A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses. But he loses them while travelling. Fortunately he has his old pair as a spare. (a) If the lenses of the old pair have a power of 2.25 diopters, what is his near point (measured from the eye) when wearing the old glasses, if they rest 2.0 cm in front of the eye

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

Answer:

The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.

Explanation:

Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.

Answer:

1) designed to measure the difference in speed of light in different directions , 1887

Explanation:

1) This experiment was designed to measure the difference in speed of light in different directions and therefore find the speed of the ether.

2) was made in 1887

3) At that time it was assumed that it was the medium in which light traveled and it is everywhere

4) the speed of the wave depends on the characteristics of the medium where it travels,

for the one in a string depends on the tension and density

for an electromagnetic wave of the permittivity and permeability of the vacuum

5) In this type of interferometer the beam is divided into two rays

6) In his interrupter, he had to accurately measure the displacement of the fringes in a telescope, for which he had to minimize vibrations, he had problems in the movement of one of the arms, changes in temperature

7) In Michelsom's second experiment, the apparatus could measure 0.01 fringes by increasing the length of the arms by 11 m

8) The new interferometer floated on a bed of mercury

9) Couldn't measure any difference in speed of light in different directions

10) Physics was forced to eliminate the concept of ETHER

11) One of the principles of relativities that the speed of light is constant in all inertial efficiency systems

12) Michelson in 1907

13) It seems that Einstein did not know the results of this experiment

Answer:

v = 6.28 m/s

Explanation:

It is given that,

A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,

Number of revolution is half. It means angular velocity is 3.14 radians.

Let v is the angular speed. So,

[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]

So, the rotation speed is 6.28 m/s.

The angular velocity is the rotation speed, which is the angle of rotation

of the windmill per second, which is 2·π radians.

Response:

The given parameter are;

The angle of rotation the windmill rotates in 0.5 seconds = One-half a

rotation.

Required:

The rotational speed (angular velocity)

Solution:

The angle of one rotation = 2·π radians

Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians

[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]

Which gives;

[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]

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Answer:

Explanation:

[tex]Power = IV[/tex]

From ohms law we know that

[tex]V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}[/tex]

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

[tex]P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}[/tex]

[tex]\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W[/tex]

The  resistor will dissipate 0.056 Watt

Answer:

in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.

Answer:

Explanation:

That Universe Consists of Matter

Answer:

M = F/3μ g - M₁/3

Explanation:

To solve this exercise we must use the equilibrium conditions translations

         ∑ F = 0

In the attachment we can see a free body diagram of each block

Block M (upper)

X axis

      fr₁ + F₂ -F = 0

      F = fr₁ + F₂              (1)

axis

     N₁-W = 0

     N₁ = Mg

the friction force has the formula

     fr₁ = μ N₁

     F = μ Mg + F₂

bottom block

X axis

     F₂ - fr₁ - fr₂ = 0

     F₂ = fr₁ + fr₂

Y axis

     N - W₁ -W = 0

     N = g (M + M₁)

we substitute

       F₂ = μ Mg + μ (M + M1) g

       F₂ = μ g (2M + M₁)

we substitute in 1

      F = μ M g + μ g (2M + M₁)

      F = μ g (3M + M₁)

we look for mass M    

      M = (F -  μ g M₁)/ 3μ g

      M = F/3μ g - M₁/3

the exercise does not have numerical data

Answer:

The moment of inertia is  [tex]I= 312.09 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

    The  mass of the platform is   m =  137 kg

     The radius is  r  =  1.53 m

    The mass of the person is  [tex]m_p = 68.7 \ kg[/tex]

    The distance of the person from the center is  [tex]d_c =1.19 \ m[/tex]

    The mass of the dog is  [tex]m_d = 25.9 \ kg[/tex]

     The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]

Generally the moment of inertia of the system is mathematically represented as

      [tex]I = I_1 + I_2 + I_3[/tex]

Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as

          [tex]I_1 = \frac{m * r^2}{2}[/tex]

substituting values

           [tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]

           [tex]I_1 = 160.35 \ kg\cdot m^2[/tex]

Also  [tex]I_2[/tex]  is the moment of inertia of the person about the axis which is mathematically represented as

          [tex]I_2 = m_p * d_c^2[/tex]

substituting values

          [tex]I_2 = 68.7 * 1.19^2[/tex]

          [tex]I_2 = 97.29 \ kg \cdot m^2[/tex]

Also  [tex]I_3[/tex]  is the moment of inertia of the dog about the axis which is mathematically represented as

