Since we cannot physically collect data from stars and most other objects in the universe, almost all of the information we obtain from the universe comes from analyzing the light, or spectra, from those objects. The study of light is known as spectroscopy. As we have seen in this simulation, every blackbody emits light with an easily identified pattern known as the blackbody curve. This is the particular way the total light emitted by a blackbody varies with its frequency. The exact form of the curve depends only on the body's temperature. Since we can treat stars as blackbodies, this is incredibly useful in astronomy that shows us that the color of a star is also indicative of its temperature. Use the simulation to determine the surface temperature of the following star: Betelgeuse is a red supergiant star in the constellation Orion. Knowing that Betelgeuse has peak intensity in the red and infrared wavelengths, adjust the intensity scale and temperature until you can determine the approximate surface temperature of the star. a. 3500K b. 4800K c. 7700K d. 11,000 K

Answer:

Option (a) is correct.

Explanation:

For us to be able to visualize an object, there are basic processes that takes place in a split second.

First Light which strike the object reflect into our eye. This reflected light passes through the pupil as the iris gives way for it. This light moves through other layers and impinges on the retina and stimulates it. This is the process of within the eye processing. The light is then converted to a signal by the optic nerve and sent to the brain for interpretation. The image produced by the retina is inverted. The brain helps to interpret this image as upright. This is the reason why we see objects just the way they are

Answer:C

Explanation:

There are two types of cells in the eyes rod and cone cells. Rod cells Provide vision during the night or dim light also called of scotopic vision whereas cone cells provide vision during day time or at bright light also called photopic vision.  

Rod cells do not support the color vision that is why it is difficult to differentiate between colors.

Also, the amount of light entering the eyes is low as result cone cells are unable to stimulate.

It is hard to see color at night because cones, which are responsible for color vision, need more light to be stimulated than rods, which allow us to see in low light but only in grayscale.

The best explanation for why it is difficult to discriminate the color of an object at night is option C. At night, the amount of light entering the eye is insufficient to stimulate the cone cells but is sufficient to stimulate the rod cells. The rods are highly sensitive to light, allowing us to see in low light conditions but do not provide color information. Cones require brighter light to function and are responsible for our color vision. Since cones do not react to low-intensity light, our vision at night is predominantly in shades of gray, serviced by the activity of rods.

Answer:

c

Explanation:

from. 50 to 60 diopters

The range of Anna's vision will be between 0-5 diopters.

It is given that

Anna is able to focus on objects within 20 cm of her eyes (her near point)

Now the distance between the cornea and retina, of Anna's eye is 20 mm

The focal length will be f= 20+2= 22cm

Now the power of eyes will be given by

[tex]P= \dfrac{1000}{f(in \ mm)}[/tex]

[tex]P= \dfrac{1000}{ 22}=4.54 \ diopters[/tex]

Thus the range of Anna's vision will be between 0-5 diopters.

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Answer:

 a₂ = m₁ / m₂ a₁

Explanation:

For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies

Let's apply Newton's second law on the star 1

     F₁ = m₁ a₁

Newton's second law in star 2

       F₂ = m₂ a₂

       | F₁ | = | F₂ |

      m₁  a₁ = m₂  a₂

      a₂ = m₁ / m₂ a₁

Answer:

D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop

Explanation:

The magnetic flux is defined as the total magnetic field times the area normal to the magnetic field lines.

Mathematically:

[tex]\phi_B=\vec{B}.\vec{A}[/tex]

where:

[tex]\vec{A}=[/tex] area vector directed normal to the surface

[tex]\vec{B}=[/tex] magnetic field vector

Final answer:

Rotating the conducting loop about an axis in the z direction will not change the magnetic flux through the loop because the angle between the magnetic field and the normal to the plane of the loop remains constant. so the correct option is D

Explanation:

The question pertains to the change in magnetic flux through a conducting loop when subjected to different actions. According to Faraday's Law of Electromagnetic Induction, the magnetic flux through a loop is given by the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field direction and the normal to the loop. In options A, B, and C, changes to the area or the strength of the magnetic field alter the magnetic flux since these factors directly affect the calculation of flux. Option D involves rotating the loop about an axis in the z direction; such a rotation does not change the angle between the magnetic field and the normal to the loop's plane, therefore would not change the flux. In contrast, option E, where the loop is rotated about an axis in the y direction, changes this angle and thus the flux.

Therefore, the action that will not change the magnetic flux through the loop is:

D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop.

Answer:

Explanation:

Given

density of cylinder is [tex]\rho _c=713 kg/m^3[/tex]

Length of first cylinder is [tex]L_1=20 cm[/tex]

radius [tex]r_1=5 cm[/tex]

For cylinder 2 [tex]L_2=10 cm [/tex]

[tex]r_2=10 cm[/tex]

[tex]h_1[/tex] and [tex]h_2[/tex] are the height above water

E

as object is floating so its weight must be balanced with buoyant force

[tex]\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1[/tex]

For 2nd cylinder

[tex]\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2[/tex]

Dividing 1 and 2 we get

[tex]\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}[/tex]

[tex]\frac{20}{10}=\frac{20-h_1}{10-h_2}[/tex]

[tex]2h_2=h_1[/tex]

[tex]\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}[/tex]                            

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5  ⇒ h/r = 2.5

⇒  h = (5/2)*r

⇒  r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = 3*35/π = 105/π   ⇒   d(r²*h)/dt = -105/π

a) if   r = (2/5)*h

⇒  d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒  (4/25)(3*h²)(dh/dt) = -105/π

⇒  dh/dt = -875/(4π*h²)

b) if  h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒  (5/2)*(3*r²)(dr/dt) = -105/π

⇒  dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒  dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm  ⇒  r = (2/5)*h = (2/5)*(125 cm)

⇒  r = 50 cm

⇒  dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒  dr/dt = -17.82*10⁻⁴ cm/s

The rate at which the depth of the water is changing is -0.178 cm/sec, and the rate at which the radius of the top of the water is changing is -0.14 cm/sec.

The subject of this question is related to Calculus and the specific topic is related rates. For this problem, in addition to the rate at which water is leaking, we need to consider the geometric relationship within the cone-shaped water tank.

First, we are given that V' = -35 cm^3/sec (we make it negative because the volume is decreasing) and we know that the volume of a cone is V = (1/3)πr²h. Our tank parameters are r = 1m and h = 2.5m, but we need everything in the same units, so we convert our radius to 100 cm.

We know through similar triangles that r/h = R/H, where r and R are the radii at any given time, and h and H are the heights at any given time respectively. Thus, r = (Rh)/H. Substituting r in our volume equation, we get: V = (1/3)π((Rh)^2)/H=h²πR²/H, and hence V=hπR². Differentiating this implicitly with t gives V' = h'πR²+2hπRr'.

Substituting for V', h and r from our given information, we get: -35=h'πR²+2(1.25)πRr'. We can solve for h' and r' separately.

(a) To find h' we set r' = 0, because we only want to know how depth h is changing. Solving for h' we find it to be -0.178 cm/sec.

(b) To find r', we set h' = 0, because we are only interested in how radius r is changing. Solving for r' we get -0.14 cm/sec.

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Final answer:

The magnitude of the electric field just outside the surface of a sphere designed to deflect solar radiation on the moon or Mars is 10⁷ N/C, calculated based on the given requirements for the electric field at a distance from the sphere.

Explanation:

The question asks for the magnitude of an electric field E just outside the surface of a sphere that is proposed to protect astronauts on the surface of the moon or Mars by deflecting high-energy charged particles emitted by the sun.

The sphere, assumed to have a uniform surface charge distribution, generates an electric field whose magnitude at a distance r from its center can be found using the formula E = kQ/r², where k is Coulomb's constant (8.99 x 10⁹ N·m²/C²), Q is the charge on the sphere, and r is the distance from the center of the sphere.

Given that the sphere's electric field's magnitude at a distance of 25 m from its center needs to be 1 x 10⁶ N/C, we find the charge Q required to produce this field.

Once Q is determined, we can calculate the electric field's magnitude just outside the sphere's surface (at r = 2.5 m, which is the radius of the 5 m diameter sphere) using the same formula.

The calculation reveals that the magnitude of E just outside the surface of the sphere is of the order 10⁷ N/C, making option c) 10⁷ N/C the correct answer.

The magnitude of the electric field just outside the surface of the sphere is [tex]\(10^7 \, \text{N/C}\),[/tex] so the correct answer is (c) [tex]\(10^7 \, \text{N/C}\).[/tex]

To find the magnitude of the electric field just outside the surface of the sphere, we'll use Gauss's Law, which states:

[tex]\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\][/tex]

Where:

- \(\vec{E}\) is the electric field vector

- \(d\vec{A}\) is a differential area vector

- \(Q_{\text{enc}}\) is the enclosed charge

- \(\varepsilon_0\) is the permittivity of free space

Since the electric field is radial, we can express it as [tex]\(E = E(r)\hat{r}\)[/tex], where [tex]\(\hat{r}\)[/tex] is a unit vector pointing radially outward from the center of the sphere.

The magnitude of the electric field \(E\) just outside the surface of the sphere is equal to the magnitude of the electric field produced by a point charge at the center of the sphere, which is given by Coulomb's Law:

[tex]\[E = \frac{k |Q|}{r^2}\][/tex]

Where:

- \(k\) is Coulomb's constant [tex](\(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\))[/tex]

- \(|Q|\) is the magnitude of the charge enclosed by the Gaussian surface

- \(r\) is the distance from the center of the sphere

Given that [tex]\(E = 1 \times 10^6 \, \text{N/C}\) at \(r = 25 \, \text{m}\), we can solve for \(|Q|\):[/tex]

[tex]\[1 \times 10^6 \, \text{N/C} = \frac{k |Q|}{(25 \, \text{m})^2}\][/tex]

[tex]\[|Q| = \frac{(1 \times 10^6 \, \text{N/C}) \cdot (25 \, \text{m})^2}{k}\][/tex]

[tex]\[|Q| = \frac{(1 \times 10^6 \, \text{N/C}) \cdot (625 \, \text{m}^2)}{8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2}\][/tex]

[tex]\[|Q| = \frac{625 \times 10^6}{8.99}\][/tex]

[tex]\[|Q| \approx 69.633 \, \text{C}\][/tex]

Now, let's calculate the electric field just outside the surface of the sphere using this charge:

[tex]\[E = \frac{k |Q|}{r^2}\][/tex]

[tex]\[E = \frac{(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \cdot (69.633 \, \text{C})}{(5 \, \text{m})^2}\][/tex]

[tex]\[E = \frac{(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \cdot (69.633 \, \text{C})}{25 \, \text{m}^2}\][/tex]

[tex]\[E = \frac{629.586 \times 10^9}{25}\][/tex]

[tex]\[E \approx 25.183 \times 10^7 \, \text{N/C}\][/tex]

Therefore, the magnitude of the electric field just outside the surface of the sphere is approximately [tex]\(2.5183 \times 10^8 \, \text{N/C}\).[/tex]

The closest option is d) [tex]\(10^8 \, \text{N/C}\).[/tex]

Answer:

The speed of the particle after travelling for 0.5 m is 100 m/s.                            

Explanation:

It is given that,

Mass of the particle, m = 0.01 kg

Net charge on the particle, q = -0.05 C  

Electric field strength, E = 2000 V/C

Distance travelled by the particle, d = 0.5 m

The work done due to motion of the particle is balanced by the change in kinetic energy as :

[tex]Fd=\dfrac{1}{2}mv^2[/tex]

v is the speed of the particle

F is the electric force

[tex]qEd=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{\dfrac{2qEd}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 0.05\times 2000\times 0.5}{0.01}}[/tex]

v = 100 m/s

So, the speed of the particle after travelling for 0.5 m is 100 m/s. Hence, this is the required solution.

Answer:

[tex]h=3.4817\ m[/tex]

Explanation:

Given:

We know that:

[tex]P=\rho.g.h[/tex]

putting respective values

[tex]65000-101325=1905 \times 9.8\times h[/tex]

[tex]h=3.4817\ m[/tex] is the reading in terms of the fluid column height on the manometer.

Considering the rise in the level of fluid this fluid is not suitable for such huge pressures for a practically convenient manometer.

To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.

The power can be determined as:

[tex]P = \sigma T^4[/tex]

Making the relationship for two states we have to

[tex]\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}[/tex]

Since the final power is 8 times the initial power then

[tex]P_2 = 8P_1[/tex]

Substituting,

[tex]\frac{1}{8} = \frac{T_1^4}{T_2^4}[/tex]

[tex]T_2 = T_1 8*(\frac{1}{4})[/tex]

[tex]T_2 = 1,68T_1[/tex]

The temperature increase would then be subject to

[tex]\Delta T = T_2-T_1[/tex]

[tex]\Delta T = 1.68T_1 -T_1[/tex]

[tex]\Delta T = 0.68T_1[/tex]

The correct option is D, about 68%

Final answer:

The correct option is (c) 100%. To increase the power radiated from an object by a factor of 8, the temperature must be doubled, corresponding to a 100% increase. This is based on the Stefan-Boltzmann law.

Explanation:

The question relates to the physics concept of radiation and how it varies with temperature. Specifically, it involves the Stefan-Boltzmann law, which states that the power radiated from an object is proportional to the fourth power of its absolute temperature.

If the power radiated from an object is to increase by a factor of 8, we use the relationship P ≈ T^4, where P is power and T is the temperature in Kelvin. To calculate the necessary temperature increase, if the initial power is P and the final power is 8P, then (T_final)^4 = 8 × (T_initial)^4.

By taking the fourth root on both sides, we find that T_final = 2× T_initial. Therefore, the temperature must be doubled, or the temperature increase is by 100%. Hence, the correct answer is option (c) 100%

The sets of driver's actions “Signalling; Changing Gears; Checking mirrors” are multi-task performances necessary for the safe vehicle operation.

Answer: Option A

Explanation:

Signaling is what drivers plan to do is important for security because other drivers can only know if he tells them. Here are some general rules about Signalling or turns.

The correct gear change is important. If he can't put the vehicle in the right gear while driving, he has less control. And also, he must check the exterior mirrors on both sides regularly. He needs to check his vehicle mirrors to make sure nobody stands or walks past him.

To solve this problem we will use the concept of the Doppler effect applied to the speed of blood, the speed of sound in the blood and the original frequency. This relationship will also be extrapolated to the frequency given by the detector and measured the change in frequencies through the beat frequency. So:

[tex]f_{blood} = f (1-\frac{v_{blood}}{v_{snd}})[/tex]

Where

[tex]f_{blood}[/tex] = Frequency of the blood flow

f = Frequency of the original signal

[tex]v_{blood}[/tex] = Speed of the blood flow

[tex]v_{snd}[/tex] = Speed of sound in blood

[tex]f''_{detector} = \frac{f_{blood}}{(1+\frac{v_{blood}}{v_{snd}})}[/tex]

[tex]f''_{detector} = f (\frac{(1-\frac{v_{blood}}{v_{snd}})}{(1+\frac{v_{blood}}{v_{snd}})})[/tex]

[tex]f''_{detector} = f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]

Now calculating the beat frequency is

[tex]\Delta f = f-f''_{detector}[/tex]

Replacing this latest value we have that,

[tex]\Delta f = f-f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]

[tex]\Delta f = f \frac{2v_{blood}}{v_{snd}+v_{blood}}[/tex]

Replacing we have,

[tex]\Delta f = (3.5*10^6)(\frac{2*(3*10^{-2})}{1.54*10^3+3*10^{-2}})[/tex]

[tex]\Delta f = 136.36Hz[/tex]

Therefore the beat frequency is 136.36Hz

Using the beat frequency relation, the expected beat frequency observed at the flow meter would be 136.36 Hz

Given the Parameters :

The expected beat frequency observed can be calculated uisng the relation :

Substituting the values into the formula :

[tex] \delta F = 3.5 \times 10^{3} \frac{2 \times 0.03}{(1540 + 0.03}[/tex]

[tex] \delta F = 3.5 \times 10^{3} \frac{0.06}{(1540.03}[/tex]

[tex] \delta F = 136.36 Hz [/tex]

Therefore, the expected beat frequency observed at the flow meter will be 136.36 Hz

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Answer:

-0.23694 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

dr = Height = 1 mile = 1609.34 m

Acceleration is given by

[tex]a=\dfrac{GM}{r^2}[/tex]

Change in acceleration is given by

[tex]da=-2\dfrac{GM}{r^3}dr[/tex]

[tex]w=ma\\\Rightarrow w=m\dfrac{GM}{r^2}\\\Rightarrow w=469\ N[/tex]

[tex]dw=mda\\\Rightarrow dw=-m2\dfrac{GM}{r^3}dr\\\Rightarrow dw=-2w\dfrac{dr}{r}\\\Rightarrow dw=-2\times 469\times \dfrac{1609.34}{6.371\times 10^{6}}\\\Rightarrow dw=-0.23694\ N[/tex]

The change in weight is -0.23694 N

The change in your weight if you were to ride an elevator from the street level where you weigh 469N to the top of the building is; -0.237 N

The formula for acceleration here is;

a = GM/r²

Where;

G is gravitational constant = 6.67 × 10⁻¹¹ m³/kg.s²

M is mass of earth = 5.972 × 10²⁴ kg

r is distance from center of earth

Since we are trying to find change in weight, let us first find the change in acceleration with respect to r;

da/dr = -2GM/r³

da =  -(2GM/r³) dr

Thus, change in weight from top to bottom is;

W_top - W_bottom = m(da)

Now, weight at bottom is gotten from the formula;

W_bottom = GmM/r²

Also, W_bottom = m(da) since we are dealing with change in weight.

Thus;

m(da)= -(2GmM/r³) dr

Recall that GmM/r². Thus;

m(da) = -2W_bottom × dr/r

where;

W_bottom = 469 N

r is radius of earth = 6371000 m

dr = 1 mile = 1609.34 m

Thus;

m(da) = -2 × 469 × 1609.34/6371000

m(da) = -0.237 N

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Answer:

a. [tex]m=\frac{2f_{s}tan\theta}{g}[/tex]

b. m=5.99kg

Explanation:

a.

In order to solve this problem, we can start by drawing a diagram of the situation. Drawing a diagram is really important since it will help use understand the problem better and analyze it as well. (See attached picture).

