A light-rail commuter train speeds up at a rate of 1.65 m/s 2 and takes 19.0 s to reach its top speed. How far down the tracks does the train go to reach its top speed if starting from rest?

Explanation:

It is known that the molecular weight of ammonia ([tex]NH_{3}[/tex]) is as follows.

   Molecular weight ([tex]NH_{3}[/tex]) = [tex]14 + 3 \times 1[/tex] = 17

(a)   Therefore, we will calculate the mass as follows.

     [tex]0.5 kmol \times (\frac{1000 mol}{1 kmol}) \times (\frac{17 g}{1 mol})[/tex]

                       = 8500 g

Now, formula to calculate weight of the system in N is as follows.

            Weight = mass × g

             = [tex]8500 g \times (\frac{1 kg}{1000 g}) \times (9.8 m/s^{2})[/tex]

             = 83.3 N     (1 [tex]kg m/s^{2}[/tex] = 1 N)

Hence, the weight of the system is 83.3 N.

(b)   Relation between specific volume and number of moles is as follows.

            [tex]v (m^{3}/kmol) = \frac{V}{n}[/tex]

Therefore, calculate the specific volume as follows.

       [tex]V_m = \frac{6 m^{3}}{0.5 k mol}[/tex]

                   = 12 [tex]m^{3}/k mol[/tex]

Also,  

               [tex]v (m^{3}/kmol) = \frac{V}{m}[/tex]  

            v = [tex]\frac{6 m^{3}}{8.5 kg}[/tex]

               = 0.705882 [tex]m^{3}/kg[/tex]

Therefore, we can conclude that the value of specific volume is 12 [tex]m^{3}/k mol[/tex]  and 0.705882 [tex]m^{3}/kg[/tex].

Answer:

a) [tex]w=83.385\ N[/tex]

b) [tex]\bar V=12\ m^3.kmol^{-1}[/tex]

[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]

Explanation:

Given:

no. of moles of ammonia in a closed system, [tex]n=0.5\ kmol=500\ mol[/tex]

volume of ammonia, [tex]V=6\ m^3[/tex]

We  know the molecular formula of ammonia: [tex]NH_3[/tex]

The molecular mass of ammonia:

[tex]M=14+3\times 1=18\ g.mol^{-1}[/tex]

Now the mass of given ammonia:

[tex]m=n.M[/tex]

[tex]m=500\times 17[/tex]

[tex]m=8500\ g=8.5\ kg[/tex]

a)

Now weight:

[tex]w=m.g[/tex]

[tex]w=8.5\times 9.81[/tex]

[tex]w=83.385\ N[/tex]

b)

Specific volume:

[tex]\bar V=\frac{6}{0.5}[/tex]

[tex]\bar V=12\ m^3.kmol^{-1}[/tex]

also

[tex]\b V=\frac{V}{m}[/tex]

[tex]\b V=\frac{6}{8.5}[/tex]

[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]

Answer:

0.044 V

Explanation:

E = Electric field = [tex]5.5\times 10^6\ V/m[/tex]

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

[tex]V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V[/tex]

The voltage across the membrane is 0.044 V

Final answer:

The voltage across an 8.00 nm-thick membrane with an electric field strength of 5.50 MV/m is calculated using the formula V = Ed, resulting in a voltage of 44.0 mV.

Explanation:

The voltage across a membrane can be determined by using the relationship between electric field strength (E), voltage (V), and the distance (d) the electric field spans. The electric field is uniform and the formula to use is V = Ed. In this case, the electric field (E) is given as 5.50 MV/m or 5.50 x 10⁶V/m, and the thickness of the membrane (d) is 8.00 nm or 8.00 x 10⁻⁹ m.

To find the voltage across the membrane, we simply multiply the electric field by the thickness of the membrane:

V = (5.50 x 10⁶ V/m) x (8.00 x 10⁻⁹ m)

Therefore:

V = 44 x 10⁻³ V

V = 44.0 mV

So, the voltage across an 8.00 nm-thick membrane with the given electric field strength is 44.0 mV.

Answer:

1/3

Explanation:

Take the displacement and divide it by the speed of light. Meters will cancel leaving you with seconds. :)

110m / 343m/s = 0.32069971 seconds or 1/3 seconds

The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer  1/3 second ( approx.). So, option (d) is correct.

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

Given The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.

