g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest at the top of an incline plane. The height of top of the incline plane is 1.2 m. Find translational speed of each cylinder upon reaching the bottom and determine which cylinder has the greatest translational speed upon reaching the bottom. Moment of inertia of hollow cylinder about its axis passing through the center is mr2 and for solid cylinder mr2/2

Answer:

it's A subduction, deep water trench

Answer: 5.08 L.

Explation down below

Answer:

Ratio of kinetic energy of object A to B = 1:5

Explanation:

Given:

Mass of object A = 200 kg

Mass of object B = 1,000 kg

Find:

Ratio of kinetic energy of object A to B

Computation:

Kinetic energy = (1/2)(m)(v²)

Kinetic energy of object A = (1/2)(200)(v²)

Kinetic energy of object A = (100)(v²)

Kinetic energy of object B = (1/2)(1,000)(v²)

Kinetic energy of object B = (500)(v²)

Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B

Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)

Ratio of kinetic energy of object A to B = 100 / 500

Ratio of kinetic energy of object A to B = 1/5

Ratio of kinetic energy of object A to B = 1:5

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

Answer:

[tex]23^{\circ}[/tex]

Explanation:

n = Refractive index of air = 1

[tex]n_1[/tex] = Refractive index of contact lens = 1.6

[tex]n_2[/tex] = Refractive index of cornea = 1.4

[tex]n_3[/tex] = Refractive index of fluid = 1.3

From Snell's law

[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]

[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]

[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]

The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].

The angle is the light traveling in the fluid will be 23⁰. Light is traveling in a particular direction with an angle.

"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for a given set of media,

The given data in the problem is;

n is the refractive index of air = 1

n₁ is the refractive index of contact lens = 1.6

n₂ is the refractive index of cornea = 1.4

n₃ is the refractive index of fluid = 1.3

According to Snell's law. The formula for Snell's law is

[tex]\rm n sin30^0 = n_1 sin\theta \\\\ \theta = sin^{- 1}(\frac{1sin30^0}{1.6} )\\\\ \theta = 18.21 ^0[/tex]

For contact lenses;

[tex]\rm n_1sin\theta = n_2 sin\theta_1 \\\\ \theta_1 = sin^{-1}\frac{1.6 \times sin 18.21^0}{1.4} \\\\ \theta_1 =20.92 ^0[/tex]

For fluid;

[tex]n_2 sin\theta_1 = n_2 sin \theta_3\\\\ \theta_3 = sin^{-1}\frac{1.4 sin 20.92^0}{1.3} \\\\ \theta_3 = 22.62 ^ 0 =23^0[/tex]

Hence the angle is the light traveling in the fluid will be 23⁰.

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Answer:

The answer is "15.56 cm".

Explanation:

[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]

Calculating object of length is x so:

[tex]u= -x[/tex]

Using formula:

[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]

This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.

The minimum horizontal velocity required is "2.6 m/s".

First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.

[tex]h=v_it+\frac{1}{2}gt^2[/tex]

where,

h = height = 3 m

vi = initial vertical speed = 0 m/s

t = time interval = ?

g = acceleration due to gravity = 9.81 m/s²

therefore,

[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]

t = 0.78 s

Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:

[tex]s = vt\\\\v = \frac{s}{t}[/tex]

where,

s = horizontal distance = 2 m

t =0.78 s

v = minimum horizontal velocity = ?

Therefore,

[tex]v = \frac{2\ m}{0.78\ s}[/tex]

v = 2.6 m/s

Learn more about equations of motion here:  

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The attached picture shows the equations of motion in the horizontal and vertical directions.

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

[tex]P=\dfrac{V^2}{R}[/tex]

Substitute all the values,

[tex]P=\dfrac{3^2}{2}\\\\P=4.5\ W[/tex]

So, the power of this circuit is equal to 4.5 watts.