In the diagram we can see the forces that are acting on the ladder. We will assume the ladder is static (this is it doesn't have any  movement) and analyze the respective forces in the x-direction and the forces in the y-direction, as well as the moments about point B.

So we start with the sum of forces about y, so we get:

[tex]\sum F_{y}=0[/tex]

N-W=0

N=W

N=mg

Next we can do the sum of forces about x, so we get:

[tex]\sum F_{x}=0[/tex]

which yields:

[tex]f_{s}-F_{W}=0[/tex]

so:

[tex]f_{s}=F_{W}[/tex]

Next the torque about point B, so we get:

[tex]\sum M_{B}=0[/tex]

so:

[tex]f_{s} L sin\theta - NLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]

From the sum of forces in the y-direction we know that N=mg (this is because the wall makes no friction over the ladder) so we can directly substitute that into our equation, so we get:

[tex]f_{s} L sin\theta - WLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]

We can now combine like terms, so we get:

[tex]f_{s} L sin\theta -\frac{WL}{2}cos\theta = 0[/tex]

we know that W=mg, so we can substitute that into our equation, so we get:

[tex]f_{s} L sin\theta -\frac{mgL}{2}cos\theta = 0[/tex]

which can now be solved for the mass m:

[tex]\frac{mgL}{2}cos\theta = f_{s} L sin\theta[/tex]

If we divided both sides of the equation into L, we can see that the L's get cancelled, so our equation simplifies to:

[tex]\frac{mg}{2}cos\theta = f_{s}sin\theta[/tex]

we can now divide both sides of the equation into g so we get:

[tex]\frac{m}{2}cos\theta = \frac{f_{s}sin\theta}{g}[/tex]

next we can divide both sides of the equation into cos θ so we get:

[tex]\frac{m}{2}= \frac{f_{s}sin\theta}{g cos\theta}[/tex]

and finally we can multiply both sides of the equation by 2 so we get:

[tex]m=\frac{2f_{s}sin\theta}{g cos\theta}[/tex]

we know that:

[tex]tan \theta=\frac{sin \theta}{cos \theta}[/tex]

so we can simplify the equation a little more, so we get:

[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]

b.

So now we can directly use the equation to find the mass of the ladder with the data indicated by the problem:

θ=32° and [tex]f_{s}=47N[/tex]

we also know that [tex]g=9.8m/s^{2}[/tex]

so we can use our equation now:

[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]

so we get:

[tex]m=\frac{2(47N)tan (32^{o})}{9.8m/s^{2}}[/tex]

which yields:

m=5.99kg

Answer:

a) m = 2fs(tanθ)/g

b) m = 5.99 kg

Explanation:

a) Expression for the ladder's mass, m, in terms of θ, g, fs, and/or L.

Data

m₁ : mass of the lader

g: acceleration due to gravity

L : ladder length

θ  : angle that makes  the  ladder  with the floor

µ = 0 : coefficient of friction between the ladder and the wall

fs : friction force between the ladder and the floor

Forces acting on the ladder

W =m*g : Weight of the ladder (vertical downward) , m: mass of the lader 

FN :Normal force that the floor exerts on the ladder (vertical upward) (point A)

fs : friction force that the floor exerts on the ladder (horizontal to the left) (point A)

N : Forces that the wall exerts on the ladder (horizontal to the right)

Equilibrium  of the forces in X

∑Fx=0

N -fs = 0

N = fs

The equilibrium equation of the moments at the point contact point of the ladder with the floor:

∑MA = 0  

MA = F*d  

Where:  

∑MA : Algebraic sum of moments in the the point (A) (contact point of the ladder with the wall)  

MA : moment in the point A ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point A ( m )

Calculation of the distances of the forces at the point A

d₁ = (L/2)*cosθ : Distance from W to the point A

d₂ = L*sinθ : Distance from N to the point A

Equilibrium of the moments at the point A

∑MA = 0  

N(d₂)-W(d₁) = 0

W( (L/2)*cosθ)= N(L*sinθ )

mg( (L/2)*cosθ)= fs(L*sinθ )

We divided by L both sides of the equation

mg (cosθ/2) =fs(sinθ)

m=2fs(sinθ)/ g( cosθ)

m = 2fs(tanθ)/ g

b) If θ=32° and the friction force on the floor is 47 N, what is the mass of the ladder?

m = 2fs(tanθ)/ g

m = 2(47)(tan32°)/(9,8)

m = 5.99 kg

Answer:

1350N

Explanation:

[tex]2.75 ms = 2.75*10^{-3}s[/tex]

The force exerted on the hand would be the momentum divided by the duration of contact.