So, distance between the  first dancer in the line and the last dancer in the line be = 120 m -10 m = 110 m

Velocity of sound in air be = 343 m/s.

Hence, The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer = distance/velocity = 110/343 second = 1/3 second ( approx.).

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Answer:

option C

Explanation:

Given,

Change on the capacitor = Q

Separation is doubled

Energy stored in the capacitor,E = ?

we know,

[tex]E = \dfrac{Q^2}{2C}[/tex]

and [tex]C = \dfrac{\epsilon_0A}{d}[/tex]

now,

[tex]E = \dfrac{Q^2}{2\epsilon A}\ d[/tex].......(1)

where d is the separation between the two plates.

now, when the separation is doubled

[tex]E' = \dfrac{Q^2}{2\epsilon A}\ (2d)[/tex]

[tex]E' = 2(\dfrac{Q^2}{2\epsilon_0 A}\ d)[/tex]

From equation (1)

  E' = 2 E

Hence, the energy stored in the capacitor is doubled if the separation is increased.

The correct answer is option C.

Answer:

a) The positive plate is at a higher potential than the negative plate.

b) Electric field between the plates = 5.76 KV/m

Explanation:

a) The positive plate is at a higher potential than the negative plate because of convention. We could define the negative plate to be "high potential." However, that convention has to be consistent. The positive plate is at the high potential, and that potential causes positive charges to flow from high potential to low potential. If we had defined the convention in the other direction, then the negative terminal would be the high potential, and that potential would cause negative charges to flow from high potential to low potential.

b) Electric field, E = potential difference across plate/distance or separation between the plates

E = V/d

V = 490V, d = 8.5cm = 0.085m

E = 490/0.085 = 5,764.706 V/m = 5.76 KV/m

Hope this helps!!!

Answer:

Frequency of the vibration due to crevice is 1500 Hz.

Explanation:

Frequency is defined as rate of vibration per second. Its S.I. unit is s⁻¹ or Hz.

According to the problem, we have to find number of crevice car makes in 1 second.

Speed of car, u = 30 m/s

Distance covered by car in 1 second, d = 30 m

For every 0.02 m, one crevice occurred by the tire of car.  

Number of crevice occurred in 1 second by the car =  [tex]\frac{30}{0.02}[/tex]

                                                                                    = 1500

Since, each crevice makes a single vibration. Thus, the frequency of these vibrations is 1500 Hz.

The frequency of the vibrations generated by the tire treads in this case, given that each crevice is 2.00 cm apart and the car moves at 30.0 m/s, is 1500 Hz.

In this scenario, it is important to understand that the frequency of the vibrations is determined by the speed of the tire and the tread pattern. Here, the crevice, which causes the vibration, occurs every 2.00 cm (or 0.02 m). This means for every meter the car moves, 0.02 m into 1 m gives a total of 50 vibrations. Therefore, at a speed of 30.0 m/s, we will have 50 vibrations/m multiplied by 30 m/s, giving a frequency of 1500 Hz or vibrations per second.

Frequency in this question refers to the number of times the vibration from the crevices occur in a unit of time (in this case, per second). This concept is crucial in understanding wave motions and vibration patterns, and is often applied in physics-related disciplines.

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Answer:

No, the deviation in the path of asteroid is not by 22°

Explanation:

Given:

velocity of asteroid, [tex]v_a=21\ km.s^{-1}[/tex]

acceleration of the rocket, [tex]a=0.035\ km.s^{-2}[/tex]

time of acceleration, [tex]t=40\ s[/tex]

Now, the final velocity of the asteroid:

using the equation of motion,

[tex]v=u+a.t[/tex]

where:

[tex]v=[/tex] final velocity

[tex]u=[/tex] initial velocity in the direction

[tex]v=0+0.035\times 40[/tex]

[tex]v=1.4\ km.s^{-1}[/tex]

Now direction of the resultant velocity:

[tex]\tan\beta=\frac{v_a}{v}[/tex]

[tex]\tan\beta=\frac{21}{1.4}[/tex]

[tex]\beta=86.186^{\circ}[/tex]

So, the deviation in the asteroid:

[tex]\theta=90-\beta[/tex]

[tex]\theta=90-86.186[/tex]

[tex]\theta=3.814^{\circ}[/tex]

Answer:

30 degrees

Explanation:

The boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.

What is vector law of addition ?