Answer:

1/Re= 1/R1 + 1/R2

Explanation:

Two resistors are connected in parallel. If R1 and R2 represent the resistance in Ohms (Ω) of each resistor, then the total resistance R is given by  [tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]. Thus, the rate of R changes when R₁ = 117 Ω and

R₂ = 112 Ω is 0.25 Ω/min

For a given resistor connected in parallel;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]

Making R from the left-hand side the subject of the formula, then:

[tex]\mathbf{R = \dfrac{R_1R_2}{R_1+R_2}}[/tex]

Given that:

Now, replacing the values in the above previous equation, we have:

[tex]\mathbf{R = \dfrac{13104}{229}}[/tex]

However, the differentiation of R with respect to time t will give us the rate at which R is changing when R1=117Ω and R2=112Ω.

So, by differentiating the given equation of the resistor in parallel with respect to time t;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex],   we have:

[tex]\mathbf{\dfrac{1}{R^2}(\dfrac{dR}{dt})=\dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})}[/tex]

[tex]\mathbf{(\dfrac{dR}{dt})=R^2 \Bigg[ \dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=(\dfrac{13104}{229})^2 \Bigg[ \dfrac{0.4}{117^2}+\dfrac{0.6}{112^2}\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=3274.44 \Bigg[ (7.7052 \times 10^{-5} )\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=0.25\ \Omega /min}[/tex]

Therefore, we can conclude that the rate at which R is changing R1=117Ω and R2=112Ω is 0.25 Ω/min

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Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

Answer:

Both sound and waves

Explanation:

Thank me later

Answer:

Diagram (3).

Explanation:

N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A ([tex]F_{A} = -F_{B}[/tex]).

The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).

The law clearly states a Force of attraction; the two objects come towards each other.

Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.

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To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

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#SPJ2

Answer:

B

Explanation:

The Lorentz force is the sum of the electric force and magnetic force.

F = qE + qvB

qE represents electric force and qvB represents magnetic force

Answer:

The correct answer is - 93.24×10^4 kg/m^3.

Explanation:

Given:

height of cylinder: 50.5 mm

diameter = 52.0

then radius will be diameter/2 = 52/2 = 26

Formula:

Density = mass/ volume

Volume = πr^2h

solution:

Now the volume of a cylinder is v = (22/7)×r^2×h

= 22/7×26×26×50.5

= 107261.59 mm^3  

Now volume in cubic meter V =10.7261 ×10^(-5) m^3

So density d = m/V = 1/(10.7261 ×10^(-5))

Or d = 93.24×10^4 kg/m^3

Answer:

B) It appears to be SHM (simple harmonic motion) which is also the projection of an object moving in a circular path on either the x or y axes.

Answer:

the answer is 86400seconds.

Explanation:

1day= 24 hours

24hours to seconds =

24×60×60= 86400 seconds.

( 60 second = 1 minutes)

( 60 minutes = 1 hour)

Answer:

[tex]4.25\ \text{m/s}[/tex]

[tex]3391.22\ \text{N}[/tex]

Explanation:

y = Height of compression = 0.38 m

m = Mass of basketball player = 101 kg

h = Height of center of gravity after jump = 0.92 m

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Energy balance of the system is given by

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]

The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]

[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]

The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          [tex]\frac{1}{f_1} = \frac{1}{q}[/tex]

           q = f₁

2) for an object located at p = 25 cm

            [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]

        [tex]\frac{1}{f_2}[/tex] = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = [tex]\frac{1}{f}[/tex]

where the focal length is in meters

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

Answer:

0.1 kg/s.

Explanation:

The density of water, d = 1000 kg/m³

Volume, V = 2 L

Time, t = 20 s

We need to find the mass flow rate from the faucet. We know that the density of an object is given by :

[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]

So, the mass flow rate is equal to 0.1 kg/s.

Answer:

B = 1.37 mT

Explanation:

Given that,

The magnitude of the electric field, E = 480 N/C

The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]

We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,

[tex]qE=qvB[/tex]

Where

B is the magnetic field

[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]

So, the magnetic field is equal to 1.37 mT.

Answer:

I think its d

Explanation:

I'm not sure I'm sorry if I'm wrong

Answer:

true it all changes

Explanation:

________________________

Answer:

electromagnetic waves only

Explanation:

I just took the test, Hope it helps!

Answer:

A: Electromagnetic waves only

Explanation: