Comets are divided into two types. Long-period comets are the comets that take more than two hundred years to finish an orbit throughout the Sun originate from the Oort Cloud.
Explanation:
The Oort Cloud has sufficient distance apart of the Sun than the Kuiper Belt, it seems that the Oort Cloud objects were made closer to the Sun than the Kuiper Belt things.Long-period comets have highly eccentric orbits.The Oort cloud is considered to own a large range commencing from among 2,000 and 5,000 AU to 50,000 AU.The chemical makeup of long-period and short-period comets is quite alike.
Final answer:
A comet with a 324-year orbit is likely from the Oort Cloud, the distant spherical region of icy bodies surrounding our solar system, known as the source of long-period comets. The Kuiper Belt, on the other hand, sources short-period comets.
Explanation:
A comet with a period of 324 years is considered a long-period comet. These comets originate from a region far beyond the inner solar system. The Oort Cloud is a vast, spherical shell of icy bodies that extends up to about 50,000 astronomical units (AU) from the Sun and contains trillions of potential comets. This makes it the most likely source of long-period comets like the one mentioned. In contrast, the Kuiper Belt is a disk-shaped region beyond Neptune, extending to about 50 AU, which is known to source short-period comets, also known as Jupiter-family comets.
Long-period comets like the one with a 324-year orbit are thought to be influenced by gravitational perturbations from nearby passing stars or galactic tides, which can send them towards the inner solar system. Once these comets enter the inner solar system, they can have dramatic interactions with planets or the Sun, sometimes leading to their disintegration or alteration in orbit. An example of such an interaction was when Comet Shoemaker-Levy 9 broke apart and collided with Jupiter in 1994.
A 210-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
Answer:
The value of the constant force is [tex]\bf{296.88~N}[/tex].
Explanation:
Given:
Mass of the merry-go-round, [tex]m = 210~Kg[/tex]
Radius of the horizontal disk, [tex]r = 1.5~m[/tex]
Time required, [tex]t = 2.00~s[/tex]
Angular speed, [tex]\omega = 0.600~rev/s[/tex]
Torque on an object is given by
[tex]\tau = F.r = I.\alpha~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where [tex]I[/tex] is the moment of inertia of the object, [tex]\alpha[/tex] is the angular acceleration and [tex]F[/tex] is the force on the disk.
The moment of inertia of the horizontal disk is given by
[tex]I = \dfrac{1}{2}mr^{2}[/tex]
and the angular acceleration is given by
[tex]\alpha = \dfrac{2\pi \omega}{t}[/tex]
Substituting all these values in equation (1), we have
[tex]F &=& \dfrac{I\alpha}{r}\\&=& \dfrac{\pi m r \omega}{t}\\&=& \dfarc{\pi(210~Kg)(1.5~m)(0.600~rev/s)}{2.00~s}\\&=& 296.88~N[/tex]
A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of refraction in the plastic? A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of refraction in the plastic? The angle of incidence is equal to the angle of refraction. The angle of incidence is greater than the angle of refraction. The angle of incidence is less than the angle of refraction. The two angles cannot be compared without additional information.
Answer:
The angle of incidence is greater than the angle of refraction
Explanation:
Refraction occurs when a light wave passes through the boundary between two mediums.
When a ray of light is refracted, it changes speed and direction, according to Snell's Law:
[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]
where :
[tex]n_1[/tex] is the index of refraction of the 1st medium
[tex]n_2[/tex] is the index of refraction of the 2nd medium
[tex]\theta_1[/tex] is the angle of incidence (the angle between the incident ray and the normal to the boundary)
[tex]\theta_2[/tex] is the angle of refraction (the angle between the refracted ray and the normal to the boundary)
In this problem, we have a ray of light passing from air into clear plastic. We have:
[tex]n_1=1.00[/tex] (index of refraction of air)
[tex]n_2=1.50[/tex] approx. (index of refraction in clear plastic)
Snell's Law can be rewritten as
[tex]sin \theta_2 =\frac{n_1}{n_2}sin \theta_1[/tex]
And since [tex]n_2>n_1[/tex], we have
[tex]\frac{n_1}{n_2}<1[/tex]
And so
[tex]\theta_2<\theta_1[/tex]
Which means that
The angle of incidence is greater than the angle of refraction
A square loop and a circular loop are each made from the same length of wire. Each loop carries the same amount of current. The loops are placed in the same uniform magnetic field. What is the ratio of the maximum torque on the square loop to the maximum torque on the circular loop? Question 2 options: π/4 2/π None of the other answers are correct. 4/π π/2
Explanation:
Let l is the length of the wire. such that, side of square is l/4. Area of square loop is, [tex]A_s=\dfrac{l^2}{16}[/tex]
Radius of the circular loop,
[tex]l=2\pi r\\\\r=\dfrac{l}{2\pi}[/tex]
Area of the circular loop,
[tex]A_a=\pi r^2\\\\A_a=\pi \times (\dfrac{l}{2\pi})^2\\\\A=\dfrac{l^2}{4\pi}[/tex]
Torque in magnetic field is given by :
[tex]\tau=NIAB\sin\theta[/tex]
It is clear that, [tex]\tau\propto A[/tex]
So,
[tex]\dfrac{\tau_s}{\tau_a}=\dfrac{A_s}{A_a}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\dfrac{l^2}{16}}{\dfrac{l^2}{4\pi}}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\pi}{4}[/tex]
So, the ratio of maximum torque on the square loop to the maximum torque on the circular loop is [tex]\pi :4[/tex].
Which statements describe the wave? Check all that
apply.
The wave is a mechanical wave.
The wave is an electromagnetic wave.
The wave moves energy through matter.
The wave moves eney through space without
matter,
The wave transfers energy parallel to the motion of
the wave.
The wave transfers energy perpendicular to the
motion of the wave.
Answer:the wave is an electromagnetic wave.
Explanation:
Answer:
A: The wave is a mechanical wave.