Vector addition is governed by the triangle law which states that when two vectors are represented by two triangle sides with their order of magnitude and direction, the resultant vector's magnitude and direction will be represented by the third triangle side.

It is given that speed of boat in still water v₁ = 4 km/h

speed of current v₂ = 2 km/h

Let the relative speed of boat with respect to current be =  v km/h

and the angle made by v₁ with v be = θ

No find the angle θ with which the boat to go straight across the river uptream by triangle law of vector addition as shown below :

[tex]\begin{aligned}\theta &=\text{sin}^{-1}\left ( \frac{2}{4} \right )\\&= 30^{0}\end{aligned}[/tex]

Therefore, the boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.

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Answer:

Final velocity will be equal to 0.321 m/sec

Explanation:

We have given mass of clay model of lion [tex]m_1=0.225kg[/tex]

Its speed is 0.85 m/sec, so [tex]v_1=0.85m/sec[/tex]

Mass of another clay model [tex]m_2=0.37kg[/tex]

It is given that second clay is motionless

So its velocity [tex]v_2=0m/sec[/tex]

Now according to conservation of momentum

Momentum before collision will be equal to momentum after collision

So [tex]m_1v_!+m_2v_2=(m_1+m_2)v[/tex], here v is velocity after collision

So [tex]0.225\times 0.85+0.37\times 0+(0.225+0.37)v[/tex]

[tex]0.595v=0.191[/tex]

v = 0.321 m/sec

So final velocity will be equal to 0.321 m/sec  

Answer

given,

height of the jump = 0.44 m

acceleration due to gravity, g = 9.8 m/s²

velocity at the height point = 0 m/s

initial speed = ?

Using equation of motion for speed calculation

v² = u² + 2 g h

0 = u² - 2 x 9.8 x 0.44

u = √8.624

u = 2.94 m/s

time taken to reach the highest point

v = u + g t

0 = 2.94 - 9.8 x t

t = 0.3 s

total time of flight will be equal to double of the time taken to reach the maximum height.

Total time = 2 x 0.3 = 0.6 s

A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.

A flea jumps straight up to a height (h) of 0.440 m. At the maximum height, the final speed (v) is zero. Given that the gravity (g) is 9.81 m/s², we can calculate the initial speed (u) using the following kinematic expression (We will assume y+ as the positive direction)

[tex]v^{2} = u^{2} + 2 \times g \times h \\\\(0m/s)^{2} = u^{2} + 2 \times (-9.81m/s^{2} ) \times 0.440m\\\\u = 2.94 m/s[/tex]

We can calculate the time (t) required to reach the maximum height using the following kinematic expression.

[tex]v = u + g \times t\\\\0m/s = 2.94m/s - 9.81m/s^{2} \times t\\\\t = 0.300 s[/tex]

The time in the air will be double the time to reach the maximum height.

[tex]t_{total} = 2 \times 0.300 s = 0.600 s[/tex]

A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.

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Answer:

[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]

Explanation:

For this case  we can use the second law of Newton given by:

[tex] \sum F = ma[/tex]

The friction force on this case is defined as :

[tex] F_f = \mu_k N = \mu_k mg [/tex]

Where N represent the normal force, [tex]\mu_k [/tex] the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

[tex] F_f = ma[/tex]

Replacing the friction force we got:

[tex] \mu_k mg = ma[/tex]

We can cancel the mass and we have:

[tex] a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}[/tex]

And now we can use the following kinematic formula in order to find the distance travelled:

[tex] v^2_f = v^2_i - 2ad[/tex]

Assuming the final velocity is 0 we can find the distance like this:

[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]

The hockey puck will travel approximately 58.5 meters across the floor before it stops.

To solve this problem, we will use the concepts of kinetic energy and the work-energy theorem.

Initially, the hockey puck has a kinetic energy of KE1 = 0.5*m*v^2 = 0.5*0.3 kg*(20 m/s)^2 =60 J

As it travels across the floor, the friction does work on it until it stops. The work done by the friction force is W = μ*m*g*d, where: μ is the coefficient of kinetic friction (0.35), m is the mass (0.3 kg), g is the gravitational constant (9.8 m/s^2), and d is the distance traveled.

