vr>vs because the rolling ball acquires rotational as well as translational kinetic energy.
The accelerating force acting on the ball as it goes along a smooth plane is mgsin. Its acceleration is therefore equal to gsin. The mgsin acts down the plane as the ball travels down the rough inclined plane, but friction develops that acts up the plane.
Since both balls' potential energy is lost at the same rate, their KEs are actually equal at the base of the planes. However, a ball sliding down a smooth plane has only translational kinetic energy, but a ball rolling down a rough plane contains both translational and rotational kinetic energy at the bottom of the plane. As a result, the ball's translational KE will be lower than its translational Kinetic energy.
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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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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
How to find the initial speed of the rock as it left the astronaut's hand?We have given with the following values,[tex]m_g=7.88*10^{18}kg\\r_g=6.32*10^4 m.\\h_{max}=1.44*10^3m.\\[/tex]
We have the expression for the initial velocity as,[tex]v=\sqrt{2gh}[/tex]
Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,[tex]g_g=\frac{GM_g}{r_g^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} =0.132m/s^2.[/tex]
Now, the velocity will become,[tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s.[/tex]
How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,[tex]v=\sqrt{ \frac{GM}{r}}=\sqrt{\frac{7.88*10^{18}*6.67*10^{-11}}{1.45*10^5 }} =3.624km/s.[/tex]
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
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1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Understanding the planetary motion equations is necessary in order to determine the solution.
How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:[tex]v=\sqrt{2gh}[/tex]
So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,[tex]a=\frac{GM}{R^2} =0.132m/s^2[/tex]
The velocity will now change to,[tex]v=\sqrt{2*9.8*0.132} =19.46m/s[/tex]
How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:[tex]v=\sqrt{\frac{GM}{R} } =3.624km/s\\where, M=7.88*10^{18}kg[/tex]
Consequently, we can say that
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
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A computer monitor accelerates electrons between two plates and sends them at high speed to form an image on the screen. If the elec- trons gain 4.1 * 10-15 J of kinetic energy as they go from one accelerat- ing plate to the other, what is the voltage between the plates?
The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential. The greater the change in voltage per unit distance, the greater the electric field.
The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J
Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb
Using,
ΔU = q × ΔV
4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV
ΔV = 3.9 × [tex]10^-^3[/tex] V
Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
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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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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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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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(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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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 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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what is the wavelength of a tubas A2(110hz) tone in air at standard conditions/
The wavelength of A2 with frequency (110hz) is 3.01 m.
What is the wavelength of a sound wave?Wavelength of a sound wave is the distance between successive similar points in the wave such as rarefactions or compressions.
A tuba is a musical instrument that produces sound waves.
Wavelength is related to frequency and velocity by the formula below:
Wavelength = velocity/frequencyThe wavelength of A2 with frequency (110hz) is 3.01 m.
In conclusion, the wavelength of a wave is inversely proportional to frequency.
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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, T in the cable is equal to 323.5 N.
What is the tension?Tension is force exerted by a cable or string on another object usually a weight suspended from the cable or string
The tension in the cable is found this:
Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°
The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80
Taking moments about the pivot:
175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1
T = 241.68/0.747
T = 323.5 N
In conclusion, the tension in the cable is determined by taking moments about the pivot.
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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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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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(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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What is the maximum speed with which a 1200- kg car can round a turn of radius 90.0 m on a flat road if the coefficient of static friction between tires and road is 0.55?
The maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.
Maximum speed of the carv(max) = √μrg
where;
μ is coefficient of frictionr is radius of the curved roadg is acceleration due to gravityv(max) = √(0.55 x 90 x 9.8)
v(max) = 22.02 m/s
Thus, the maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.
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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
Name two units for measuring the diameter of nucleus atom.
The two units for measuring the diameter of nucleus atom are femtometre and metre.
How do you measure the size of the nucleus ?
Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.
femtometre (fm), which equals [tex]10^{-15}[/tex] metre.
A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.
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As shown in (Figure 1), a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated.
a) Using the information on the figure, find the index of refraction of material X .
b) Find the angle the light makes with the normal in the air.
Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
Explanation: To find the answer, we need to know the Snell's law.
What is Snell's law of refraction? Using this, how to solve the problem?The Snell's law for refraction can be written as,[tex]\frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}[/tex]
where, i is the incident angle, r is the refracted angle, n is the refractive index.
As we know that the refractive index of water is 1.33For the first case, incident angle from the picture is 65°, and the refracted angle is 48°. Thus, the refractive index of the medium X will be,[tex]\frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09\\[/tex]
In the second case, incident angle is 48° and we have to find the refracted angle r for the air.As we know that the refractive index of air is 1.Thus, the refracted angle will be,[tex]\frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
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The light makes an angle of 81.25° with the normal in the air and the index of refraction of material X will be 1.09.
In order to determine the solution, we must understand Snell's law.
What is the refraction law of Snell? How can the issue be resolved with this?One way to express Snell's law for refraction is as follows:[tex]\frac{sin(i)}{sin(r)}=\frac{n_r}{n_i}[/tex]
where the refractive index is n and the incidence angle is i. The refracted angle is r.
As is well known, water has a refractive index of 1.33.The incidence angle from the image in the first case is 65°, while the refracted angle is 48°. Consequently, the medium X's refractive index will be,[tex]n_X=\frac{n_w*sin 48}{sin65} =1.09[/tex]
The incident angle in the second scenario is 48°, and we must determine the refracted angle r for the air.Since we now know that air has a refractive index of 1, so that the refracted angle is,[tex]sin(r)=n_w*sin48=0.988\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
As a result, we can infer that the material X will have an index of refraction of 1.09 and that the angle the light makes with the normal in the air is 81.25°.
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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]
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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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
How much work must be done to stop a 975- kg car traveling at 105 km/h ?
Express your answer to two significant figures and include the appropriate units.
The amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.
How to calculate work done?The amount of work done by a moving object can be calculated using the following formula:
W (Kinetic energy) = ½ mv²
Where;
m = massv = velocityAccording to this question, a car of 975 kg is traveling at 105 km/h. This speed in m/s is 29.17m/s.
K.E = ½ × 975 × 29.17²
K.E = 414,808.34J
Therefore, the amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.
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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 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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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:
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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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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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 >>>>>
In physics, when a baseball player catches a ball, which one of newtons laws is it an example of?
A. 1st law
B. 2nd law
C. 3rd law
According to Howard Gardner's theories, who among the following would need to have good linguistic intelligence to be successful?
A psychologist will need to have good linguistic intelligence in other to be successful.
Who is a Psychologist?This is referred to as a professional who specializes in the handling of mental health challenges in individuals.
It is best for such professional to have a good linguistic intelligence as the right words being said to the patient will solve the problem thereby bringing in more success.
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