C: The wave moves energy through matter.
F: The wave transfers energy perpendicular to the motion of the wave.
Hope it works!
Explanation:
Two small nonconducting spheres have a total charge of 90.0 C.
(a) When placed 28.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive. What is the charge on each?
(b) What if the force were attractive?
Answer: (a) Smaller charge is [tex]2.7 \times 10^{-5} C[/tex] and larger charge is [tex]11.7 \times 10^{-5} C[/tex].
(b) Smaller charge is [tex]-11.4 \times 10^{-5}[/tex] and larger charge is [tex]9.1 \times 10^{-5}[/tex].
Explanation:
(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.
Therefore, total charge on the two non-conducting spheres will be calculated as follows.
[tex]Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}[/tex]
= [tex]9 \times 10^{-5} C[/tex]
Therefore, force between the two spheres will be calculated as follows.
F = [tex]k\frac{Q_{1}Q_{2}}{r^{2}}[/tex]
12 N = [tex]\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}[/tex]
[tex]Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}[/tex]
or, [tex]Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}[/tex]
[tex]9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}[/tex]
[tex]Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0[/tex]
[tex]Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C[/tex]
This means that smaller charge is [tex]2.7 \times 10^{-5} C[/tex] and larger charge is [tex]11.7 \times 10^{-5} C[/tex].
(b) When force is attractive in nature then it means both the charges are of opposite sign.
Hence, total charge on the non-conducting sphere is as follows.
[tex]Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}[/tex]
[tex]Q_{1} - Q_{2} = 9 \times 10^{-5} C[/tex]
Now, force between the two spheres is calculated as follows.
F = [tex]k\frac{Q_{1}Q_{2}}{r^{2}}[/tex]
12 N = [tex]\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}[/tex]
[tex]Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}[/tex]
[tex]Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}[/tex]
[tex]Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}[/tex]
[tex]Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}[/tex]
Hence, smaller charge is [tex]-11.4 \times 10^{-5}[/tex] and larger charge is [tex]9.1 \times 10^{-5}[/tex].
(a). The charges on one sphere is [tex]0.89\mu C[/tex] and [tex]1.18 \mu C[/tex] on other sphere.
(b). If the force were attractive then charge on both sphere will be in opposite sign.
Electrostatic force :The electrostatic force is given as,
[tex]F=k\frac{Q_{1}Q_{2}}{r^{2} }[/tex]
Where ,
[tex]r[/tex] is distance between charges.[tex]k[/tex] is coulombs constant,[tex]k=9*10^{9} Nm^{2}/C[/tex][tex]Q_{1}[/tex] and [tex]Q_{2}[/tex] are charges on both non conducting spheres .Given that, [tex]F=12N,r=28cm=0.28m,Q_{1}+Q_{2}=90[/tex]
substitute values in above relation.
[tex]12=\frac{9*10^{9} *Q_{1}(90-Q_{1})}{(0.28)^{2} } \\\\Q_{1}^{2}-(90*10^{-6})Q_{1} +1.05*10^{-10}=0\\ \\Q_{1}=8.88*10^{-5}=0.89 \mu C\\ \\Q_{2}=1.18*10^{-6} =1.18\mu C[/tex]
If the force were attractive then charging on both sphere will be in opposite sign.
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How are linear measurements usually expressed in the building industry?
In the building industry, linear measurements are expressed in units like miles, feet, and inches, or in the metric system, meters and millimeters. The choice of unit depends on the precision needed for the task, with smaller units used for finer details and larger ones for general dimensions.
Linear Measurements in the Buildings Industry
In the building industry, linear measurements are typically expressed in units that provide the most practical and precise information for the specific dimension being measured. Linear dimensions refer to measurements that can be expressed using linear units such as miles, feet, or inches; and in the metric system, meters or millimeters (mm). These measurements are essential for architects and builders to communicate the sizes of different components of a structure, like the length of a wall or the size of a window. Typically, for smaller measurements such as the dimensions of timber or the size of rooms, feet or meters are used, while millimeters might be preferred for finer details.
For example, when measuring a specific length like the width of a staircase, the builders would commonly use millimeters in the metric system to ensure precision. Similarly, to describe to a European the dimensions of "two-by-four" lumber used in the US, conversions from inches to centimeters and from feet to meters are required. The key is to use the most appropriate unit of measurement that ensures clarity and precision for constructing a building accurately.
When dealing with linear measurements, especially in scientific contexts or in detailed building plans, precise units like millimeters, centimeters, and meters are often used. The context will dictate whether a larger unit such as kilometers or smaller units like micrometers are more appropriate.
The position vector of a particle is directed along the positive y axis. What is the direction of the net force acting on the particle if the net torque is directed along the negative x direction?
a. Posivite z direction
b. Posivite x direction
c. negative x direction
d. negative y direction
e. negative z direction
Answer:
(e) The direction of the net force is negative Z direction.
Explanation:
Let the direction of the particle (positive y) represents direction of current
Let direction of the net torque (negative x) represents direction of the field
Let direction of net force be direction of force
Apply Fleming's right hand rule, which states that when the thumb, the middleave finger and forefinger are held mutually at right angle to each other, with the thumb pointing in the direction of current (particle, positive y ), and the forefinger pointing in the direction of the field (torque, negative x), then the middle finger will be pointing in the direction of the force.
To find direction of the net force:
Y direction is already represented, negative X direction is also represented and finally force will be in Z - direction
Since we know that torque = force x position of particle
Torque is negative, thus force will be negative.
Therefore, the direction of the net force is negative Z direction
A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel-drive truck is 1 m/s2 , determine (a) the reaction at each of the front wheels, (b) the force between the boulder and the pallet.
Answer:
(a) reaction at each front wheel is 5272N (upward)
(b) force between boulder and pallet is 4124N (compression)
Explanation:
Acceleration of the truck [tex]a_{t[/tex] = 1 m/[tex]s^{2}[/tex] (to the left)
when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,
[tex]a_{A}[/tex] = 0.5 m/[tex]s^{2}[/tex] (upward) , [tex]a_{B}[/tex] = 0.5 m/[tex]s^{2}[/tex] (upward)
Let T be tension in the cable
pallet and boulder: ∑fy = ∑(fy)eff = 2T- ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])g = ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])[tex]a_{B}[/tex]
= 2T- (400 + 50)*(9.81 m/[tex]s^{2}[/tex]) = (400 + 50)*(0.5 m/[tex]s^{2}[/tex])
T = 2320N
Truck: [tex]M_{R}[/tex] = ∑([tex]M_{R}[/tex])eff: = [tex]-N_{f}[/tex] (3.4m) + [tex]m_{T}[/tex] (2.0m) - T (0.6m)= [tex]m_{T} a_{T}[/tex] (1.0m)
Nf = (2.0m)(2000 kg)(9.81 m/[tex]s^{2}[/tex] )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/[tex]s^{2}[/tex]) = 11541.2N - 409.4N - 588.2N = 10544N
∑fy (upward) = ∑(fy)eff: [tex]N_{f}[/tex] + [tex]N_{R}[/tex] - [tex]m_{T}[/tex]g = 0
10544 + [tex]N_{R}[/tex] - (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 0
[tex]N_{R}[/tex] = 9076N
∑fx (to the left) = ∑(fx)eff: [tex]F_{R}[/tex] - T = [tex]m_{T} a_{T}[/tex]
[tex]F_{R}[/tex] = 2320N + (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 4320N
(a) reaction at each front wheel:
1/2 [tex]N_{f}[/tex] (upward): 1/2 (10544N) = 5272N (upward)
(b) force between boulder and pallet:
∑fy (upward) = ∑(fy)eff: [tex]N_{B}[/tex] + [tex]M_{B}[/tex]g - [tex]m_{B}[/tex][tex]a_{B}[/tex]
[tex]N_{B}[/tex] = (400kg)(9.81 m/[tex]s^{2}[/tex]) + (400kg)(0.5 m/[tex]s^{2}[/tex]) = 4124N (compression)
Answer:
a) The reaction at each of the fron wheels is 5266.1 N
b) The force between the boulder and the pallet is 4120 N
Explanation:
The acceleration of truck is:
[tex]a_{T} =a_{A} +a_{B}[/tex]
Where
aA = acceleration of pallet = ?
aB = acceleration of boulder = ?
aA = aB
aT = acceleration of truck = 1 m/s²
[tex]1=a_{A} +a_{A}\\a_{A}=a_{B}=0.5m/s^{2}[/tex]
From diagram 1 and 2, the system of external forces is:
∑Fy = ∑(Fy)ef (eq.1)
From diagram 1:
∑Fy = 2T - g(mA + mB)
Where T = tension force
mA = mass of pallet = 50 kg
mB = mass of boulder = 400 kg
From diagram 2:
∑(Fy)ef = aB(mA + mB)
Substituting into equation 1:
[tex]2T-g(m_{A} +m_{B} )=a_{B} (m_{A} +m_{B} )\\T=\frac{a_{B}(m_{A} +m_{B}) +g(m_{A} +m_{B} ) }{2} =\frac{0.5(50+400)+9.8(50+400)}{2} =2317.5N[/tex]
From diagram 3 and 4, represents the system of external forces:
∑MR = ∑(MR)ef (eq. 2)
From diagram 3:
∑MR = -N(2 + 1.4) + mTg(2) - T(0.6)
Where
N = normal force
mT = mass of truck = 2000 kg
From diagram 4:
∑(MR)ef = mTaT
Substituting into equation 2:
[tex]-N(2+1.4)+m_{T} g(2)-T(0.6)=m_{T} a_{T} \\-N(3.4)+(2000*9.8*2)-(2317.5*0.6)=2000*1\\N=\frac{(2000*9.8*2)-(2317.5*0.6)-2000}{3.4} =10532.2N[/tex]
From diagram 3 and 4:
∑Fy = ∑(Fy)ef
[tex]N+N_{R} -m_{T} g=0\\10532.2+N_{R}-(2000*9.8)=0\\N_{R}=9067.8N[/tex]
a) The reaction at each of the front wheels is:
Rf = N/2 = 10532.2/2 = 5266.1 N
b) From diagram 5 and 6:
∑Fy = ∑(Fy)ef
[tex]N_{B} +m_{B} g=m_{B} a_{B} \\N_{B}-(400*9.8)=(400*0.5)\\N_{B}=4120N[/tex]
Two particles are separated by a certain distance. The force of gravitational interaction between them is F0. Now the separation between the particles is tripled. Find the new force of gravitational interaction F1. Express your answer in terms of F0.
Answer:
F'=1/9*F0
Explanation:
F0 is the gravitational force between the particles. When the distance is triplicated we have that
[tex]F'=G\frac{m_1m_2}{(3r)^{2}}[/tex]
where r is the distance before the particles are separated, m1 and m2 are the masses their masses and G is the Canvendish's constant.
By some algebra we have
[tex]F'=\frac{1}{9}G\frac{m_1m_2}{r^2}=\frac{1}{9}F_0[/tex]
hope this helps!!
Answer:
F₁ = [tex]\frac{1}{9}[/tex]F₀
Explanation:
Newton's law of universal gravitation states that the force of attraction or repulsion, F, between two particles of masses M₁ and M₂ is directly proportional to the product of these particles and inversely proportional to the square of the distance, r, between the two particles. i.e
F ∝ M₁M₂ / r²
F = GM₁M₂ / r² --------------------(i)
Where;
G is the constant of proportionality.
From equation (i), since the force is inversely proportional to the square of the distance, holding other variables constant, the equation can be reduced to;
F = k / r²
This implies that;
Fr² = k -------------------(ii)
Now, according to the question;
F = F₀
Substitute this into equation (ii) as follows;
F₀ r² = k ----------------(iii)
Also, when the distance of separation, r, is trippled i.e r becomes 3r;
F = F₁
Substitute these values into equation (ii) as follows;
F₁(3r)² = k
9F₁r² = k ---------------(iv)
Substitute the value of k in equation (iii) into equation (iv) as follows;
9F₁r² = F₀ r² --------------(v)
Cancel r² on both sides of equation (v)
9F₁ = F₀
Now make F₁ subject of the formula
F₁ = [tex]\frac{1}{9}[/tex]F₀
Therefore, the new force F₁ = [tex]\frac{1}{9}[/tex]F₀
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.
Complete Question:
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.
a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.
b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.
c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.
d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?
Answer:
force, 1.318 ₓ 10⁻⁴
direction, 18.435°
Explanation:
The attached file gives a breakdown step by step solution to the questions
A gas contracts from an initial volume of 3.83 L to a final volume of 2.33L against an external pressure of 810. mmHg. During the contraction the gas absorbs 10. J of heat. The gas is the system. Which of the following is true: 1. Heat flows into the system from the surroundings and the sign of a sys will be positive and the sign of w sys is positive. 2. The system is doing work on the surroundings. 3. Need more information such as an equivalence statement to answer this question. 4. Heat flows from the system into the surroundings and the sign of a sys will be positive and the sign of w sys is negative. 5. The change in the energy of the system AE is negative.
Answer:
HEAT FLOWS INTO THE SURROUNDING FROM THE SYSTEM AND THE SIGN OF THE SYSTEM WILL BE POSITIVE.
Explanation:
This is so of an endothermic reaction and the contraction of the gas is an endothermic one because the total heat content of the products formed is more than that of the reactants and the gas therefore absorbs heat from the surrounding during the contraction. An endothermic reaction is one in which heat is absorbed from the surroundings and the change in heat content ΔH of is positive.
The reaction for the contraction of the gas produces + 10 J of heat.
Light travels through a liquid at 2.25e8 m/s. What is the index of refraction of the liquid?
Answer:
Refractive index = 1.33
Explanation:
Refractive index is of a material explains how light passes through a medium. It is the ratio of speed of light in vacuum to the speed of light in the medium.
We know the speed of light in water is [tex] 2.25* 10^8 [/tex] and there is a change iñ speed of light in every medium it passes (except when it passes through vacuum). The change in speed of light is known to be = [tex] 2.99*10^8[/tex]
Given speed of light = 2.25*10^8 m/s
Formula for refractive index:
[tex]n = \frac{c}{v}[/tex]
Where,
change in speed of light, c= [tex] 2.99*10^8[/tex]
speed of light, v = 2.25*10^8 m/s
[tex] Refractive index, n = \frac{2.99*10^8}{2.55*10^8} [/tex]
= 1.328
= 1.33
The liquid is water, and the index of refraction of water is 1.33
As we know,
Speed of light, [tex]v = 2.25\times 10^8 \ m/s[/tex]Change in speed of light, [tex]c = 2.99\times 10^8[/tex]The refractive index will be:
→ [tex]n = \frac{c}{v}[/tex]
By substituting the values,
[tex]= \frac{2.99\times 10^8}{2.55\times 10^8}[/tex]
[tex]= 1.33[/tex]
Thus the approach above is right.
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Block A weighs 1.20 N and block Bweighs 3.60 N. The coefficient of kinetic frictionbetween all surfaces is 0.300. Find the magnitudeof the horizontal force F necessary to drag blockB to the left at constant speed a) if A rests on Band moves with it and b) if A is held at rest.
Answer:
Explanation:
a ) In this case A rests on B and both rest on horizontal surface
Both moves together .
Total reaction due to weight of both of them
R = 1.2 + 3.6 = 4.8 N
friction force by horizontal surface = μ R , μ is coefficient of friction .
= .3 x 4.8 = 1.44 N
Force equal to frictional force will be required to put both of them in uniform motion .
b ) If A is held stationary , friction force will arise at both , the upper and lower surface of B .
At upper surface friction force = μ x weight of A
= .3 x 1.2 = .36 N
At lower surface friction force = μ x weight of A +B
= .3 x 4.8
= 1.44
Total frictional force on B
= 1.8 N N .
So 1.8 N force will be required to put B in uniform keeping A stationary.
To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest. In both scenarios, the friction force between block B and the surface is given by the equation F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. The normal force is the sum of the weights of the blocks in scenario a) and the weight of block B in scenario b).
Explanation:To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest.
a) When block A rests on block B and moves with it, the friction force between the blocks and the surface is the force needed to overcome the friction. The friction force is given by the equation:
F_friction = μ * N
Here, μ is the coefficient of kinetic friction and N is the normal force between the blocks and the surface. In this case, the normal force N is the sum of the weights of both blocks, so N = m_A * g + m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_A * g + m_B * g)
b) When block A is held at rest, the friction force between block B and the surface is the force needed to overcome the friction. The friction force is the product of the coefficient of static friction and the normal force:
F_friction = μ * N
Here, μ is the coefficient of static friction and N is the normal force between block B and the surface. The normal force N is equal to the weight of block B, so N = m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_B * g)
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Monochromatic light of wavelength 491 nm illuminates two parallel narrow slits 6.93 μm apart. Calculate the angular deviation of the third-order (for m = 3) bright fringe (a) in radians and (b) in degrees.
To solve this problem we will apply the concepts related to the double slit experiment. Here we test a relationship between the sine of the deviation angle and the distance between slit versus wavelength and the bright fringe order. Mathematically it can be described as,
[tex]dsin\theta = m\lambda[/tex]
Here,
d = Distance between slits
m = Any integer which represent the order number or the number of repetition of the spectrum
[tex]\lambda[/tex] = Wavelength
[tex]\theta[/tex] = Angular deviation
Replacing with our values we have,
[tex](6.93*10^{-6}) sin\theta = (3)(491*10^{-9})[/tex]
[tex]\theta = sin^{-1} (\frac{(3)(491*10^{-9}}{6.93*10^{-6}) })[/tex]
Part A)
[tex]\theta = 0.2141rad[/tex]
PART B)
[tex]\theta = 0.2141rad(\frac{360\°}{2\pi rad})[/tex]
[tex]\theta = 12.27\°[/tex]
One end of a string 4.32 m long is moved up and down with simple harmonic motion at a frequency of 75 Hz . The waves reach the other end of the string in 0.5 s. Find the wavelength of the waves on the string. Answer in units of cm.