The work done by the friction is equal to the initial kinetic energy (work-energy theorem), so we can solve for d: 60 J = 0.35 * 0.3 kg * 9.8 m/s^2 * d. Simplifying this equation gives d = 60 J / (0.35 * 0.3 kg * 9.8 m/s^2) = 58.5 m

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Answer:

Yes it will slide down the ramp

Explanation:

Let m be the mass of the box and gravitational acceleration g = 9.81m/s2, we can calculate the gravity that affects the box

W = mg

When the box is at 35 degree incline, this gravity is split into 2 components, 1 parallel to the incline (Wsin35) and the other one perpendicular with the incline (Wcos35).

The one perpendicular with the incline has an equal and opposite normal force of Wcos35

This normal force would dictate the static friction force where coefficient = 0.5. So the static friction is 0.5mgcos35

The box would slide when the parallel component of gravity wins over static friction force

mgsin35 > 0.5mgcos35

Since mg is positive we can cancel them out on both sides

sin35 > 0.5cos35

0.57 > 0.5*0.82

0.57 > 0.41

This is true so we can conclude that the box slides down the ramp

Answer:

The right option is option E. None of the answer choices given are totally correct.

Explanation:

All insulators normally have an equal amount of positive and negative charges distributed on their surface.

The amber rod (an insulator) is called negative because after the coming together with fur (another insulator), the amber rod rubs off electrons from the fur onto itself and has an overall more negatively charged particles than positively charged particles on its surface.

The fur in turn becomes positive because it has more positive charges than negative on its surface.

So, the convention allows the now rubbed off amber rod to be called negative.

So, it is evident that none of the answer choices are totally correct, the right answer is more of a mix of some of the answer choices and more!

Hope this helps!!

Answer:

Explanation:

Given

mass of vehicle [tex]m=1500\ kg[/tex]

Speed of vehicle [tex]u=50\ km/hr\approx 13.89\ m/s[/tex]

Kinetic Energy Possessed by mass

[tex]K_i=\frac{1}{2}mu^2[/tex]

[tex]K_i=\frac{1}{2}\times 1500\times (13.89)^2[/tex]

[tex]K_i=144.69\ kJ[/tex]

when vehicle is slowed down to speed of [tex]v=10\ km/hr\approx 2.78\ m/s[/tex]

Kinetic Energy at this speed

[tex]K_f=\frac{1}{2}mv^2[/tex]

[tex]K_f=\frac{1}{2}\times 1500\times (2.78)^2[/tex]

[tex]K_f=5.78\ kJ[/tex]

Energy Recovered [tex]=K_i-K_f[/tex]

Energy Recovered[tex]=144.69-5.78=138.9\ kJ[/tex]        

Explanation:

Implicit techniques for the discovery of extra-solar planets are used by astronomers. Evidence from the radial velocity of a change in the rotation of the star indicates the presence of a planet. If, with reference to Earth, the inclination of the planet's orbit is viewed as an edge-on, the detection of light originating from the star throughout its planetary transit would then be a verification.

The spring does 0.1 Joules of work when the object is released and travels back to its initial position.

We have,

The work done by the spring can be calculated using the formula for the potential energy stored in a spring:

Potential Energy (PE) = 0.5 * k * x²,

where k is the spring constant and x is the displacement from the equilibrium position.

Given:

Spring constant (k) = 20 N/m,

Displacement (x) = 0.10 m (10 cm).

Calculate the potential energy when the spring is stretched to 10 cm:

PE = 0.5 * k * x²

= 0.5 * 20 N/m * (0.10 m)²

= 0.5 * 20 * 0.01 J

= 0.1 J.

The spring stores 0.1 Joules of potential energy when stretched to 10 cm.

When the mass is released and returns to its initial position, this potential energy is converted back into kinetic energy as the mass accelerates.

Therefore,

The spring does 0.1 Joules of work when the object is released and travels back to its initial position.

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Final answer:

The spring does 0.1 Joules of work when the mass travels back to its initial position.

Explanation:

To find the work done by a spring when an object is released and travels back to its initial position, we can use the formula:

Work = (1/2) * k * x^2

Where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring constant is 20 N/m and the displacement is 10 cm (which is 0.1 m). Plugging these values into the formula, we get:

Work = (1/2) * 20 N/m * (0.1 m)^2 = 0.1 J

So, the spring does 0.1 Joules of work when the mass travels back to its initial position.

Answer:

[tex]Q = T1 - T2 (KA)/L[/tex]

Explanation:

Where Q is the co-efficient of heat transfer.