As the hand is coming to rest, final velocity would be 0

[tex]F = \frac{\Delta P}{\Delta t} = \frac{m(0 - v)}{\Delta t} = \frac{1.65*(2.25 - 0)}{2.75 * 10^{-3}} = -1350 N[/tex]

The magnitude of the force would be 1350N

The fraction of the volume of an iceberg that is visible when it floats in saltwater or freshwater can be calculated using Archimedes' principle. For the given densities, the iceberg would sink in both saltwater and freshwater.

When an iceberg floats in water, a fraction of its volume is submerged, which is determined by the density of water and the density of the iceberg. To calculate the fraction of the iceberg that is submerged, you can use Archimedes' principle. Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

In this case, the density of the iceberg is given as 917 kg/m³. Let's calculate the fraction of the volume of the iceberg that would be visible when floating in (a) the ocean (salt water, density 1024 kg/m³) and (b) a river (fresh water, density 1000 kg/m³).

Fraction submerged = (density of iceberg - density of seawater) / density of iceberg

Using the given densities:

(917 kg/m³ - 1024 kg/m³) / 917 kg/m³ = -0.1166

The negative value indicates that the iceberg would not float in saltwater. In other words, it would sink.

Fraction submerged = (density of iceberg - density of fresh water) / density of iceberg

Using the given densities:

(917 kg/m³ - 1000 kg/m³) / 917 kg/m³ = -0.0902

Again, the negative value indicates that the iceberg would sink in freshwater as well.

Answer:

a. get warmer.

Explanation:

When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.

It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.

Answer:

0.1587

Explanation:

Given data

μ = 0.5 mg  

standard deviation σ =0.1 mg

n=100

we know that

[tex]P(\overline X<0.49)[/tex]

[tex]Z= (\frac{\overline X-\mu}{\sigma/\sqrt{n} })[/tex]

putting the values we get

[tex]Z= (\frac{0.49-0.5}{0.1/\sqrt{100} })[/tex]

Z=-1

Area under the curve for z =-1 is 0.1587 (from z score table)

P(X<0.49) = 0.1587

P(X<0.37) = 0.0013

To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

[tex]140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N[/tex]

[tex]120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N[/tex]

[tex]170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N[/tex]

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

[tex]F = 756 - 622.3[/tex]

[tex]F = 133.7N[/tex]

Through the Newtonian relationship of the Force we have to:

[tex]F= ma[/tex]

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{133.7}{63.5}[/tex]

[tex]a = 2.1m/s^2[/tex]

PART B) For the maximum magnitude of the acceleration downward we have that:

[tex]F = 622.3 - 533[/tex]

[tex]F = 89.3N[/tex]

Through the Newtonian relationship of the Force we have to:

[tex]F= ma[/tex]

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{89.3}{63.5}[/tex]

[tex]a = 2.1m/s^2[/tex]

[tex]a = 1.04 m/s^2[/tex]

Answer:

1- Period is the time for which one full rotation is completed. Regardless of their positions on the platform, periods of all three are the same. T2 = T.

2- Similarly, T3 = T.

3- The platform is making rotational motion. So, the relation between the angular velocity and the linear velocity is

[tex]v = \omega R[/tex]

For all the people, angular velocity is the same. Their linear velocities are different.

[tex]\omega = \frac{v_1}{R} = \frac{v_2}{3R/5}\\v_2 = \frac{3v}{5}[/tex]

4- Similarly,

[tex]\omega = \frac{v_1}{R} = \frac{v_2}{R/2}\\v_2 = \frac{v}{2}[/tex]

5- Radial acceleration in constant circular motion is

[tex]a_{rad} = \frac{v^2}{R}[/tex]

For the second person:

[tex]a_2 = \frac{v_2^2}{3R/5} = \frac{(3v/5)^2}{3R/5} = \frac{9v^2/25}{3R/5} = \frac{3v^2}{5R} = 3a/5[/tex]

6- Similarly,

[tex]a_3 = \frac{v_3^2}{R/2} = \frac{(v/2)^2}{R/2} = \frac{v^2/4}{R/2} = \frac{v^2}{2R} = a/2[/tex]

Explanation:

As a result, the period is same for every object on the rotating platform, as they all complete their revolutions at the same time. Their speed and radial acceleration is different according to their distance to the center.

The rotational kinematics relations allow to find the results for the questions about the movement of the three people on the turntable are:

      1 and 2) All periods are equal, T₂ = T and T₃ = T.

       3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]  

       4) The linear velocity of the 3rd person is: v₃ = ½ v

       5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]  

      6) The linear acceleration of the 3rd person: a₃ = ½ a

Rotational kinematics studies the rotational motion of bodies looking for relationships between angular position, angular velocity, and angular acceleration.