          [tex]I_3 = m_d * d_d^2[/tex]

substituting values

          [tex]I_3 = 25.9 * 1.45^2[/tex]

          [tex]I_3 = 54.45 \ kg \cdot m^2[/tex]

Thus  

        [tex]I= 160.35 + 97.29 + 54.45[/tex]

        [tex]I= 312.09 \ kg \cdot m^2[/tex]

Answer:

t = 134834.31 years

Explanation:

First we find the speed of the ship:

v = 0.890 c

where,

v = speed of the ship = ?

c = speed of light = 3 x 10⁸ m/s

Therefore, using the values, we get:

v = (0.89)(3 x 10⁸ m/s)

v = 2.67 x 10⁸ m/s

Now, we find the distance in meters:

Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)

s = 11.35 x 10²⁰ m

Now, for the time we use the following equation:

s = vt

t = s/v

t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)

t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)

t = 134834.31 years

Answer:

E. None of these statements are true.

Explanation:

We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.

for A, the tension on the rope does not depend on if both teams pull are in equilibrium.

for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.

for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.

for D,  just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.

we go with option E.

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

(a)  the final speed of the water after contraction is 15 m/s.

(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.

The given parameters;

Let the initial area of the pipe = A₁

Apply the continuity equation to determine the final speed of the water after contraction as follows;

[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]

The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;

[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]

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Answer:

The intensity is  [tex]I_2 = 1654 \ W/m^2[/tex]

Explanation:

From the question we are told that

    The intensity of the unpolarized light is  [tex]I_o = 4000 \ W/m^2[/tex]

    The  angle between the ideal polarizing sheet is  [tex]\theta = 0.429 \ rad = 0.429 * 57.296 = 24.58^o[/tex]

Generally the intensity of  light emerging from the first polarizer is mathematically represented as

               [tex]I_2 = \frac{I_o}{2}[/tex]

substituting values

               [tex]I_1 = \frac{4000}{2}[/tex]

                [tex]I_1 = 2000 \ W/m^2[/tex]

Then the intensity of  incident light emerging from the second polarizer is mathematically represented by Malus law as

                 [tex]I_2 = I_1 cos^2 (\theta )[/tex]

substituting values

                [tex]I_2 = 2000 * [cos (24.58)]^2[/tex]

                [tex]I_2 = 1654 \ W/m^2[/tex]

Answer:

Explanation:

Find:

Computation:

F = (9/5)C + 32

3 times that of the Celsius

If C = x

So F = 3x

So,

3x = (9/5)x + 32

15x = 9x +160

6x = 160

x = 26.67

So, C = 26.67° and F = 80°

1/5 times that of the Celsius

If C = x

So F = x/5

So,

x/5 = (9/5)x + 32

x = 9x + 160

x = -20

So, C = -20° and F = -4°

Answer:

 t = 8.98 10⁻⁷ m

Explanation:

This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.

* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180

* The beam when passing to the middle its wavelength changes

           λ = λ₀ / n

if we take this into account, the constructive interference equation for normal incidence is

            2t = (m + ½) λ₀ / n

let's apply this equation to our case

for λ₀ = 479 nm = 479 10⁻⁹ m

             t = (m + ½) 479 10⁻⁹ / 1.33

             (m + ½) = 1.33 t / 479 10⁻⁹

for λ₀ = 798 nm = 798 10⁻⁹ m

             t = (m' + ½) 798 10⁻⁹ /1.33

            (m' + ½) = 1.33 t / 798 10⁻⁹

as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions

               (m + ½) = 1.33 t / 479 10⁻⁹

             ((m-1) + ½) = 1.33 t / 798 10⁻⁹

             (m + ½) = 1.33 t / 798 10⁻⁹ +1

resolve

             1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1

             1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹

             1.33t = (1 .33t + 798 10⁻⁹) 479/798

             1.33t = (1 .33t + 798 10⁻⁹) 0.6

             1.33 t = 0.7983 t + 477.6 10⁻⁹

             t (1.33 - 0.7983) = 477.6 10⁻⁹

             t = 477.6 10⁻⁹ /0.5315

             t = 8.98 10⁻⁷ m

Answer:

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Explanation:

When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.

In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:

[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]

Where:

[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.

If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:

[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]

[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]

And change in heat energy ([tex]Q[/tex]), measured by joules, by:

[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]

Where:

[tex]n[/tex] - Molar quantity, measured in moles.

The final temperature of the monoatomic ideal gas is now cleared:

[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]

Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:

[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]

[tex]T_{2} = 420.279\,K[/tex]

The final pressure of the system is calculated by the following relationship:

[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]

If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:

[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]

[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Answer:

The planet that is farther from the star must have a time period greater.