T-1 is initial temperature of exchanger

T-2 is final temperature of sink where has to to be dissipated

k is the co-efficient of thermal conductivity

A is the total area of exchanger surface...

and L is the total length of exchanger...

Answer:

Total weight of the hot tub and water is 5676.6 pounds-force

Explanation:

rho = 62.4lbm/ft^3 × 1ft^3/7.481gal = 8.34lbm/gal

Mass of water = rho × volume = 8.34lbm/gal × 600 gallons = 5004lbm = 5004×0.45359kg = 2269.8kg

Total mass of hot tub and water = 320kg + 2269.8kg = 2589.8kg

Local acceleration due to gravity = 32ft/s^2 = 32ft/s^2 × 1m/3.2808ft = 9.75m/s^2

Total weight of hot tub and water = 2589.8kg × 9.75m/s^2 = 25250.55N = 25250.55/4.4482 lbf = 5676.6 lbf

Answer:

Explanation:

Given

W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)

According to work energy theorem work done by all the forces is equal to change in kinetic energy of object

[tex]W=\frac{1}{2}mv^2---1[/tex]

where m=mass of object

v=velocity of object

When the object is already have velocity v then the final speed is given by work energy theorem

[tex]W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2[/tex]

From 1 and 2 we get

[tex]\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2[/tex]

[tex]2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2[/tex]

[tex]v_f^2=2v^2[/tex]

[tex]v_f=\sqrt{2}v[/tex]                

Answer:

a) E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol

b) E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol

c) E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol

Explanation:

E = hf where E = energy, H = Planck's constant = 6.62607004 × 10⁻³⁴ J.s and f = 1/time period

a) Period = 1 fs = 1 × 10⁻¹⁵ s

f = 1/(10⁻¹⁵ ) = 10¹⁵ Hz

E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁵ = 6.63 × ⁻10¹⁹ J = 6.63 × 10⁻¹⁶ KJ

In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol

E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol

b) Period = 10 fs = 10 × 10⁻¹⁵ = 10⁻¹⁴ s

f = 1/period = 1/10⁻¹⁴ = 10¹⁴ Hz

E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁴ = 6.63 × 10^-20 J = 6.63 × 10⁻¹⁷ KJ

In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol

E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol

c) Period = 1s

f = 1/period = 1.0 Hz

E = 6.62607004 × 10⁻³⁴ × 1 = 6.63 × 10⁻³⁴ J = 6.63 × 10⁻³¹ KJ.

In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol

E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol

Answer:

(i) E =  6.626 X 10⁻¹⁹ J    = 400 KJ/mol

(ii) E = 6.626 X 10⁻²⁰ J   = 40 KJ/mol

(iii) E = 6.626 X 10⁻³⁴ J   = 4 X 10⁻¹³ KJ/mol

Explanation:

Energy associated with excitation of a quantum is given as;

E = hf

where;

E is the energy of excitation

h is Planck's constant = 6.626 X 10⁻³⁴Js⁻¹

f is the threshold frequency in s⁻¹

Thus, E = h/t

Part (i)

E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁵)

E = 6.626 X 10⁻¹⁹ J  

In (KJ/mol) = 6.626 X 10⁻²² KJ X 6.022 X10²³ = 400 KJ/mol

Part (ii)

E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁴)

E = 6.626 X 10⁻²⁰ J  

In (KJ/mol) = 6.626 X 10⁻²³ KJ X 6.022 X10²³ = 40 KJ/mol

Part (iii)

E = (6.626 X 10⁻³⁴)/(1)

E = 6.626 X 10⁻³⁴ J  

In (KJ/mol) = 6.626 X 10⁻³⁷ KJ X 6.022 X10²³ = 4 X 10⁻¹³ KJ/mol

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

[tex]W=mg[/tex]: weight of the ladder, with m = 20 kg (mass) and [tex]g=9.8 m/s^2[/tex] (acceleration of gravity)

[tex]W_M=Mg[/tex]: weight of the person, with M = 54 kg (mass)

[tex]N_1[/tex]: normal reaction exerted by the wall on the ladder

[tex]N_2[/tex]: normal reaction exerted by the floor on the ladder

[tex]F_f = \mu N_2[/tex]: force of friction between the floor and the ladder, with [tex]\mu[/tex] (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

[tex]W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}[/tex]

where

[tex]Wsin 21^{\circ}[/tex] is the component of the weight of the ladder perpendicular to the ladder

[tex]W_M sin 21^{\circ}[/tex] is the component of the weight of the man perpendicular to the ladder

[tex]N_1 sin 69^{\circ}[/tex] is the component of the normal  force perpendicular to the ladder

And solving for [tex]N_1[/tex], we find the force exerted by the wall on the ladder:

[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N[/tex]

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of [tex]N_2[/tex].

