The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
Speed of the satelliteThe speed of the satellite is calculated as follows;
v = √GM/r
where;
M is mass of Earthr is radius of satellitev = √[(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) / (3.57 x 6.37 x 10⁶)]
v = 4,188 m/s
Thus, the speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
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1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.
(a) Find the position of the object.
(b) What is the magnification of the mirror?
2.A contact lens is made of plastic with an index of refraction of 1.45. The lens has an outer radius of curvature of +2.04 cm and an inner radius of curvature of +2.48 cm. What is the focal length of the lens?
The position of the object is = -68cm
The magnification of the mirror= 0.3
Calculation of object distanceThe image distance = 20.5cm
The focal length= R/2 = 31.5/2= 15.75
The object distance= ?
Using the lens formula,1/f = 1/v-1/u
1/u = 1/v- 1/f
1/u = 1/20.5 - 1/15.75
1/u = 0.0489- 0.0635
1/u = -0.0146
u = -68cm
The magnification of the mirror is image size/object size
= 20.5cm/-68cm
= 0.3
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A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .
What is the average power delivered by the engine?
Express your answer to two significant figures and include the appropriate units.
From the calculations, the power expended is 43650 W.
What is the power expended?Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
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White light is spread out into its spectral components by a diffraction grating. If the grating has 1975 lines per centimeter, at what angle does red light of wavelength 640 nm appear in first-order spectrum? (Assume that the light is incident normally on the grating.)
The angle of the red light is mathematically given as
[tex]\theta = 7.24 \textdegree[/tex]
What angle does red light of wavelength 640 nm appear in the first-order spectrum?Generally, the equation for the grating element is mathematically given as
d= 1 / N
Therefore
d= 1/1965
d= 5.089 * 10^{-6} m
Generally, the equation for the differential formula is mathematically given as
[tex]d sin \theta = m\lambda[/tex]
Therefore
[tex]sin \theta = \lambda / d[/tex]
[tex]sin \theta= (640 * 10 ^ {-9} m)/(5.089 * 10 ^ {-6} m)[/tex]
[tex]\theta = 7.24 \textdegree[/tex]
In conclusion, The angle of the red light
[tex]\theta = 7.24 \textdegree[/tex]
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An Airbus A380-800 passanger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? The average density of the air around the airplane is ρ^air = 1.21 kg/m^3.
The mass of the airplane of area of two wings 395m², and the average speed of lower and upper surface of the wings are, 259 and 288m/s is 386.8×10^3 kg.
To find the answer, we need to know about the Bernoulli's principle.
How to find the mass of the airplane?The Bernoulli's principle states that, the sum of pressure energy, kinetic energy and potential energy of an incompressible, non-viscous, fluid in streamlined flow is a constant.It can be expressed as,P+ [tex]\frac{1}{2}[/tex] ρv²+ρgh=a constant.
Instead of ρ we take d as density.
We have given that,[tex]A= 395 m^2\\v_l=295 m/s\\v_u=288m/s\\density=1.21kg/m^3[/tex]
We equate the principle for lower and upper surfaces of the wing like,[tex]P_1+\frac{1}{2}v_1^2d+dgh_1=P_2+ \frac{1}{2}v_2^2d+dgh_2\\here\\h_1=h_2\\thus\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\P_1-P_2=\frac{1}{2}*1.21(288^2-259^2)=9597.12 atm\\[/tex]
Thus, the mass of the airplane from the above equation will be,[tex]\frac{F}{A}=9597.12 atm\\\\ F=9597.12*395m^2=37.9*10^5 N\\\\mg=37.9*10^5 N\\\\m=\frac{37.9*10^5 N}{9.8}\\\\ m=386.8*10^3kg[/tex]
Thus, we can conclude that, the mass of the airplane is 386.8×10^3 kg.
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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:P+ [tex]\frac{1}{2}[/tex]dv²+dgh = constant.
We substitute d for to represent density.
We've done that,[tex]v_1=259m/s\\v_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dv_1^2+dgh= P_2+\frac{1}{2}dv_2^2+dgh\\\\P_1+\frac{1}{2}dv_1^2=P_2+\frac{1}{2}dv_2^2\\\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\\\P_1-P_2=9597 atm[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]F/A= 9597 atm\\mg=9597*395 =38*10^5N\\m=\frac{38*10^5}{9.8} = 387*10^3kg[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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Q2. A force of 1000N is experienced by a charge in a field of strength of 250NC. Find the value of the charge establishing the field.
