The horizontal force required to move the crate at a steady speed across the floor will be 58.86 N.
What is the friction force?
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
The given data in the problem is;
Mass,m = 24 kg
The coefficient of kinetic friction, μ=0.25
The friction force is defined as the product of the coefficient of friction and normal reaction.
f=μN
If there is no other force, the horizontal force is equal to the friction force;
F=f
The horizontal force required to move the crate at a steady speed across the floor is found as;
F = μN
The normal reaction force is;
N=W
N=mg
F = μmg
F=0.25 × 24-kg × 9.81 m/s²
F = 58.86 N
Hence horizontal force is required to move the crate at a steady speed across the floor will be 58.86 N.
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The bars in the drawing above are magnets. According to chapter 21 which of the following picture(a) is true? Explain. Then explain why the other 2 pictures are incorrect. Make sure to explain all 3 sets.
Magnets are materials which are able to attract others to themselves by producing a magnetic field.
Based on the properties of magnetic lines of forces;
Picture 1 is correct because the magnetic lines of forces originate from the North pole and are directed towards the south pole.Picture 2 is incorrect because unlike poles attractPicture 3 is incorrect because like poles repel.In conclusion, the law of magnetism states that like poles repel whereas unlike poles attract.
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The rms voltage is given by the following.
ΔVrms = ΔVmax/[tex]\sqrt{2}[/tex] = 0.707ΔVmax
Substituting the maximum voltage, we obtain
ΔVrms = ΔVmax/[tex]\sqrt{2}[/tex]
= 0.707ΔVmax = 0.707(165 V) = 117 V.
An rms voltage of 117 V corresponds to a maximum voltage of 165 V.
What is the maximum current Imax in the resistor?
The maximum current in the resistor is determined by dividing root mean square current by 0.7071.
Maximum current in the resistor
The maximum current in the resistor is calculated as follows;
I₀ = I(rms)/0.7071
where;
I₀ is the peak current or maximum current in the resistorI(rms) is the root mean - square currentThus, the maximum current in the resistor is determined by dividing root mean square current by 0.7071.
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Two forces F1=20N and F2=30N are acting on an object as shown in figure: F1=20N F2=30N (i) Find the net force acting on the object? (ii) State the direction of the net force acting on the object? (iii) If the body still does not move under the application of these forces, what can be the possible reason for this? Identify the name of this extra force and its direction. Two forces F1 = 20N and F2 = 30N are acting on an object as shown in figure : F1 = 20N F2 = 30N ( i ) Find the net force acting on the object ? ( ii ) State the direction of the net force acting on the object ? ( iii ) If the body still does not move under the application of these forces , what can be the possible reason for this ? Identify the name of this extra force and its direction .
The net force is the 10N force and its direction is towards F2 force.
What is resultant force?The resultant force or the net force is the actual force that acts on a body. We are told that the two forces act in opposite directions hence the net force is F2 - F1 = 30 N - 20 N = 10 N.
ii) The direction of this net force is going to be towards the 30N force.
iii) If the body does not move under the application of these forces then the F1 force must be reinforced by another force which balances it against the F2 force.
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A car moves at a speed of 45 km/h, when traveling 1500 meters, how long did it take to go that distance?
A proton has a positive electric charge of q = 1.6 × 10–19 coulombs. what is the electric potential at a point 5.3 × 10–11 m from the proton? 4.4 × 10–18 volts 2.7 × 101 volts 3.0 × 10–9 volts 5.1 × 101 volts
B. The electric potential at the given point from the proton is 2.7 x 10¹ volts
Electric potential of the proton
The electric potential of the proton is calculated as follows;
V = kq/r
where;
k is Coulomb's constantq is charger is distanceV = (9 x 10⁹ x 1.6 x 10⁻¹⁹)/(5.3 x 10⁻¹¹)
V = 27.2 volts
V = 2.7 x 10¹ volts
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A roller coaster has a mass of 450 kgIt sits at the top of a hill with height 49 m. If it drops from this hill, how fast is it going when it reaches the bottom? Assume there is no air resistance or friction.)