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

[tex]\sum F_y = 0\\N_2 - W - W_M =0[/tex]

And substituting and solving for N2, we find:

[tex]N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N[/tex]

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

[tex]\sum F_x = 0\\F_f - N_1 = 0[/tex]

And re-writing the equation,

[tex]\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257[/tex]

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

[tex]\mu = \frac{N_1}{N_2}[/tex]

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}[/tex]

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)[/tex]

And substituting, we get

[tex]N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N[/tex]

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

[tex]\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332[/tex]

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Answer:

F=5618278.8 N

Explanation:

Given that

m = 970 kg

Initial speed ,u = 52 km/h

u = 14.44 m/s          ( 1 km/h = 0.277 m/s)

Distance d= 1.8 cm  = 0.018 m

The final speed ,v = 0 m/s

We know that

v² = u² + 2 a d

a=Acceleration

0² = 14.44² + 2 x a x 0.018

[tex]a=-\dfrac{14.44^2}{2\times 0.018}\ m/s^2[/tex]

a=5792.04 m/s²

We know that

Force = Mass x acceleration

F=  m a

F= - 5792.04 x 970 N

F= - 5618278.8 N

Therefore the magnitude of the force will be 5618278.8 N.

F=5618278.8 N

Answer:

Q1. corect is B, Q2. it is A, Q3.  E and Q4. A  

Explanation:

Q1 For this exercise we can use Newton's second law where acceleration is centripetal.

              F = m a

              a = v² / r.

              G m M / r² = m v² / r

               G M / r = v²

The velocity has a constant magnitude whereby we can divide the length of the circular orbit (2π r) between the period

               G M / r = (2π r / T)²

                r³ = G M T2 / 4π²

Let's calculate

              T = 103 day (24 h / 1 day) (3600 s / 1h) = 8,899 10⁶ s

              r³ = 6.67 10⁻¹¹  1.99 10³⁰ (8,899 10⁶) 2 / 4π²

              r = ∛ (266.25 10³⁰)

              r = 6.4 10¹⁰  m

The distance matches the value in part B

Q2 Astronauts have measured the acceleration of gravity, so we can use the second law with a body on the planet's surface

            F = m g

            G m M_p / R_p² = m g

             G M_p / R_p² = g

              M_p = g R_p² / G

They indicate that the radius of the planet is half the radius of the Earth

              R_p = ½ R_earth

              R_p = ½ 6.37 10⁶

              R_p = 3.185 10⁶ m

Let's calculate

             M_p = 8.2 (3,185 10⁶)² / 6.67 10⁻¹¹

             M_p = 1.25 10²⁴ kg

The correct answer is A

Q3 We use Newton's second law again, with part Q1, where M is the mass of the planet and m is the mass of the moon

                r³ = G M T² / 4π²

               T = 63 days (24h / 1day) (3600s / 1h) = 5.443 10⁶ s

               r³ = 6.67 10⁻¹¹ 1.25 10²⁴ (5.443 10⁶)² / 4π²

               r = ∛ (62.56807 10²⁴)

               r = 3.97 10⁸ m

The correct answer is E

Q4 To calculate this part let's use the conservation of mechanical energy,

Starting point The surface of the moon

          Em₀ = K + U = ½ m v2 - G m M / r

Final point. Infinity with zero speed

           [tex]Em_{f}[/tex] = 0

              Em₀ = Em_{f}

              ½ m v² - G m M / R = 0

              v² = 2 G M / r

              M = v2 r / 2G

              r = 2 G M / v²

Since we don't know the radius of the moon, we will also use the equation in part 2

              M = g r² / G

               r = √ GM / g

Let's replace

              2G M / v² = √ G M / g

              4 G M / v⁴ = 1 / g

              M = v⁴ / (g 4G)

              M = 3000⁴ / (2.7  4 6.67 10-11)

              M = 1.12 10²³ kg

corract is  A

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees  downward from the horizontal?