Answer:
4 C
Explanation:
The strength of an electric field can be defined as: [tex]E = \frac{f}{q}[/tex] where f=force and q=charge, and e=strength
Since we're given the strength and the force we can simply rearrange the equation so that we solve for Q:
Original Equation:
[tex]E=\frac{f}{q}[/tex]
Multiply both sides by q
[tex]E*q = f[/tex]
Divide both sides by E
[tex]q=\frac{f}{e}[/tex]
So now plug the known values into the equation
[tex]q=\frac{1000 N}{250 N/C}[/tex]
Simplify:
[tex]q = 4 c[/tex]
How big is this restoring force compared with the tensile force stretching the spring?
A. Bigger
B. Not enough info
C. Smaller
D. Same size
The restoring force on the spring is found to have exactly the same magnitude as the stretching force. Option D
What is the restoring force?The restoring force is the force that seeks to restore the spring to its equilibrium position. It has the same magnitude as the stretching force but acts in opposite direction.
Thus, the restoring force on the spring is found to have exactly the same magnitude as the stretching force.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
Tension in the cableApply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
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If the range of a projectile is and 256√3 m in the maximum height reached is 64 m. calculate the angle of projection
The angle of projection given that the range is 256√3 m and the maximum height reached is 64 m is 30°
Data obtained from the questionRange (R) = 256√3 mMaximum height (H) = 64 mAcceleration due to gravity (g) = 9.8 m/s²Angle of projection (θ) = ?How to determine the angle of projectionR = u²Sine(2θ) / g
256√3 = u²Sine(2θ) / 9.8
Cross multiply
256√3 × 9.8 = u²Sine(2θ)
Divide both sides by Sine(2θ)
u² = 256√3 × 9.8 / Sine(2θ)
H = u²Sine²θ / 2g
64 = [256√3 × 9.8 / Sine(2θ)] × [Sine²θ / 2 × 9.8]
64 = [256√3 / Sine(2θ)] × [Sine²θ / 2]
Recall
Sine²θ = SineθSineθ
Sine2θ = 2SineθCosθ
Thus,
64 = [256√3 / 2SineθCosθ] × [SineθSineθ / 2]
64 = 256√3 × Sineθ / 4Cosθ
Recall
Sineθ / Cosθ = Tanθ
Thus,
64 = 256√3 / 4 × Tanθ
Divide both side by 256√3 / 4
Tanθ = 64 ÷ 256√3 / 4
Tanθ = 0.5774
Take the inverse of Tan
θ = Tan⁻¹ 0.5774
θ = 30°
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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 Force on the left hand pole, F' = 0.167N
What is the force on the left hand pole?Force is an agent which produces a change in the motion or state of an object.
Force is a vector quantity.
The general force is calculated as follows:
F = mg/sinθ
m = 17.1 g = 0.0171 kg
g = 9.81 m/s²
θ = 45°
F = 0.0171 * 9.81/sin45
F = 0.237 N
Force on the left hand pole, F' = Fcosθ
F' = 0.237 * cos 45
F' = 0.167N
In conclusion, the force on the left hand pole is the horizontal component of force.
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Compare and contrast visible light, infrared light, and ultraviolet light.
Answer:
NE BİLİM
Explanation:
PUAN İÇİN YAZIYORUM SEN BENİ GÖRMEZDEN GEL ANLDIN MI DOSTUM bu arada Suriye >>>>>
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 36.5°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.
(a)the frog's launch speed (in m/s)
(b)the time the frog spends in the air (in s)
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
What's the expression of range of a projectile motion?Range = U²× sin(2θ)/gU= initial velocity, θ= angle of projectile and g= acceleration due to gravity U=√{Range×g/sin(2θ)}Here, range= 2.20m, = 36.5°U= √{2.20×9.8/sin(73)}U= √{2.20×9.8/sin(73)} = 22.5m/s
What's the expression of time of flight in projectile motion?Time of flight= (2×U×sinθ)/g So, T= (2×22.5×sin36.5°)/9.8= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
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It's a snowy day and you're pulling a friend along a
level road on a sled. You've both been taking physics,
so she asks what you think the coefficient of friction
between the sled and the snow is. You've been
walking at a steady 1.5 m/s, and the rope pulls up
on the sled at a 35 ° angle. You estimate that the
mass of the sled, with your friend on it, is 57 kg and
that you're pulling with a force of 75 N
The coefficient of friction between the sled and the snow is 0.119.