To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,
[tex]\text{Length of the string} = L = 4.32 m[/tex]
[tex]\text{Frequency of the wave} = f = 75 Hz[/tex]
[tex]\text{Time taken to reach the other end} = t = 0.5 s[/tex]
Velocity of the wave,
[tex]V = \frac{L}{t}[/tex]
[tex]V = \frac{4.32 m}{0.5s}[/tex]
[tex]V = 8.64m/s[/tex]
Wavelength of the wave,
[tex]\lambda = \frac{V}{f}[/tex]
[tex]\lambda = \frac{8.64m/s}{75Hz}[/tex]
[tex]\lambda = 0.1152m[/tex]
[tex]\lambda = 11.52cm[/tex]
Therefore the wavelength of the waves on the string is 11.53 cm
11.52cm
Explanation:The velocity, v, of a wave undergoing a simple harmonic motion is related to the wavelength, λ, and frequency, f, of the wave as follows;
v = f x λ -----------------(i)
But;
The velocity is also given as the ratio of the length, l, of the body producing the wave to the time taken, t, to undergo the motion. i.e
v = [tex]\frac{l}{t}[/tex] --------------(ii)
Now, substitute the value of v in equation (ii) into equation (i) as follows;
[tex]\frac{l}{t}[/tex] = f x λ ----------------(iii)
From the question,
l = 4.32m
t = 0.5s
f = 75Hz
Substitute these values into equation (iii) as follows;
[tex]\frac{4.32}{0.5}[/tex] = 75 x λ
Make λ subject of the formula;
λ = [tex]\frac{4.32}{0.5*75}[/tex]
λ = 0.1152m
Convert the value to cm by multiplying by 100
λ = 0.1152 x 100cm = 11.52cm
Therefore, the wavelength of the waves on the string is 11.52cm
A budding magician holds a 2.24-mW laser pointer, wondering whether he could use it to keep an object floating in the air with the radiation pressure. This might be an idea for a new trick! Assuming the laser pointer has a circular beam 3.00 mm in diameter and the magician rigs up a totally reflecting sail on which to shine the laser, what is the maximum weight the magician could suspend with this technique?
Answer:
W = 1.49 10⁻¹¹ kg
Explanation:
For this problem, let's use Newton's equation of equilibrium
F - W = 0
F = W (1)
Strength can be found from the definition of pressure
P = F / A
F = P A
The radiation pressure for a reflective surface is
P = 2 I / c)
We substitute in equation 1
2 I / c A = W
The intensity is defined by the ratio of the power between the area
I = P / A
P = I A
We substitute
2 P / c = W
W = 2 2.24 10-3 / 3 108
W = 1.49 10⁻¹¹ kg
Leading edge flaps can be used to decrease (or eliminate) the leading edge suction peak at a desired lift coefficient. When airfoils are designed for cruise performance, however, a better strategy is to design an airfoil that produces the correct lift with no suction peak using a cambered airfoil (i.e. without including leading edge flaps). To see that this is possible, we will consider the NACA 44XX airfoils. Also, p is the location of the maximum camber and is second digit/10. Apply thin airfoil theory to answer the following questions: (a) Determine the angle of zero lift for the 44XX airfoils (b) Determine the angle at which the suction peak is eliminated. We will call this the design angle of attack for the 44XX airfoils (c) What is the design lift coefficient for the 44XX airfoils (i.e. the lift coefficient at the design angle of attack)?
For the NACA 44XX series airfoils, using thin airfoil theory, the angle of zero lift is determined to be -1.6 degrees. The design angle of attack, which eliminates the suction peak, coincides with this angle. The design lift coefficient is 0.04, equivalent to the maximum camber of the airfoil.
The question provided revolves around the aerodynamic properties of the NACA 44XX series airfoils, specifically in terms of their performance at cruise without the need for leading edge flaps. Given the NACA numbering system, where the second digit represents the location of the maximum camber (p) as a fraction of the chord, we can determine several characteristics for the 44XX airfoils.
(a) The angle of zero lift for the 44XX airfoils can be found using thin airfoil theory, which tells us that for a symmetrical airfoil (e.g., NACA 0012), the angle of zero lift is 0 degrees. For cambered airfoils, the angle of zero lift becomes negative and is proportional to the camber. Specifically, the angle of zero lift is -4p degrees for cambered airfoils according to thin airfoil theory. Therefore, for the 44XX series with p value of 0.4 (from the second digit), the angle of zero lift is -4 * 0.4 = -1.6 degrees.(b) The design angle of attack, at which the suction peak is eliminated, would ideally coincide with the angle of zero lift, as this is where the airfoil is designed to operate without a strong suction peak. Thus, for the NACA 44XX airfoils, the design angle of attack would also be -1.6 degrees.(c) The design lift coefficient (CL design) for the NACA 44XX airfoils at the design angle of attack can be approximated using the lift curve slope for a thin airfoil, which is typically about 2pi per radian. Considering the zero angle of lift coefficient for a 4% cambered airfoil is equivalent to the maximum camber (which is 0.04 for this series), the lift coefficient at the design angle of attack would be the lift coefficient at zero angle of attack plus the product of the lift curve slope and the design angle of attack in radians, albeit the design angle of attack for these airfoils is the angle of zero lift, resulting in a design lift coefficient equivalent to the camber value, which is 0.04 or 0.4 in non-dimensional terms.