Answer:

[tex]W_{work}=2.67*10^{-3}J[/tex]

Explanation:

Given data

Charge q=28 nC

Electric field E=5.00×10⁴ V/m.

Distance d=2.70 m

Angle α=45°

To find

Work done by electric force

Solution

[tex]W_{work}=F_{force}*D_{distance}Cos\alpha \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J[/tex]

Answer:

v = 30.15 m/s

a = 60.07 m/s2

Explanation:

Velocity is derivative of position with respect to time

[tex]v_x = x' = 3*0.25t^2 = 0.75t^2[/tex]

[tex]v_z = z' = 5*0.75t^4/2 = 1.875t^4[/tex]

Acceleration is derivative of velocity with respect to time

[tex]a_x = v_x' = 2*0.75 t = 1.5t[/tex]

[tex]a_z = v_z' = 4*1.875t^3 = 7.5t^3[/tex]

At t = 2 seconds

[tex]v_x = 0.75*2^2 = 3m/s[/tex]

[tex]v_z = 1.875*2^4 = 30m/s[/tex]

[tex]v = \sqrt{v_x^2 + v_z^2} = \sqrt{3^2 + 30^2} = \sqrt{909} = 30.15 m/s[/tex]

[tex]a_x = 1.5*2 = 3 m/s^2[/tex]

[tex]a_z = 7.5*2^3 = 60 m/s^2[/tex]

[tex]a = \sqrt{a_x^2 + a_z^2} = \sqrt{3^2 + 60^2} = \sqrt{3609} = 60.07 m/s[/tex]

Answer:

I. The hollow sphere has the greater angular momentum.

Explanation:

Given that the mass and radius of hollow sphere and solid sphere are same. Let the mass and radius of two spheres be m and r respectively. The two spheres are rotating having same angular velocity ω .

Moment of inertia of solid sphere, I₁ = [tex]\frac{2}{5}\times{m}r^{2}[/tex]

Moment of inertia of hollow sphere, I₂ = [tex]\frac{2}{3}\times{m}r^{2}[/tex]

Since, I₁ and I₂ are not equal. Therefore, the statement iv is wrong.

The relation between angular momentum, moment of inertia and angular velocity is :

L = Iω

Let L₁ and L₂ be the angular momentum of solid sphere and hollow sphere respectively.

L₁ = I₁ω     and   L₂ = I₂ω

As ω is same for both spheres but I₂ is greater than I₁, hence L₂ is greater than L₁.

Therefore, statement I is correct that the hollow sphere has the greater angular momentum.

The angular momentum of the hollow sphere is greater than that of solid sphere.

The moment of inertia of solid sphere is given as follows;

[tex]I_{ss} = \frac{2}{5} mr^2 = 0.4mr^2[/tex]

The moment of inertia of hollow sphere is given as follows;

[tex]I_{hs} = \frac{2}{3} mr^2 = 0.67 mr^2[/tex]

The angular momentum of each sphere is calculated as follows;

[tex]L = I\omega \\\\L_{ss} = 0.4mr^2 \omega \\\\L_{hs} = 0.67 mr^2 \omega[/tex]

Thus, we can conclude that the angular momentum of the hollow sphere is greater than that of solid sphere.

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Answer:

C

Explanation:

Correct answer: C. Fly the approach to runway 26. The winds from the storm could suddenly gust up and you don’t want to be in a short final or even in the flare when a tailwind from a storm gusts up. No other traffic and calm winds currently means that you can land on any runway you want. Go for the safe bet and land on 26.

In this scenario, the pilot should fly a left hand approach to runway 08.

In this scenario, with a calm wind, the pilot should fly a left hand approach to runway 08.

Since the wind is calm, there is no need for the pilot to adjust the approach. Flying in a clockwise pattern to runway 08 is the normal procedure.

Choosing any other option would not be appropriate or necessary.

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Answer:

(a) [tex]I_{1}=3.2*10^{-3}W/m^{2}[/tex]

(b) [tex]\beta =95dB[/tex]

Explanation:

Given data

Distance r₁=50 m

Distance r₂=2 m

Intensity I₂=2.0 W/m²

To find

(a) The Sound Intensity I₁

(b) The Sound Intensity level β

Solution

For (a) the Sound Intensity I₁

[tex]\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}[/tex]

For (b) the Sound Intensity level β

The Sound Intensity level β is calculated as follow

[tex]\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB[/tex]

Answer:

Explanation:

initial speed, u = 19 m/s

(a) Let it rises upto height h.