To find the answer, we need to know about the friction.
How to find the coefficient of friction between the sled and the snow?Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.To solve the problem, we have to draw the free body diagram of the given system.We have given with the following values,[tex]a=0\\\alpha =35^0\\T=75N\\m=57kg[/tex]
Here, acceleration will be equal to zero, because the velocity is given as constant.
Thus, from the diagram, we can write the balancing equations as follows,[tex]ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha[/tex]
Substituting N in f and f in the equation of ma, then we get,[tex]ma= Tcos\alpha -k(mg-Tsin\alpha )[/tex]
Substituting values, we get the coefficient of friction as,[tex]0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119[/tex]
Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.
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The sled's coefficient of friction with the snow is 0.119.
We must understand the friction in order to choose the solution.
How can I determine the sled and snow's coefficient of friction?A force that works parallel to the surface of contact and opposes the relative motion is present whenever one body moves over the surface of another body. Friction is the name for this opposing force.We must create the given system's free body diagram in order to solve the issue.The values that we have provided are[tex]\alpha =35\\T=75N\\m=57kg\\a=0[/tex]
Because the velocity is specified as constant in this case, the acceleration will be equal to zero.
Consequently, we can express the balancing equations as follows using the diagram:[tex]ma=Tcos\alpha -f\\ where,f=kN\\N+Tsin\alpha =mg\\ thus,\\N=mg-Tsin\alpha[/tex]
When we substitute N for f and f in the equation for ma, we obtain,[tex]ma=Tcos\alpha -k(mg-Tsin\alpha )[/tex]
By substituting values, we obtain the friction coefficient as.[tex]k=0.119[/tex]
As a result, we may say that there is 0.119 coefficient of friction between the sled and the snow.
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A person standing at the edge of a cliff throws one ball straight up and another ball straight down, each at the same initial speed. Neglecting air resistance, which ball hits the ground below the cliff with the greater speed?
A 500 N force accelerates an object at 20 m s-2. What is its mass?
Answer: The mass of the object is 25kg.
The given question deals with Newton's second law of motion and its applications.
Explanation: Given force, F=500N
acceleration, a=20 m/[tex]s^{2}[/tex]
From Newton's 2nd law of motion , we have
F=ma where m=mass of the object
⇒500=m×20
⇒m=500/20=25
∴ Mass of the object is 25 kg .
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(b) The particle displacement y of air molecules due to a sound wave is given by y 4m and w = = 0.008 cos wt sin kz. Where k - m 50m rad/s. calculate i. The distance between two consecutive nodes ii. The amplitude after 0.56s
The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.
What's the distance between consecutive nodes of the displacement of air molecules?Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.So, distance between consecutive nodes = wavelength = 2π÷k= 2π/(4π÷m)
= m/2
What's the amplitude after 0.56s of the displacement of air molecules?Displacement after 0.56 s = 0.008×cos(50π×0.56s)
=1.75×10^(-4) m
Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.
Calculate:
I) the distance between 2 consecutive nodes
ii) the amplitude after 0.565s
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Light of intensity I0 is polarized vertically and is incident on an analyzer rotated at an angle from the vertical. Find the angle if the transmitted light has intensity
I = (0.750)I0, I = (0.500)I0, I = (0.250)I0, and I = 0.
(Enter your answers in degrees.)
a. θ = 41. 4°
b. θ = 60°
c. θ = 75. 5°
d. θ = 90°
How to determine the angleFrom the given information, we would be using the Malus' law
It is given as;
I = I0 cos²θ
Where I0 is the intensity of the polarized light after passing through P
a. To find the angle, compare with the given equation
I = (0.750)I0
I = I0 cos θ
then
cos θ = 0. 750
θ = [tex]cos^-^1(0. 750)[/tex]
θ = 41. 4°
b. I = (0.500)I0
cos θ = 0. 500
θ = [tex]cos^-^1(0. 500)[/tex]
θ = 60°
c. I = (0.250)I0
cos θ = 0. 250
θ = [tex]cos^-^1 (0. 250)[/tex]
θ = 75. 5°
d. I = 0
cos θ = 0
θ = [tex]cos^-^1 (0)[/tex]
θ = 90°
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A 80-kg base runner begins his slide into second base when he is moving at a speed of 3.0 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
______J
(b) How far does he slide?