(a) Angle of zero lift: 0 degrees. (b) Design angle: ~1.376°. (c) Design lift coefficient: ~0.047. Utilized thin airfoil theory.
To address these questions, we can utilize thin airfoil theory, which provides approximate solutions for the aerodynamic characteristics of airfoils at small angles of attack. Let's go through each question:
(a) Determine the angle of zero lift for the NACA 44XX airfoils:
The angle of zero lift, also known as the zero-lift angle of attack [tex](\(\alpha_{L=0}\))[/tex], can be calculated using thin airfoil theory. For symmetrical airfoils like the NACA 44XX series, the zero-lift angle of attack is simply 0 degrees.
(b) Determine the angle at which the suction peak is eliminated. We will call this the design angle of attack:
The suction peak is eliminated when the airfoil is operating at its design lift coefficient [tex](\(C_{L_{\text{design}}}\))[/tex]. In thin airfoil theory, the lift coefficient is directly proportional to the angle of attack [tex](\(\alpha\))[/tex]. For cambered airfoils, this design lift coefficient is achieved at an angle of attack where the camber is aligned with the oncoming flow, resulting in no suction peak.
For NACA 44XX airfoils, the maximum camber is located at [tex]\(p = \frac{4}{10} = 0.4\)[/tex]. The angle of attack at which the camber aligns with the flow is called the camber line slope angle [tex](\(\alpha_{camber}\))[/tex]. For NACA 44XX series airfoils, the camber line slope angle can be approximated by:
[tex]\[\alpha_{camber} \approx 0.06 \times p = 0.06 \times 0.4 = 0.024 \text{ radians}\][/tex]
To convert this to degrees:
[tex]\[\alpha_{camber} \approx 0.024 \times \frac{180}{\pi} \approx 1.376^\circ\][/tex]
So, the design angle of attack where the suction peak is eliminated is approximately [tex]\(1.376^\circ\).[/tex]
(c) What is the design lift coefficient for the NACA 44XX airfoils?
At the design angle of attack, the lift coefficient [tex](\(C_L\))[/tex] can be calculated using thin airfoil theory. For symmetrical airfoils, [tex]\(C_L\)[/tex] is given by:
[tex]\[C_L = 2\pi \alpha\][/tex]
Substituting the design angle of attack [tex]\(\alpha_{camber}\)[/tex], we get:
[tex]\[C_{L_{\text{design}}} = 2\pi \times 1.376^\circ = 0.047\][/tex]
So, the design lift coefficient for the NACA 44XX airfoils is approximately [tex]\(0.047\).[/tex]
Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magnetic field of the two wires is zero if these currents are in the same direction.Assume that the positive x axis is directed from the 73.0-A wire to the 25.0-A wire perpendicular to the wires, with the origin on the 73.0-A wire.B). Find all locations where the net magnetic field of the two wires is zero if these currents are in the opposite direction.Assume that the positive x axis is directed from the 73.0-A wire to the 25.0-A wire perpendicular to the wires, with the origin on the 73.0-A wire.
Answer:
a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm
That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.
b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm
That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.
Explanation:
The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m
The magnetic field in a current carrying wire at a distance r from the wire is given by
B = (μ₀I/2πr)
μ₀ = magnetic constant = (4π × 10⁻⁷) H/m
a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.
According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.
The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.
Hence,
For the 73.0 A wire, I₁ = 73.0 A, r₁ = x
For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)
B = B₁ - B₂ = 0
(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0
(I₁/r₁) = (I₂/r₂)
(I₁/x) = [I₂/(0.4-x)]
(73/x) = [25/(0.4-x)]
73(0.4-x) = 25x
29.2 - 73x = 25x
73x + 25x = 29.2
98x = 29.2
x = (29.2/98) = 0.298 m
b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?
According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).
The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.
Hence,
For the 73.0 A wire, I₁ = 73.0 A, r₁ = x
For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)
B = B₁ - B₂ = 0
(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0
(I₁/r₁) = (I₂/r₂)
(I₁/x) = [I₂/(x-0.4)]
(73/x) = [25/(x-0.4)]
73(x-0.4) = 25x
73x - 29.2 = 25x
73x - 25x = 29.2
48x = 29.2
x = (29.2/48) = 0.608 m
Hope this Helps!!!
Final answer:
The net magnetic field can be found zero at certain points between or outside two long parallel wires carrying currents, either in the same or opposite directions, by equating the magnetic fields produced by each and solving for the distance.
Explanation:
The problem involves finding the locations where the net magnetic field is zero due to currents carried by two long parallel wires. The Biot-Savart Law or Ampère's Law can be used for such problems; however, in high school problems, we typically use the formula for the magnetic field due to a long straight wire, B = (μ0 * I) / (2π * r), where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance from the wire.
For wires with currents in the same direction, the magnetic fields due to each wire will point in opposite directions in the region between the wires and in the same direction outside the wires. To find locations where the net magnetic field is zero, we can equalize the magnetic fields coming from each wire and solve for the distance r from one of the wires where the fields cancel out.
For wires with currents in opposite directions, the magnetic fields due to the wires will be in the same direction in the region between the wires and in opposite directions outside of the wires. Again, we can find locations of zero net magnetic field by equalizing the magnitudes of the magnetic fields and solving for r.
Two slits separated by a distance of d = 0.190 mm are located at a distance of D = 1.91 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a monochromatic and coherent light source with a wavelength of ? = 648 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima.
At what angle from the beam axis will the first (m=1) maximum appear? (You can safely use the small angle approximation.)
Answer:
[tex]\theta = 0.195^0[/tex]
Explanation:
wavelength [tex]\lambda = 648 nm \ = 648*10^{-9}m[/tex]
d = 0.190 mm = 0.190 × 10⁻³ m
D = 1.91 m
By using the formula:
[tex]dsin \theta = n \lambda\\\\\theta = sin^{(-1)}(\frac{n \lambda}{d})\\\\\\\theta = sin^{(-1)}(\frac{1*648*10^{-9}}{0.190*10^{-3}})[/tex]
[tex]\theta = 0.195^0[/tex]
The first maximum will appear at an angle [tex]\theta = 0.195^0[/tex] from the beam axis
Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force between them?