Use third equation of motion

v² = u² - 2 gh

where, v is the final velocity and it is zero.

0 = 19 x 19 - 2 x 9.8 x h

h = 18.4 m

(c) Let the ball takes time t to reach to the maximum height.

use first equation of motion

v = u - gt

0 = 19 - 9.8 x t

t = 1.94 s

(c) The time taken by the ball to reach to the ground = 2 x time to reach to maximum height

T = 2 x t = 2 x 1.94 = 3.88 s

(d) When the ball reaches the ground, let the velocity is v.

Use third equation of motion

v² = u² - 2 gh

where, v is the final velocity

v² = 0 + 2 x 9.8 x 18.4

v = 19 m/s

(a) The ball reaches a height of approximately 18.68 meters. (b) It takes approximately 1.94 seconds to reach its highest point  (c) The time taken to hit the ground is 1.94 seconds. When the ball returns to the level from which it started, its velocity is -19.0 m/s.

(a) To find the height that the ball reaches, we can use the kinematic equation:
Δy = v2 / (2g)
where Δy is the change in height, v is the initial velocity, and g is the acceleration due to gravity. Plugging in the given values, we have:
Δy = (19.0 m/s)2 / (2 * 9.8 m/s2)
Calculating, we find that the ball rises to a height of approximately 18.68 meters.
(b) The time it takes for the ball to reach its highest point can be calculated using the equation:

t = v / g
where t is the time, v is the initial velocity, and g is the acceleration due to gravity. Substituting the given values, we get:
t = (19.0 m/s) / (9.8 m/s2)
Simplifying, we find that it takes approximately 1.94 seconds for the ball to reach its highest point.
(c) Since the time it takes for the ball to reach its highest point is the same as the time it takes for it to fall back down, the time it takes for the ball to hit the ground after reaching its highest point is also 1.94 seconds.
(d) When the ball returns to the level from which it started, its velocity is equal in magnitude but opposite in direction to its initial velocity. Therefore, the velocity when it returns is -19.0 m/s.

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a) The average true power is 318.3 W

b) The reactive power is 132.6 W

c) The apparent power is 344.8 W

d) The power factor is 0.92

Explanation:

a)

For a circuit made of a resistor and a capacitor, the average (true) power is given by the resistive part of the circuit only.

Therefore, the average true power is given by:

[tex]P=I^2R[/tex]

where

I is the current

R is the resistance

In this problem, we have

V = 67 V (rms voltage)

[tex]R=12 \Omega[/tex] is the resistance of the load

[tex]X=5\Omega[/tex] is the reactance of the circuit

First we have to find the impedance of the circuit:

[tex]Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega[/tex]

Then we can find the current in the circuit by using Ohm's law:

[tex]I=\frac{V}{Z}=\frac{67}{13}=5.15 A[/tex]

Therefore, the average true power is

[tex]P=I^2R=(5.15)^2(12)=318.3 W[/tex]

b)

The reactive power of a circuit consisting of a resistor and a capacitor is the power given by the capacitive part of the circuit.

Therefore, it is given by

[tex]Q=I^2X[/tex]

where

I is the current

X is the reactance of the circuit

In this circuit, we have

[tex]I=5.15 A[/tex] (current)

[tex]X=5 \Omega[/tex] (reactance)

Therefore, the reactive power is

[tex]Q=(5.15)^2(5)=132.6W[/tex]

c)

In a circuit with a resistor and a capacitor, the apparent power is given by both the resistive and capacitive part of the circuit.

Therefore, it is given by

[tex]S=I^2Z[/tex]

where

I is the current

Z is the impedance of the circuit

Here we have

I = 5.15 A (current)

[tex]Z=13 \Omega[/tex] (impedance)

Therefore, the apparent power is

[tex]S=I^2 Z=(5.15)^2(13)=344.8 W[/tex]

d)

For a circuit with a resistor and a capacitor, the power factor is the ratio between the true power and the apparent power. Mathematically:

[tex]PF=\frac{P}{S}[/tex]

where

P is the true power

S is the apparent power

For this circuit, we have

P = 318.3 W (true power)

S = 344.8 W (apparent power)

So, the power factor is

[tex]PF=\frac{318.3}{344.8}=0.92[/tex]

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