Explanation:

We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:

[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex]     (1)

Where:

r: is the radius of the circular orbit of the planet and the star

T: is the period

G: is the gravitational constant

M: is the mass of the planet

From equation (1) we have:

[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex]   (2)          

Where k is a constant

From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.

I hope it helps you!

Answer:

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Explanation:

Given:

Mass of truck(m) = 2,100 kg

Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s

Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s

Find:

Change in kinetic energy (ΔKE)

Computation:

Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]

Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]

Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]

Change in kinetic energy (ΔKE) = 1,050[121.9648]

Change in kinetic energy (ΔKE) = 128063.04

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Explanation:

When an ideal gas undergoes a slow isothermal expansion, following phenomenon occur

1. Work done bu the gas = Energy absorbed as heat.

2. Work done by environment = Energy absorbed as heat.

3. Increase in internal energy= Heat absorbed= work done by gas = work done by environment.

Hence all option are correct.

Increase in internal energy is equal to the heat absorbed or work done by gas or environment. All the statements are correct.

If an ideal gas undergoes a slow isothermal expansion,

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Answer:

The working principle of a water pump mainly depends upon the positive displacement principle as well as kinetic energy to push the water.

Explanation:

it mainly depends upon the positive displacement principle and also kinetic energy to push water. hope this hepls!

Answer:

Will be equal to alpha x r; less than UsN

Answer:

d) (CATA + CBTB) / (CA + CB)

Explanation:

According to the given situation, the final temperature of both objects is shown below:-

We assume T be the final temperature

while m be the mass

So it will be represent

m CA (TA - T) = m CB (T - TB)

or we can say that

CATA - CA T = CB T - CBTB

or

(CA + CB) T = CATA + CBTB

or

T = (CA TA + CBTB) ÷ (CA + CB)

Therefore the right answer is d

The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].

The given parameters;

The final temperature of both objects is calculated as follows;

heat lost by object A is equal to heat gained by object B

[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]

Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].

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Answer:D

Explanation:I got it right

Answer:

They needed to be able to unite even though they spoke different languages.

Explanation:

Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

Answer:

Resistance of each bulb = 360 ohms

Explanation:

Let each bulb have a resistance r .

Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.

[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]

[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]

R(equivalent) = r/15

Now, As per Ohms Law :

V = I * R(equivalent)

120 V = 5 A * r/15

r = 360 ohms

Answer:

[tex]\theta=10.20^{\circ}[/tex]  

[tex]\Delta l=0.10 ft[/tex]    

Explanation:

First of all, we analyze the system of blocks before starting to move.

[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]  

[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]  

[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]  

[tex]tan(\theta)=0.18[/tex]  

[tex]\theta=arctan(0.18)[/tex]  

[tex]\theta=10.20^{\circ}[/tex]  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]

Where:

[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]

[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]

[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]

[tex]\Delta l=0.10 ft[/tex]    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

(a) The inclined angle for which both blocks begin to slide is 10.3⁰.

(b) The compression of the spring is 0.22 ft.

The given parameters;

The normal force on block A and B:

[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]

The frictional force on block A and B:

[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]

The net force on the blocks when they starts sliding;

[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]

The change in the energy of the blocks is the work done in compressing the spring;

[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]

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Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

Answer:

The kinetic energy is 25000 J

Explanation:

At the top of the hill, the potential energy = 50000 J

the potential energy = mgh

where m is the mass

g is the acceleration due to gravity

h is the vertical height at the top of the hill

Note the mass of the roller coaster and acceleration due to gravity will always remain constant, so that halfway down the hill, only the height changes by half its initial value.

This means that at halfway down the hill, the potential energy of the roller coaster is

PE = [tex]mg\frac{h}{2}[/tex] = 50000/2 = 25,000 J

We also know that the total mechanical energy of a system is given as

ME = KE + PE = constant

where

ME is the mechanical energy of the system

PE is the potential energy of the system

KE is the kinetic energy of the system

Let us now analyse.

At the top of the hill, all the mechanical energy of the roller coaster is equal to its potential energy due to the height on the hill above ground, since the roller coaster is not moving (kinetic energy is energy due to motion). Halfway down, the mechanical energy of the roller coaster is due to both the kinetic energy and the potential energy, since the roller coaster is moving down, and is still at a given height above the ground. Having all these in mind, we can proceed and say that at halfway down the hill, ignoring friction,

ME = KE + PE = constant

50000 = KE + 25000

therefore

KE = 25000 J