The speed of the roller coater at the bottom of the hill is 31 m/s.
Speed of the roller coater at the bottom of the hill
Apply the principle of conservation of mechanical energy as follows;
K.E(bottom) = P.E(top)
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
v is the speed of the coater at bottom hillh is the height of the hillg is acceleration due to gravityv = √(2 x 9.8 x 49)
v = 31 m/s
Thus, the speed of the roller coater at the bottom of the hill is 31 m/s.
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Part A
How did Einstein’s and Newton’s theories differ in terms of explaining the cause of gravity?
Newton thinks that object falls due to gravity whereas Einstein thinks gravity is not the reason behind it.
How did Einstein’s and Newton’s theories differ in terms of explaining the cause of gravity?Newton concluded that objects fall because they are pulled by Earth's gravity. Einstein's thinks that these objects do not fall due to gravity. According to Einstein, these objects and Earth just freely move in a curved spacetime
So we can conclude that Newton thinks that object falls due to gravity whereas Einstein thinks gravity is not the reason behind it.
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A cup of tea at room temperature of 24°C is heated until it has twice the internal energy. Calculate the final temperature of the tea.
The final temperature of the tea cup is 100°C.
What is internal energy?The Internal energy is the energy of a substance due to to the constant random motion of its particles.
The symbol for internal energy of a substance is U and it is measured in Joules.
ΔU = q + W
q is the heat, q = mcΔTW is the mechanical work.In conclusion, the final temperature of the tea cup at room temperature of 24 °C which is heated until it has twice the internal energy is 100°C.
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A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
What is the tension in rope A?
How much tension does rope B carry?
And what is the tension in rope C?
The tension in rope A is 740.88 N, the tension carried by rope B is 751.4 N and the tension carried by rope C is -124 N.
Angle made by B with respect to horizontaltan θ = Δy/Δx
tan θ = (10 - 4)/(5 - 4)
tan θ = 6
θ = arc tan (6) = 80.53
Net force in vertical directionC sin(0) + B sin(80.53) = A sin(90)
C sin(0) + B sin(80.53) = (75.6 x 9.8)
0 + 0.986 B = 740.88
0.986 B = 740.88
B = 740.88/0.986
B = 751.4 N
Net force in horizontal directionC cos(0) + B cos(80.53) = (75.6 x 9.8) cos(90)
C + 0.165 B = 0
C + 0.165(751.4) = 0
C + 124 = 0
C = - 124 N
Thus, the tension in rope A is 740.88 N, the tension carried by rope B is 751.4 N and the tension carried by rope C is -124 N.
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A force F1 of magnitude 5.90 units acts on an object at the origin in a direction = 36.0° above the positive x-axis. A second force F2 of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force F1 + F2.
The magnitude and direction of the resultant force will be 7.65 units and 40.2°.
What is force?Force is explained as the multiplication of mass and acceleration. Its unit is N. It is the external agent which helps to change the shape and size of an object.
Force is used to push and pull the object and sometimes changes the direction of motion. The net force acting on an object is equal to the sum of all forces acting upon it.
F = F1 + F2 + F3
Given data;
Megnitude of force 1, F₁ = 5.90 units
Megnitude of force 2, F₂ = 5.00 units
α is the angle between the two forces=36.0°
Θ is the angle of the resultant force
The megnitude of the resultant force is;
[tex]\rm F =\sqrt{ F_{1}+F_{2}+2F_{1}F_{2}cos \alpha }\\\\ F = \sqrt{5.90 + 5.00 +2 \times 5.90 \times 5 \times cos 36^0 } \\\\ F = 7.65 \ units[/tex]
The direction of the resultant force is;
[tex]\rm \theta = tan^{-1}(\frac{F_{2}}{F_{1}} )[/tex]
[tex]\rm \theta = tan^{-1}(\frac{5.00}{5.90})\\\\ \theta = 40.2 ^0[/tex]
Hence the magnitude and direction of the resultant force will be 7.65 units and 40.2°.