A): 90 N, repulsive
B): 90 N, attractive
C): 23 N, attractive
D): 45 N, attractive
If one could answer this ASAP, it will be greatly appreciated
Answer:
Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?
Using Coulomb's law:
F = 1/(4πE) x Q1 x Q2/ r^2
Where
k= 1/(4πE) = 9 x 10^9Nm2/C2
Therefore,
F = 9x 10^9 x 2.50 x 10^-5 x2.50 x
10^-5/. ( 0.5)^2
F= 5.625/ 0.25
F= 22.5N approximately
F= 23N.
To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.
Hence To = 23N, attractive. C ans.
Thanks.
1.
An object (denoted by a dark circle) is placed in front of a plane mirror as shown below. Light
from the object emanates in a variety of directions. For each light ray incident to the mirror,
accurately draw the corresponding reflected ray. Use a protractor, straightedge, and the law
of reflection.
What’s the answer???
Answer:
Explanation:
solution is found below
A "home-made" solid propellant rocket has an initial mass of 9 kg; 6.8 kg of this is fuel. The rocket is directed vertically upward from rest, burns fuel at a constant rate of 0:225 kg=s, and ejects exhaust gas at a speed of 1980 m=s relative to the rocket. Assume that the pressure at the exit is atmospheric and that air resistance may be neglected. Calculate the rocket speed after 20 s and the distance traveled by the rocket in 20 s. Plot the rocket speed and the distance traveled as functions of time.
Answer:
v = 1176.23 m/s
y = 741192.997 m = 741.19 km
Explanation:
Given
M₀ = 9 Kg (Initial mass)
me = 0.225 Kg/s (Rate of fuel consumption)
ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)
v = ? if t = 20 s
y = ?
We use the equation
v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)
⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t
then we have
v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)
v = 1176.23 m/s
then we apply the formula
y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt
⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt
⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)
For t = 20 s we have
y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)
⇒ y = 741192.997 m = 741.19 km
The graphs are shown in the pics.
Following are the solution to the given points:
Given:
[tex]\to v_o =1980\ \frac{m}{s}\\\\[/tex]
[tex]\to a=g= -9.8 \frac{m}{s} \ \text{(Diection downwords )}[/tex]
Solution:
Using formula:
[tex]\to V= V_o +at\\\\[/tex]
[tex]\to v^2-v^2_0= 2ad \\\\[/tex]
For point a:
[tex]\to V= V_o +at\\\\[/tex]
[tex]=1980+ (-9.8) (20) \\\\[/tex]
[tex]= 1980-196 \\\\ =1784\ \frac{m}{s}\\\\[/tex]
The velocity at the conclusion of a [tex]20[/tex] second is [tex]\bold{1784\ \frac{m}{s}}\\\\[/tex].
For point b:
Using formula:
[tex]\to v^2-v^2_0= 2ad \\\\[/tex]
[tex]\to d=\frac{v^2_0 -V^2}{2a}\\\\[/tex]
[tex]= \frac{(1980)^2 - (1784)^2}{2\times 9.8} \\\\ =\frac{3920400-3182656}{19.6}\\\\= \frac{737744}{17.6}\\\\ =37640\ m \\\\ = 37.64\ km\\\\[/tex]
In [tex]20[/tex] seconds, the total distance traveled is [tex]401737 \ \ m (or \ 401.737\ \ km)[/tex].
For point c:
Please find the attached file.
The amount of pressure exerted by a solid is based on
the mass of the object.
the mass of the object and the area of the bottom surface.
the density of the object and the frictional coefficient of the surface it's
contacting.
the mass of the object and the density of the material it's contacting.
Answer:the mass Of The object,and there area Of The bottom surface
Explanation:
There mass Of The object, and the area Of The bottom surface
The amount of pressure exerted by a solid is based on the mass of the object and the area of the bottom surface. therefore the correct answer is option B.
What is pressure?The total applied force per unit of area is known as the pressure.
Both the external force being applied and the area to which it is being applied affect the pressure.
The mathematical expression for the pressure
Pressure = Force /Area
the pressure is expressed by the unit pascal or N /m²
A specific force produces less pressure when its region of influence is expanded. It is also true that increasing pressure is produced as a force's region of influence is reduced. For instance, because the force is acting across a smaller area at the pointy end of a high-heeled shoe, it applies more pressure there than it does at the flat end.
Thus, The mass of the object and the size of the bottom surface determine how much pressure a solid will exert. Therefore, choice B is the appropriate answer.
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Unpolarized light with intensity I0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at 42.0 ∘ to that of the first. Determine the intensity of the beam after it has passed through the second polarizer.
Answer:
[tex]0.276I_0[/tex]
Explanation:
When unpolarized light passes through a polarizer, only the component of the light vibrating in the direction parallel to the axis of the polarizer passes through: therefore, the intensity of light is reduced by half, since only 1 out of 2 components passes through.
So, after the first polarizer, the intensity of light passing through is:
[tex]I_1=\frac{I_0}{2}[/tex]
Where [tex]I_0[/tex] is the initial intensity of the unpolarized light.
Then, the light (which is now polarized) passes through the second polarizer. Here, the intensity of the light passing through the second polarizer is given by Malus Law:
[tex]I_2=I_1 cos^2 \theta[/tex]
where:
[tex]\theta[/tex] is the angle between the axes of the two polarizers
In this problem the angle is
[tex]\theta=42^{\circ}[/tex]
So the intensity after of light the 2nd polarizer is:
[tex]I_2=I_1 (cos 42^{\circ})^2=\frac{I_0}{2}(cos 42^{\circ})^2=0.276I_0[/tex]
The intensity of the beam after it has passed through the second polarizer should be 0.276 I0.