In the case where the angular accleration is zero, the expression for the velocity is:

        [tex]w = \frac{\Delta \theta }{\Delta t}[/tex]  

Where w is the angular velocity and Δw and Δt are the variation in angle t over time.

1 and 2)

Indicates that people are on a turntable, the period is when we have a complete rotation θ = 2π rad in time, therefore the period and the angular velocity are related.

          [tex]w= \frac{2\pi }{T} \\T = \frac{2\pi }{w}[/tex]  

In the apparatus of parks the angular velocity is constant and we see that it does not depend on the radius, therefore the period for all the people is the same.

         T = T₁ = T₂

3) They indicate that the speed of the 1 person who is in the position r=R on the plate is v, let's calculate the speed for the 2 person who is in the position  r = [tex]\frac{3}{5} \ R[/tex]  

Linear and angular variables are related.

        v = w r

Let's substitute for the 1st person.

        v = w R

For the 2nd person.

        v₂= w ( [tex]\frac{3}{5} R[/tex])

We solve these two equations.

         [tex]v_2 = \frac{3}{5} \ v[/tex]  

4) We carry out the same calculation for the 3rd person.

         v₃ = w ½ R

We solve the two equations.

          v₃ = ½ v

5) Ask for radial acceleration.  

The relationship between radial and angular acceleration is.

           a = α R

We substitute for the 1st person.

          a = α R

For the second person.

          a₂ = α ( [tex]\frac{3}{5} R[/tex])

We solve the two equations

         a₂ = [tex]\frac{3}{5} \ a[/tex]

6) Ask the radial acceleration of the 3rd person.

We substitute.

         a₃ = α (½ R)

We solve.

        a₃ = ½ a

In conclusion, using the rotational kinematics relations we can find the results for the questions about the movement of the three people on the turntable are

      1 and 2) All periods are equal, T₂ = T and T₃ = T.

       3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]  

       4) The linear velocity of the 3rd person is: v₃ = ½ v

       5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]  

      6) The linear acceleration of the 3rd person: a₃ = ½ a

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Answer:

Explanation:

1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.

2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.

3) TRUE, W = q.VBA.

4) FALSE; eV is a unit for work (or) energy.

5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.

6) FALSE; in the direction of electric field the potential decreases.

7) FALSE; equipotential surface is perpendicular to the electric field lines.

8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.

9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.

10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.

Answer:

U. With no variation.

Explanation:

Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.

Answer:

[tex]1.04\times 10^7\ J.[/tex]

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, [tex]W=P\times\Delta V[/tex]

Pressure, P=1.01 × 105 Pa.

Final volume, [tex]V_f=8.50\ m^3.[/tex]

Initial volume, [tex]V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.[/tex]

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, [tex]Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.[/tex]

Also, according to First law of thermodynamics:

[tex]\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.[/tex]

Hence, this is the required solution.

The energy needed to convert all of the water into steam in this isothermal system is 11.30 MJ. This is calculated using the formula Q = mLv, substituting the given values for mass and latent heat of vaporization.

The heat required to convert water from liquid into steam, in an isothermal process, can be found using the formula Q = mLv, where 'm' is the mass of the water, and 'Lv' is the latent heat of vaporization.

Given that we have 5.00 kg of water and the heat of vaporization for water under atmospheric pressure is 2.26 × 106 J/kg, we can substitute these values into our formula:

Q = 5.00 kg × 2.26 × 106 J/kg.

This gives us Q = 11.30 × 106 J, or 11.30 MJ. Therefore, 11.30 MJ of heat is added to the system to convert all the water into steam at atmospheric pressure.

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The spring constant of the spring is approximately 234.50 N/m.

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. In this case, we know that the mass of the block is 6.7 kg and the displacement of the spring is 0.28 m. Therefore, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement. Plugging in the given values, we get:

F = kx
F = (6.7 kg)(9.8 m/s^2)
F = 65.66 N
65.66 N = k(0.28 m)
k = (65.66 N)/(0.28 m)
k ≈ 234.50 N/m

Therefore, the spring constant of the spring is approximately 234.50 N/m.

Answer:

0.5 N

Explanation:

[tex]r_{1}[/tex] = radius of piston 1

[tex]r_{2}[/tex] = radius of piston 2

Given that :

[tex]r_{2} = 200 r_{1}[/tex]

[tex]F_{1}[/tex] = Force applied on piston 1

[tex]F_{2}[/tex] = Force applied on piston 2 = Weight being lifted = 20000 N

Using pascal's law

[tex]\frac{F_{1}}{\pi r_{1}^{2} } = \frac{F_{2}}{\pi r_{2}^{2} } \\\frac{F_{1}}{r_{1}^{2} } = \frac{20000}{(200)^{2}r_{1}^{2} } \\\\F_{1}= \frac{20000}{(200)^{2}} } \\F_{1}= 0.5 N[/tex]

Answer:

a) -6,750 N  b) -1.35*10⁶ N  c) 11.5 times d) 2,295 times

Explanation:

a) In order to answer the question, we can apply Newton's 2nd Law to the person, as follows:

Fnet = m*a

Assuming that the deceleration is uniform, we can use any of the kinematic equations.