________m
The mechanical energy lost due to friction is 360 J.
The distance the runner slides is, s = 0.655 m
What is the mechanical energy of the runner?
The mechanical energy lost due to friction acting on the runner is equal to the change in kinetic energy.
Change in Kinetic energy = 1/2m(v -u)²
Change in Kinetic energy = 80 * (0 - 3.0)²/2
Change in Kinetic energy = 360 J
Mechanical energy lost = 360 J
Distance he slide is determined using the formula below as follows:
Acceleration of runner = coefficient of friction * acceleration due to gravity
acceleration, a = 0.7 * 9.8 = 6.86
v² = u² - 2as
s = v² + u²/2a
s = 0 + 3³/2 * 6.86
s = 0.655 m
In conclusion, the mechanical energy lost due to friction is the loss in kinetic energy.
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For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.
polystyrene
flint glass
(1) The critical angles for the air is 90⁰,
(2) The critical angles for the polystyrene is 39.1 ⁰ and
(3) The critical angles for the flint glass is 37.2 ⁰.
Critical angles for the different mediumθc = sin⁻¹( 1/η)
where;
η is the refractive indexRefractive index of air = 1
Refractive index of polystyrene = 1.5865
Refractive index of flint glass = 1.655
Critical angles for airθc = sin⁻¹( 1/1)
θc = 90⁰
Critical angles for polystyreneθc = sin⁻¹( 1/1.5865)
θc = 39.1 ⁰
Critical angles for flint glassθc = sin⁻¹( 1/1.655)
θc = 37.2 ⁰
Thus, the critical angles for the air is 90⁰, the critical angles for the polystyrene is 39.1 ⁰ and the critical angles for the flint glass is 37.2 ⁰.
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In the given figure, weight of stone inside water
is 9N and water displaced by stone is 2N then,
i)What is the actual weight of stone?
ii) Which principle is the
experiment based on?
The actual weight of the stone is 11 N. It is based on the Archimedes principles.
What is Archimedes principle?Archimedes principle states that the up thrust by water on an object is equal to the weight of water displaced.Upthrust by water on an object= actual weight of object - weight inside waterWhat is the actual weight of the object, if its weight inside water is 9N and weight of water displaced is 2N?Actual weight= weight inside water+ weight of water displaced
= 9N + 2N = 11N
Thus, we can conclude that the actual weight of the object is 11N.
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A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.05 m/s. If the roof is pitched at 34.0° below the horizon and the roof edge is 2.10 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.
(a) the time the baseball spends in the air (in s)
(b) the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m)
(a) The time the baseball spends in the air is 0.92 s.
(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
Time spent in air by the baseball
h = vt - ¹/₂gt²
-2.1 = (4.05 x sin 34)t - ¹/₂(9.8)(t²)
-2.1 = 2.26t - 4.9t²
4.9t² - 2.26t - 2.1 = 0
t = 0.92 s
Horizontal distance traveled by the baseballR = Vx(t)
R = (4.05 x cos 34)(0.92)
R = 3.1 m
Thus, the time the baseball spends in the air is 0.92 s.
The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
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"You can't see the forest for the trees" might seem an appropriate analogy for astronomers attempting to determine the shape of the Milky Way galaxy when we are in fact located inside the galaxy. Discuss techniques used by astronomers to determine the type of galaxy in which we live, why it is so difficult to determine the shape of our galaxy, and where our Sun is located in our galaxy?
Some of the techniques used by astronomers to determine the type of galaxy in which we live are:
radio, optical, infraredx-ray astronomyWhat is Astronomy?This refers to the study of heavenly bodies and space and other things that space is made up of.
Hence, we can see that the reason why it is so difficult to determine the shape of our galaxy is that astronomers can only infer its presence from the motions of stars in the galaxy, and a precise shape is difficult to be determined.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The vertical component of force exerted by the hi.nge on the beam will be,142.10N.
To find the answer, we need to know more about the tension.
How to find the vertical component of the force exerted by the hi.nge on the beam?Let's draw the free body diagram of the system.To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.[tex]F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\[/tex]
To find the answer, we have to find the tension,[tex]Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N[/tex]
Thus, the vertical component of the force exerted by the hi.nge on the beam will be,[tex]F_V=(29*9.8)-(169.43*sin57)=142.10N[/tex]
Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.