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a bus strats from from rest. if the accleration is 2m/s,find the distance travelled and the velocity after 2 seconds
Answer:
Velocity: 4 m/s
Distance: 4 m
Explanation:
initial velocity: u = 0 m/s
Acceleration: a = 2 m/s²
Time taken: t = 2 seconds
Distance traveled: d = (what we need to find)
Final velocity: v = (what we need to find)
---------------------------------------
v = u + at
v = 0 + 2 × 2
v = 4 m/s
---------------------------------------
d = 0 × 2 + 1/2 × 2 × (2)²
d = 1/2 × 2 × 4
d = 4 m
Which of these must be the same on both
sides of a balanced chemical equation?
A the number of atoms
B
the number
C the number
D the number of solids, liquids, and gases
of compounds
of molecules
Answer:
the total number of atoms of each element present is the same on both sides of the equation. (if I remember correctly)
PLS HELP:
Find the magnitude of an electric field located 7.00 cm from a source charge of - 11 nC. (1nC = 1 x 10-9 C). Will the field lines point toward or away from the source charge? Show your work and include an electric field diagram assuming the source is a point charge
Answer:
E=kQ/r*2
=(9. 10*9)(11.10*-9)/(0,07)*2
=20204 N. C*-1
The field lines will point towards the source charge because it is negatively charged.
Hi! Whoever can answer this question the best I will give the brainliest answer!!
Describe the structure of a battery and how it works.
Thank you!! ✨
For the circuit shown in the figure, the switch S is initially open and the capacitor is uncharged. The switch is then closed at time t = 0. How many seconds after closing the switch will the energy stored in the capacitor be equal to 50.2 mJ?
Hi there!
To start, we can use the equation for the charge of a capacitor versus time.
[tex]q(t) = C\epsilon (1 - e^{-\frac{t}{\tau}})[/tex]
[tex]q(t)[/tex] = Charge on capacitor vs. time
[tex]\epsilon[/tex] = EMF of battery (40 V)
**Note: 'Cε' is equal to the MAXIMUM CHARGE on a capacitor.
Where the time constant is given as:
[tex]\tau = RC[/tex]
[tex]\tau[/tex] = Time Constant (s)
[tex]R[/tex] = Resistance (0.5 MΩ)
[tex]C[/tex] = Capacitance (90 μF)
Let's first convert the given units to the necessary 'J', 'Ω' and 'F' to make things easier:
[tex]J = 50.2 mJ[/tex]
[tex]50.2 mJ * \frac{0.001J}{1mJ} = 0.0502 J[/tex]
[tex]R = 0.5 M\Omega\\\\0.5M\Ohm * \frac{1000000\Omega}{1M\Ohm} = 500000\Omega[/tex]
[tex]F = 90 \mu F[/tex]
[tex]90 \mu F * \frac{0.000001 F}{1\mu F} = 0.00009 F[/tex]
Now, let's calculate the time constant.
[tex]\tau = RC = (500000)(0.00009) = 45 s[/tex]
Time to use the equation for energy stored in a capacitor:
[tex]E_C = \frac{1}{2}\frac{Q^2}{C}[/tex]
Our 'Q' varies with time, so:
[tex]E(t)= \frac{1}{2}\frac{( C\epsilon (1 - e^{-\frac{t}{\tau}}))^2}{C}\\\\E(t) = \frac{1}{2}\frac{( C^2\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)}{C}\\\\E(t) = \frac{1}{2}( C\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)[/tex]
Let's now plug values into our equation.
[tex]0.0502 = \frac{1}{2}(0.00009)(40^2) (1 - e^{-\frac{t}{45}})^2[/tex]
Simplifying:
[tex]0.697 = (1 - e^{-\frac{t}{45}})^2[/tex]
Take the square root of both sides.
[tex]0.835 = 1 - e^{-\frac{t}{45}}\\\\ e^{-\frac{t}{45}} = 0.165[/tex]
Take the natural log of both sides.
[tex]-\frac{t}{45} = ln(0.165) \\\\\frac{t}{45} = 1.802\\\\t = \boxed{81.08 s}[/tex]
A ball is rolling across the floor at a constant velocity. What is the value of the change in its kinetic energy, ΔEk?
Positive
Zero
Negative
B. When the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.
What is change in kinetic energy?