Calculation of the intensity:
Since
after the first polarizer, the intensity of light should be
I1 = I0/2
Here,
Io should be initial intensity of the unpolarized light
Now
The intensity should be
= I0/2(cos 42)
= 0.276 I0
Here theta be 42 degrees
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Every time your cats paw hits her toy string, it swings away from her. How does this demonstrate Newton’s third law of motion?
Answer:
This demonstrates Newton's Third Law because for every action there is an equal and opposite force.
Explanation:
The cat's paw hitting her toy string would be an action therefore when is swings to and from her it it creating and equal and opposite force. In other words, the cat hits it and when the string comes back to her it's coming back the same force just opposite of the direction she hit.
Please use Gauss’s law to find the electric field strength E at a distance r from the center of a sphereof radius R with volume charge density ???? = cr 3 and total charge ????. Your answer should NOT contain c. Be sure to consider regions inside and outside the sphere.
Answer:
See the explaination for the details.
Explanation:
Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
Please kindly check attachment for the step by step explaination of the answer.
An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 atm, but inside the pressure is held at 1.0 atm. What net force pushes out on the window
Answer:
Net forces which pushes the window is 30342.78 N.
Explanation:
Given:
Dimension of the office window.
Length of the window = [tex]3.1[/tex] m
Width of the window = [tex]2.1[/tex] m
Area of the window = [tex](3.1\times 2.1) = 6.51\ m^2[/tex]
Difference in air pressure = Inside pressure - Outside pressure
= [tex](1.0-0.954)[/tex] atm = [tex]0.046[/tex] atm
Conversion of the pressure in its SI unit.
⇒ [tex]1[/tex] atm = [tex]101325[/tex] Pa
⇒ [tex]0.046[/tex] atm = [tex]0.046\times 101325 =4660.95[/tex] Pa
We have to find the net force.
We know,
⇒ Pressure = Force/Area
⇒ [tex]Pressure=\frac{Force }{Area}[/tex]
⇒ [tex]Force =Pressure\times Area[/tex]
⇒ Plugging the values.
⇒ [tex]Force =4660.95\times 6.51[/tex]
⇒ [tex]Force=30342.78[/tex] Newton (N)
So,
The net forces which pushes the window is 30342.78 N.
A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop. The cars ride on the inside of the loop at the top, and the speeds are fast enough to ensure that the cars remain on track. The biggest loop is 40.0m high. Suppose the speed at the top is 14.4m/s and the corresponding centripetal acceleration is 2g.
(a) What is the radius of the arc of the teardrop at the top?
(b)If the total mass of a car plus the riders is M, what force does the rail exert on the car at the top?
(c) Suppose the roller coaster had a circular loop of radius 21.4 m. If the cars have the same speed, 14.4 m.s at the top, what is the centipetal acceleration at the top?
Answer:
a)radius of the arc of the teardrop at the top is 10.58m
b)T = mg
c) the centipetal acceleration at the top is 14.5m/s
Explanation:
Part A
Given that roller coaster is of tear drop shape
So the speed at the top is given as
v = 14.4 m/s
acceleration at the top is given as
a = 2 g
[tex]a = 2(9.8) = 19.6 m/s^2[/tex]
now we know the formula of centripetal acceleration as
[tex]a = \frac{v^2}{R}\\[/tex]
[tex]19.6 = \frac{14.4^2}{R}\\R = 10.58 m[/tex]
Part B
now
the total mass of the car and the ride is M
Let the force exerted by the track be n
By Newton law
[tex]n +Mg =\frac{Mv^2}{r} \\\\n=\frac{Mv^2}{r} -Mg\\\\=M(\frac{v^2}{r}-g )\\\\=M(2g-g)\\\\T=Mg[/tex]
Part C
If the radius of the loop is 21.4 m
speed is given by same v = 14.4 m/s
now the acceleration is given as
[tex]a = \frac{v^2}{R}[/tex]
[tex]a = \frac{14.4^2}{21.4} \\\\= 9.69 m/s^2[/tex]
Now for normal force at the top is given by force equation
[tex]F_n + mg = ma\\F_n = m(a-g)[/tex]
The force exerted by the rail is less than zero because acceleration is less than 9.69m/s²
So the normal force would have to point away from the centre, For safe ride this normal force must be positive i.e [tex]a \prec g[/tex]
[tex]\frac{v^2}{r} \prec \sqrt{g} \\\\v = \sqrt{rg} \\\\v = \sqrt{21.4 \times 9.8} \\\\v = 14.5m/s[/tex]
Does a 0.14 kg baseball moving at 41 m/s or a 0.058-kg tennis ball moving at 67 m/s
have more kinetic energy?
Answer:
The tennis ball has more kinetic energy
Explanation:
Recall the formula for kinetic energy: [tex]K=\frac{1}{2} \,m\,v^2[/tex] , so we can estimate it for each case and compare the results:
For the baseball:
[tex]K=\frac{1}{2} \,m\,v^2\\K=\frac{1}{2} \,0.14\,(41)^2\,\,J\\K=117.67\,\,J[/tex]
For the tennis ball:
[tex]K=\frac{1}{2} \,m\,v^2\\K=\frac{1}{2} \,0.058\,(67)^2\,\,J\\K=130.181\,\,J[/tex]
Therefore, the tennis ball has more kinetic energy
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.17 mm and place your screen 8.95 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.57 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength λ expressed in nanometers?
Answer:
laser's wavelength λ = 597.4 nm
Explanation:
Given:
Slit spacing, d = 1.17mm
Tenth bright fringe y = 4.57cm
Distance from slits, D = 8.95m
n = 10
λ = (d * y) / (D * n)
λ = (1.17x10⁻³ * 4.57x10⁻²) / (8.95 x 10)
λ = 5.3469x10⁻⁵ / 8.95x10¹
λ = 0.5974 x 10⁻⁵⁻¹
λ = 0.5974 x 10⁻⁶ m
λ = 597.4 x 10⁻⁹ m
λ = 597.4 nm