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The hi.nge will apply a force of 142.10N on the beam in the vertical direction.
We must learn more about the tension in order to find the solution.
How can I determine the vertical component of the force the hi.nge has on the beam?Let's create the system's free body diagram.We must balance the total vertical force to zero in order to get the vertical component of the force applied to the beam by the height.[tex]F_V=mg-Tsin\alpha[/tex]
We must identify the tension in order to find the solution.[tex]Tlsin\alpha =mg\frac{l}{2}sin\beta \\T=\frac{mgsin90}{2sin57} =169.43N[/tex]
Consequently, the force that the height exerts on the beam will have a vertical component that is,[tex]F_v=(29*9.8)-(169.43*sin57)=142.10N[/tex]
This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.
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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam = 8.662×101 N
What is the magnitude of the force that the beam exerts on the hi.nge?
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
How to find the magnitude of the force that the beam exerts on the hi.nge?Let's draw the free body diagram of the system using the given data.From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.[tex]N_x=86.62N[/tex]
We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.[tex]N_y=F_V=mg-Tsin59\\[/tex]
To find Ny, we need to find the tension T.For this, we can equate the net horizontal force.[tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]
Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,[tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]
Thus, the magnitude of the force that the beam exerts on the hi.nge will be,[tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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The hi.nge will be subjected to a force of 261.12N from the beam.
We must understand the tension in order to choose the solution.
How can the amount of force the beam applies on the height be determined?Let's use the provided information to create the system's free body diagram.We need to calculate the force the beam is exerting on the height using the diagram.For this, it is assumed that the horizontal component of force is 86.62N, the same as the horizontal component of the normal reaction that the beam exerts on the height.We need to identify the vertical component of the normal reaction the beam exerts on the height. We must equalize the total force acting in the vertical direction to achieve this.[tex]N_y=F_v=mg-Tsin59[/tex]
Finding the tension T is necessary to determine Ny. Thus, we can use the net horizontal force to equate this.[tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]
As a result, the normal reaction that the beam has on the height becomes, with a vertical component,[tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]
As a result, the force the beam applies on the height will be of the order of,[tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]
Thus, we can infer that the force the beam applies to the height is 261.12N in size.
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(These are multiple choice questions)
1) The magnitude of the charge of five electrons is
A. 1.0×10-19 C.
B. 1.6×10-19 C.
C. 6.0×10-19 C.
D. 8.0×10-19 C.
2) Total resistance (in ohms-Ω) between the points A and B of the following circuit is
A. 70 Ω
B. 30 Ω
C. 25 Ω
D. 20 Ω
3) Given the electron configuration of a neutral atom 1s22s22p63s23p5. The atom
A. can become a negative ion easily.
B. is a halogen.
C. has 7 electrons in the outer most shell.
D. all of the above.
4) Identify the number of protons, neutrons and electrons in an atom of
A. 136 protons, 92 neutrons and 92 electrons
B. 92 protons, 136 neutrons and 92 electrons
C. 92 protons, 138 neutrons and 92 electrons
D. 230 protons, 92 neutrons and 92 electrons
5)Which of the following is not an application of Total Internal Reflection?
A. mirage
B. optical fiber
C. prismatic binocular
D. hologram
The application of total Internal Reflection occurs in a mirage. Option A
What is the charge?1) We know that a charge can be positive or negative. If the charge is negative, we call it an electron. If the charge is positive, we call it a proton. Now we know that the magnitude of charge on each electron is 1.6×10-19 C. Hence, the magnitude of charge on five electrons = 5 * 1.6×10-19 C = 8.0×10-19 C. Option D
2) The details of question 2 are not shown hence the question can not be answered.
3) Looking at the electronic configuration of the element; 1s22s22p63s23p5, it is clear to see that it has the ns2 np5 outermost configuration that is common to halogens thus it;
can become a negative ion easily. is a halogen.has 7 electrons in the outer most shell.4) The atom is composed of the protons, neutrons and electrons. Given the nuclide identified as 230/92U we have 92 protons, 138 neutrons and 92 electrons. Option C
5) The application of total Internal Reflection occurs in a mirage. Option A
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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.