In this particular case, examining the givens that we have (the final speed (which is 0), the initial speed (impact speed), and the distance of deceleration (1 m) the most useful equation is the following:

vf²-v₀² = 2*a*d

Replacing by the givens, and solving for a, we get:

a = (15m/s)² / 2* (1 m) = -112.5 m/s²

With this value of a, we can get the net force F:

F = 60 kg* (-112.5 m/s²) = -6,750 N

b) For this part, same reasoning applies, the only difference being the deceleration distance, which for this case is only 5 mm.

We can apply the same kinematic equation:

vf²-v₀² = 2*a*d

Once again, replacing by the givens, and solving for a, we get:

a = (15m/s)² / 2* (0.005 m) = -22,500 m/s²

With this value of a, we can get the net force F:

F = 60 kg* (-22,500 m/s²) = -1.35*10⁶ N

c) If we compare the forces that  we got above with the weight of the person (which is the same to compare the acceleration with g), we have, for the first case (person restrained) a value of  approximately 12 g, but for the unrestrained case, we got a value of 2,295 g!

Answer:

[tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]

Explanation:

Light which travels from the crackers reaches the detector at [tex]c=3\times 10^{8}\ m/s[/tex]

[tex]\Delta x_1[/tex] = Distance at which event 1 leaves a char mark = 3.4 m

[tex]\Delta x_2[/tex] = Distance at which event 2 leaves a char mark = 2.1 m

The speed of light in a medium is a universal constant

[tex]c=\dfrac{\Delta x_1}{\Delta t_1}\\\Rightarrow \Delta t_1=\dfrac{\Delta x_1}{c}\\\Rightarrow \Delta t_1=\dfrac{3.4}{3\times 10^8}\\\Rightarrow \Delta t_1=1.13333\times 10^{-8}\ s[/tex]

[tex]c=\dfrac{\Delta x_2}{\Delta t_2}\\\Rightarrow \Delta t_2=\dfrac{\Delta x_2}{c}\\\Rightarrow \Delta t_2=\dfrac{2.1}{3\times 10^8}\\\Rightarrow \Delta t_2=0.7\times 10^{-8}\ s[/tex]

The pulse will be detected at [tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]

In the reference frame of the detector, both light pulses will be detected approximately 1.13 x 10^-8 seconds after they occur.

In the reference frame of the detector, event 1 leaves a mark at a distance of 3.40 m and event 2 leaves a mark at a distance of 2.10 m. Since the events are simultaneous in the reference frame of the detector, the time it takes for each light pulse to be detected will be the same for both events. To calculate the time, we can use the speed of light as the distance traveled divided by the speed of light. Therefore, the time it takes for each light pulse to be detected is t = distance/speed of light.

For event 1, the distance is 3.40 m and for event 2, the distance is 2.10 m. The speed of light is approximately 3.00 x 10^8 m/s. Plugging in these values, we get:

t = (3.40 m) / (3.00 x 10^8 m/s) = 1.13 x 10^-8 s

Therefore, each light pulse will be detected approximately 1.13 x 10^-8 seconds after the events occur.

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Answer:

C/V

Explanation:

Mathematically it is known that one farad is defined as the capacitance across which, when charged with one coulomb, there is a potential difference of one volt. In other words, the Farad is Coulomb/Volt or C/V

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

       ΔE = [tex]E_{nf}[/tex] - E₀ = - k²e² / 2m (1 / [tex]n_{f}[/tex]²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

       E = h f = h c / λ

       h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

       1 / λ = Ry (1/ [tex]n_{f}[/tex]² - 1 / n₀²)

Let's apply these equations to our case

     λ = 821 nm = 821 10⁻⁹ m

     E = h c / λ

     E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

     E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

       E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

       [tex]E_{nf}[/tex] - E₀ = -13.606 (1 /  [tex]n_{f}[/tex]² - 1 / n₀²)   [eV]

Let's look for the energy of some levels

n         [tex]E_{n}[/tex] (eV)          [tex]E_{nf}[/tex] - E[tex]E_{ni}[/tex] (eV)

1         -13,606           E₂-E₁ = 10.20

2        -3.4015           E₃-E₂ = 1.89

3        -1.512              E₄- E₃ = 0.662

4        -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value