The change in kinetic energy of an object is the dereference between the final kinetic energy and the initial kinetic energy.
ΔK.E = K.Ef - K.Ei
ΔK.E = 0.5m(vf² - vi²)
where;
K.Ef is the final kinetic energyK.Ei is the initial kinetic energyvf is final velocityvi is initial velocityAt constant velocity, the initial velocity and final velocity are equal.
ΔK.E = 0.5m(0) = 0
Thus, when the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.
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A solid, homogeneous sphere with of mass of M = 2.35 kg and a radius of R = 12.9 cm is resting at the top of an incline as shown in the figure.
The height of the incline is h = 1.29 m, and the angle of the incline is θ = 20.1°. The sphere is rolled over the edge very slowly. Then it rolls down to the bottom of the incline without slipping.
What is the total kinetic energy (translational plus rotational kinetic energy) of the sphere when it reaches the bottom of the incline?
The total kinetic energy of the sphere when it reaches the bottom of the incline is 44.57 J.
Total kinetic energy of the sphereThe total kinetic energy of the sphere = translational + rotational kinetic energy.
Translational kinetic energyK.E = ¹/₂mv²
K.E(bottom) = P.E(top) = mgh
K.E = (2.35)(9.8)(1.29)
K.E = 29.71 J
Rotational kinetic energyK.E(rot) = ¹/₂Iω²
K.E(rot) = ¹/₂I(v²/R²)
where;
I is rotational kinetic energyI = ¹/₂MR²
K.E(rot) = ¹/₂( ¹/₂MR²)(v²/R²)
K.E(rot) = ¹/₄Mv²
K.E(rot) = ¹/₄M(2gh)
K.E(rot) = ¹/₂Mgh
K.E(rot) = ¹/₂(29.71) = 14.855 J
Total kinetic energy = 29.71 + 14.855 = 44.57 J
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D
Question 2
What are the coefficients needed to balance this chemical equation?
___O₂ → CO₂ +___ H₂O
C₂H5OH +_
O 6, 2, 6, 4
O 6, 6, 2,4
O 1, 3, 2, 3
1, 2, 2, 1
2 pts
Hello and Good Morning/Afternoon:
Original Question: C₂H₅OH + __O₂ → __CO₂ + __ H₂O
To balance this equation:
⇒ must ensure that there is an equal number of elements on both sides of the equation at all times
Let's start balancing:
On the left side of the equation, there are 2 carbon molecule⇒ but only so far one on the right side
C₂H₅OH + __O₂ → 2CO₂ + __ H₂O
On the left side of the equation, there are 6 hydrogen molecules⇒ but only so far two on the right side
C₂H₅OH + __O₂ → 2CO₂ + 3H₂O
On the right side of the equation, there are 7 oxygen molecules⇒ but only so far three on the left side
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
Let's check and make sure we got the answer:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
2 Carbon ⇔ 2 Carbon
6 Hydrogen ⇔ 6 Hydrogen
7 Oxygen ⇔ 7 oxygen
Thefore the coefficients in order are:
⇒ 1, 3, 2, 3
Answer: 1,3,2,3
Hope that helps!
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. A wheel of diameter 0.2 m starts from rest and accelerated with constant angular acceleration to an angular velocity of 900 rev min¹ in 5 s. (a) Find the position at the end of 1 s of a point originally at the top of the wheel. (b) Compute and show in a diagram the magnitude and the direction of the acceleration at the end of 1 s.
(a)The position at the end of 1 s of a point originally at the top of the wheel will be 9.425 radians.
(b)The magnitude and the direction of the acceleration at the end of 1 s will be 1.885 m/sec² clockwise.
What is angular acceleration?Angular acceleration is defined as the pace of change of angular velocity concerning time.
Given data;
Diameter of the wheel,d = 0.2 m
Initial velocity,ω₀ = 0
Angular acceleration,α = c
n = 900 rev min¹
ω = 94.2478 rad /sec
Time of rolling,t = 5 s
From Newton's first equation of motion;
ω = ω₀ + αt
94.2478 rad /sec = 0 + α(5 sec)
α = 18.85 rad/sec²
The position at the end of 1 s of a point is found as;
Θ = ω₀t+1/2(αt²)
Θ = 0+1/2[(18.85 rad/sec²) × (1 sec)²]
Θ = 9.425 radian
The magnitude and the direction of the acceleration at the end of 1 s are found as;
a = αr
a= 18.85 rad/sec² × 0.1 m
a = 1.885 m/sec²
Hence the position at the end of 1 s and magnitude and the direction of the acceleration at the end of 1 s will be 18.85 rad/sec²and the magnitude and the direction of the acceleration at the end of 1 s will be 1.885 m/sec² clockwise.
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Three ropes A, B and C are tied together in one single knot K. (See figure.)
If the tension in rope A is 65.3 N, then what is the tension in rope B?
The tension in the rope B is determined as 10.9 N.
Vertical angle of cable B
tanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
Angle between B and Cθ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The magnitude of the force on the left-hand pole is mathematically given as
f'=0.167N
[tex]Q=45 \textdegree[/tex]
What is the magnitude of the force on the left-hand pole.?Generally, the equation for Force is mathematically given as
[tex]F=\frac{mg}{sin\theta}[/tex]
Therefore
[tex]F=\frac{17.1*10^{-3}*9.8}{sin45}[/tex]
F=0.237N
Considering horizontal
f'-fcos=0
Therefore
f'=0.237*cos45
f'=0.167N
In conclusion, the slope
[tex]tan\theta=lv/ln\\\\tan\theta=30/30\\\\\theta=tan^{-1}\\\\[/tex]
[tex]Q=45 \textdegree[/tex]
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Two loudspeakers, 3.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.15 m . Assume the speed of sound is 340 m/s .
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.15 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
For the given question the answer to the first part is that frequency of sound will be 404.76 Hz. and answer to the second part will be that the initial frequency for which that place will have the highest level of sound intensity
Given, the speed of sound 340 m/s
a) L1 + L2 = 3.5 m
and Ld = | L1 - L2 | = ( 1.75 + 0.21 ) - (1.75 - 0.21 ) = 1.96 - 1.54 = 0.42 m
Ld = λ/2
λ = 2Ld = 2×0.42 = 0.84 m
and finally,
f = v/λ
f = 340/0.84
f = 404.76 hertz
Frequency came out to be 404.76 hertz in this case
b) For the first frequency
0.42 = λ
f = v/λ
f = 340 / 0.42
f = 809.52 Hertz
Frequency came out to be 809.52 Hertz in this case.
To conclude with we can say that the Frequency of the sound in case on came out to be 404.76 hertz which is approximately 405 Hz after applying all the concepts and calculations, in second case first frequency for which that location will be a maximum of sound intensity came out to be 809.52 Hertz after applying all the concepts and calculations.
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What is standard one kilogram ?
Answer:
As of the 2019 redefinition of the SI base units, the kilogram is defined in terms of the second and the metre, both based on fundamental physical constants.
It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures. It was originally defined as the mass of one liter (10-3 cubic meter) of pure water. At the Earth's surface, a mass of 1 kg weighs approximately 2.20 pounds ( lb ).
Explanation:
Latency refers to the 27 seconds of time it takes for someone to refocus on driving after they finish texting or talking on the phone. true false
The term "latency" describes the 27 seconds that pass after someone has finished texting or conversing on the phone before they can concentrate again on driving. The given statement is true.
What is latency?A measure of delay is latency. The amount of time it takes for some data to travel across a network is known as latency.
The amount of time it takes for information to travel from its source to its destination and back is often quantified as a round trip delay.
The 27 seconds that pass after someone has done texting or talking on the phone before they can focus on driving again are referred to as "latency." The assertion is accurate.
Hence, the given statement is true.
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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.
The tension in the cable is mathematically given as
T =266.659 N
What is the tension in the cable.?Generally, the equation for the angle of the boom with horizontal is mathematically given as
A = tan-1(5 /10)
[tex]A= 26.57 \textdegree[/tex]
The angle of cable with horizontal
B = tan-1(4 / 10)
B= 21.80 degrees
Hence
175.5 * cos26.57 + 94.7 *cos26.57 * 1/2 = T (sin(26.57+21.80)) * 1
T =266.659 N
In conclusion, the tension in the cable.
T =266.659 N
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Lindsay is planning a flight from St. Catherines to Hamilton, which lies due west of St. Catharines.
Her aircraft flies at a speed of 200 km/h in still air. A wind of 50.0 km/h is blowing from [W 60.0° N].
In what direction must she aim the aeroplane to fly directly to Hamilton? (KU, TI)
Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.
Using Sine rule to solve this question
Sine rule => SinA/a = SinB/b = SinC/c = constant
The magnitude of wind is 50 with an angle of 60 degrees.
The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.
One side is 50 km/hr at an angle of 60 degrees.
sin 60°/200 = sin θ / 50
50 × sin 60° = 200 × sin θ
√3/2 = 4 × sin θ
√3/8 = sin θ
sin θ = 0.2165
θ = sin⁻¹(0.2165)
θ = 12.5°
So Lindsay have to fly the plane in the direction of [W 12.5° S].
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For a vector with origin at coordinate (0, 0) (horizontal, vertical) and end at (6, 8) northeast of the origin,
what is the horizontal component of the vector?
The horizontal component of the vector is determined as 6i.
Horizontal component of the vector
The horizontal component of the vector is calculated as follows;
V = (0 i, 0 j) + (6i, 8j)
V = (6i, 8j)
where;
6i is the horizontal component of the vector8j is the vertical component of the vectorThus, the horizontal component of the vector is determined as 6i.
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A slinky spring 3m long rests on a horizontal bench with one end fixed. When the free end is suddenly pushed forward, the compression pulse travels to the fixed end and back in 2 secs. What is the speed of the longitudinal waves along the spring? What is the wavelength of the longitudinal waves produced when the free end is moved to and fro with a frequency of 2.5Hz? A slinky spring 3m long rests on a horizontal bench with one end fixed . When the free end is suddenly pushed forward , the compression pulse travels to the fixed end and back in 2 secs . What is the speed of the longitudinal waves along the spring ? What is the wavelength of the longitudinal waves produced when the free end is moved to and fro with a frequency of 2.5Hz ?
The speed of the longitudinal waves along the spring is 1.5 m/s.
The wavelength of the longitudinal waves produced when the free end is moved to and fro is 0.6 m.
What is wavelength?The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.
Wavelength λ = c/f
where c is the speed of light in vacuum = 3*10⁸ m/s
A slinky spring 3m long rests on a horizontal bench with one end fixed. When the free end is suddenly pushed forward, the compression pulse travels to the fixed end and back in 2 secs.
λ = v/f
v =fλ =λ /T where T is the time period = 2sec
Put the values we get
v = 3/2 =1.5 m/s
Thus, the speed of wave is 1.5 m/s.
Using the same expression for frequency 2.5Hz, wavelength is
λ = v/f
λ = 1.5 /2.5 = 0.6 m
Thus, the wavelength is 0.6m.
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Mithun wants to test if a convex lens produces an inverted image of an object placed at a distance away from the lens. Which of the following should he use directly as an object for the test?
a. A
b. B
c. C
d. D
Answer:
(D) any of these objects can be used as long as all points on the object lie farther away from the lens than the focal length of the lens.
Points lying between the focal point and the lens will be erect and will not form an object visible on a screen.
What is the maximum height that the water would reach in the internal network of a building, without the use of a pump, if the pressure at the sidewalk level is 1.3 kgf/cm2? (density of water 1000 kg/m3)
The maximum height to be reached by the water at the given pressure is 13 m.
Maximum height reached by the water
The maximum height to be reached by the water is calculated as follows;
P = ρgh
where;
P is the pressure of the waterρ is the density of the waterg is acceleration due to gravityh is the maximum height1 kgf/cm² = 98066.5 Pa = 98066.5 N/m²
1.3 kgf/cm² = 127486 N/m²
h = P/ρg
h = (127486)/(1000 x 9.8)
h = 13 m
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