players or objects) and having inertia, continues in it's straight course (again, unless acted upon by outside forces). If the stone hits another stone, it transfers some of its momentum in an elastic collision to that stone and the original stone is deflected in a calculable manner.
Collisions are used in the game to either clear opponent's stones from the house or out of their defensive positions, or to make adjustments to one's stones present in the house, all based on the momentum of the moving stone, and its transference.
Waves with a wavelength of 10 meters will begin to touch bottom when the depth of water is _______ meters.
Answer:
When the depth is less than 0.5 m
Explanation:
Waves will begin to touch bottom when depth of the water is less than 1/20 of the wavelength. This type of wave is known as a shallow wave.
wavelength = 10 m
It will begin to touch bottom when its depth is less than
1/20 x 10 = 0.5 m
Solve the equation 30=10x for x.
Answer:
x = 3
Explanation:
30 = 10x
30/10 = x
3 = x
Answer:
X=3
Explanation: 30 = 10x for x
Divide both sides by 10
30/10 = 10x/10
3 = x
A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s² . She is driving at 20 m/s when suddenly she sees an obstacle in the road 50 m in front of her.
What is the distance she passes after noticing the obstacle before fully stopping?
Answer:
43.3 m
Explanation:
given :
the initial velocity of the car , u = 20m/s
The reaction time is 0.5s
the distance that the car will travel during the reaction time
= initial velocity x reaction time
= 20 m/s x 0.5 s
= 10 m
Now we turn our attention to the stopping distance AFTER she reacts and steps on the brakes:
recall that one of the forms that the equations of motions can be expressed is:
v² = u² + 2as
where v = final velocity = 0 because the car comes to a stop
u = initial velocity = 20m/s
a = acceleration = -6.0 m/s² (note negative sign because it is a deceleration)
s = distance traveled (we are asked to find this)
substituting the known values into the equation:
v² = u² + 2as
0² = 20² + 2(-6.0) s
0 = 400 - 12s
1.2s = 400
s = 400/12
s = 33.33 m
Total distance travelled
= distance during reacting time + distance while decelerating
= 10 + 33.3
= 43.3 m
Answer:
43.3 m
Explanation:
Given :
Reaction time = 0.5 sDeceleration = -6.0 m/s²Speed = 20 m/sObstacle Distance = 50 mDistance Travelled during Reaction Period
S = utS = 20 × 0.5S = 10 mUsing the formula v² - u² = 2aS :
Rearrange it so it is equal to 'S'S = v² - u² / 2aS = 0² - 20² / 2(-6) [v = 0, because speed is 0 when car stops]S = -400/-12S = 100/3 = 33.3 mDistance travelled after fully stopping :
10 m + 33.3 m43.3 mstate the basic law of charges
Answer:
Things that have the same charge push each other away (they repel each other). This is called the Law of Charges. … Things that have more electrons than protons are negatively charged, while things with fewer electrons than protons are positively charged. Things with the same charge repel each other.
Explanation:
Marisol drives north at 64.3 km/h. How far does Marisol travel after 5.8 h?
Answer:
372.94 km
d= st
= (64.3 km/h)(5.8h)
=372.94 km
Suppose you wanted to break a meter stick over your knee. The cross-section of a meter stick is rectangular. Will it be easier to break with the long side against your knee or with the short side against your knee? Cross-section of a meter stick : (i.e. looking down the end of a meter stick)
Answer:
It will be easier to break the meter rule with the long side against my knee.
Explanation:
To break the meter rule involves the principle of bending moment. The long side will require less force to generate the same amount of bending moment that will have to be generated to break the meter rule. The short side on the other hand will require more force to generate this mount of bending moment. This is because the shorter has a very small surface area, which concentrates the force on your knee. The pressure is then dissipated as more pressure to your knee. Th longer side has a lesser surface area so, most of the force is used in breaking the meter rule.
If a negative point charge is placed at PP without moving the original charges, the net electrical force the charges ±Q±Q will exert on it is
Answer:
hello your question is incomplete below is the complete question
point p is equidistant from two point charges +-Q of equal magnitude if a negative point charge is places at P without moving the original charges the net electrical force the charges +-Q will exert on it is A. directly upward B. directly downward C. zero D. directly to the right E. directly to the left why is it A?????
Answer : Directly upward
Explanation:
The direction of the resultant force acting on the charge placed at P will be in the upward direction. this is because the net electric force at point p due to the opposite charges will be acted in the downward direction. If the negative charge is placed at point P, then the direction of the resultant force acting on the charge placed at P will be in the opposite direction to that of the field. i.e in the upward direction