Tension in rope A:
7.42×102 N
Tension in rope B:
7.52×102 N
What is the tension in rope C?
Tension in the rope C is 1.24× 10² N.
To find the answer, we need to know about the horizontal component of tension in the rope B.
What's the angle made by the rope B by horizontal?From the figure of the answer, in the triangle PQR, tan(θ)= PQ/QR = 6/1= 6 θ= tan inverse of 6 = 80.5°What's the horizontal component of the tension in rope B?Horizontal component= tension in rope B × cos80.5°
= 7.52×10² N × cos80.5°
= 1.24×10² N
What's the tension in the rope C?From the figure, we have found that the tension in rope C = horizontal component of the tension in rope BSo, tension in rope C= 1.24×10² NThus, we can conclude that the tension in the rope C is 1.24× 10² N.
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Define a dipole. Hence, write the expression for calculating the electric moment of a dipole
Answer:
Explanation:
An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa
What word means the same thing as force of gravity?
A. Density
B. Weight
C. Mass
D. Volume
Answer:
B. Weight
Explanation:
Weight is the force of gravity exerted on a body
OR
It's the force exerted on a body by the influence of the earth's gravitational force
I AM OFFERING 100 POINTS!!! I REALLY NEED HELP!!!!!!
Part C
Now prepare the cold sand and cold water samples from part A:
Fill a 100-milliliter container with 50 grams of sand. Fill a 100-milliliter container with 50 grams of cold tap water. Fill the last 100-milliliter container with 100 grams of cold tap water. Use the scale to measure the masses.
*Image should be there*
2. Pour all the ice cubes into a tub, and fill it with cool tap water to a depth of 2 inches. Place the sand and water samples in the ice water. Cover the entire tub.
*Image should be there*
3. Every 15 minutes, remove the cover and check the temperatures of the samples using the three thermometers. Wait 30 seconds before recording the thermometer reading. Once the temperatures of the three samples are no more than a degree apart, record the temperatures.
*Image should be there
Answer:
Every 15 minutes, remove the cover and check the temperatures of the samples using the three thermometers. Wait 30 seconds before recording the thermometer reading. Once the temperatures of the three samples are no more than a degree apart, record the temperatures.
(three 100-milliliter containers (each with a thermometer) in an ice bath inside an uncovered basin, with one container holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)
Next, prepare the hot water.
-Fill each of the three 200-milliliter containers with 100 grams of hot tap water. Measure the mass using the mass scale.
(100 grams of hot water in a 200-milliliter container atop a mass scale)
-Prepare a hot-water bath by boiling water in a pot. Use the heat mitts to pour the hot water from the pot into the second tub. Fill the tub to a depth of about 2 inches. Carefully place the three containers of hot water into the bath without submerging them. Cover and wait for five minutes until the temperatures stabilize.
(three 200 mL containers (each holding 100 grams hot water) in a covered tub containing 2 inches of boiling water)
Prepare to mix the cold samples with hot water:
-Have three empty 300-milliliter mixing containers and three thermometers ready. Timing is important. Uncover the cold-water bath, and pour each cold sample into a different mixing container.
(three 300-milliliter containers (each with a thermometer), one holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)
-Uncover the hot-water bath, and use a heat mitt to remove the three containers of hot water from the hot-water bath. Pour 100 grams of hot water into each mixing container.
(three 300 mL containers (each with a thermometer), one holding a 150-gram mixture of sand and hot water, one holding a 150-gram mixture of cold water and hot water, and one holding a 200-gram mixture of cold water and hot water)
Explanation:
What do an electron and a neutron have in common?
Answer:
To give light yo people and the neutron are to give you health care
An object of height 8.50 cm is placed 20.0 cm to the left of a converging lens with a focal length of 12.0 cm. Determine the image location in cm, the magnification, and the image height in cm.
The image distance is 33.3 cm while the image height is 14.2 cm.
What is a converging lens?A converging lens will always have a positive focal length hence, we have to find the object distance as follows;
1/f = 1/v + 1/u
1/12 = 1/v + 1/20
1/v = 1/12 - 1/20
1/v = 0.08 - 0.05
v =33.3 cm
Now;
Magnification = 33.3 cm/20.0 cm =1.67
M = Image height/Object height
1.67 = Image height/8.50 cm
Image height = 1.67 * 8.50 cm
Image height = 14.2 cm
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