5. Sandor fills a bucket with water and whirls it in a vertical circle to demonstrate that the
water will not spill from the bucket at the top of the loop. If the length of the rope from his
hand to the centre of the bucket is 1.24 m, what is the minimum tension in the rope (at the
top of the swing)? How slow can he swing the bucket? Explain your answer.

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  [tex]P_{avg} = F * v_{avg}[/tex]

Explanation:

From the question we are told that

    The that the average power is  mathematically represented as

            [tex]P_{avg} = \frac{W }{\Delta t }[/tex]

Where W is  is the Workdone which is  mathematically represented as

         [tex]W = F * s[/tex]

      Where F is  the applies force and  s  is the displacement  due to the force  

        So  

                [tex]P_{avg} = \frac{F *s }{\Delta t }[/tex]

Now this  displacement can be represented mathematically as  

            [tex]s = v_{avg} * \Delta t[/tex]

Where [tex]v_{avg }[/tex] is the average  velocity and [tex]\Delta t[/tex] is the time  taken  

So  

            [tex]P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }[/tex]

=>         [tex]P_{avg} = F * v_{avg}[/tex]

Answer:

Pavg =  Fvavg

Explanation:

Since the P (power) done by the F (force) is:

P = Fs/t

and we are looking for the velocity, so then it would be:

P = Fv

with the average velocity the answer is:

Pavg =  Favg

If an object is moving at a constant speed, and the force F is also constant, this formula can be used to find the average power. If v  is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity v in this case meaning the instantaneous velocity, of course).

Answer:

Explanation:

As a result of impact of time widening, a clock moving as for an observer seems to run all the more gradually than a clock that is very still in the observer's casing.  

At the point when observed from earth, the pendulum on the spaceship takes more time to finish one oscillation.  

Hence, the clock related with that pendulum will run more slow (gives fewer oscillations as observed from the earth)  than the clock related with the pendulum on earth.

Ans => B fewer oscillations

Answer:

The correct options are:

B) They have the same wave speed as visible light

D) They have a lower frequency than gamma rays.

Explanation:

B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s

D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).

We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.

Answer:

  m = 4

Explanation:

The expression that explains the constructive interference of a diffraction pattern is

         a sin θ = m λ

where a  is the width of the slit and λ the wavelength

         sin θ = m λ / a

The maximum value is for when the sine is 1, let's substitute

         1 = m λ/a  

         m = a /λ

let's reduce the magnitudes to the SI system

        a = 2.1 um = 2.1 10⁻⁶

        lam = 475 nm = 475 10⁻⁹ m

let's calculate

        m = 2.1  10⁻⁶ / 475 10⁻⁹

        m = 4.42

with m must be an integer the highest value is

         m = 4

Answer:

R = 1138.9 Ω

Explanation:

Hello,

In this case, for the given power (P) and current (I), we can compute the resistance (R) via:

R = P / I²

Thus, we obtain:

R = 1.64x10⁵ W / (12.0 A)²

R = 1138.9 Ω

Best regards.

Answer:

moves into a region of higher potential

Potential difference = 835   V

Ki = 835 eV

Explanation:

given data

initial speed = 400000 m/s

solution

when proton moves against a electric field  so that it will move into higher potential  region

and

we know Work done by electricfield  W is express as

W = KE of proton   K

so

q × V   =  0.5 × m × v²     ......................1

put here va lue

1.6 × [tex]10^{-19}[/tex] × V   =   0.5 × 1.67 × [tex]10^{-27}[/tex] × 400000²

Potential difference V = 1.336 × 10-16 / 1.6  × 10-19      

Potential difference = 835   V

and

KE of proton in eV is express as

Ki  =   V numerical

Ki = 835 eV

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/([tex]\frac{1}{4}[/tex]year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

Answer:

Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.

Answer:

depending on how high it goes at 100m it has taken 2 secondes

Explanation:

At the highest point, the arrow is changing from moving up to moving down. At that exact point, its speed AND its velocity are both ZERO.

And air resistance actually makes no difference.

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex]       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

[tex]|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s[/tex]

The interaction time to avoid that the water balloon breaks is 0.029s

Answer:

a) 40 V

b) 69.23 V

c) 69.23 V

Explanation:

See attachment for solution

Answer:

C) It is either ferromagnetic or paramagnetic

Explanation:

The complete question is given below

We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?

A) It must be ferromagnetic.

B) It must be paramagnetic.

C) It is either ferromagnetic or paramagnetic.

D) It must be diamagnetic.

A ferromagnetic material will respond towards a magnetic field. They are those materials that are attracted to a magnet. Ferromagnetism is associated with our everyday magnets and is the strongest form of magnetism in nature. Iron and its alloys is very good example of a material that readily demonstrate ferromagnetism.

Paramagnetic materials are weakly attracted to an externally applied magnetic field. They usually accelerate towards an electric field, and form internal induced magnetic field in the direction of the external magnetic field.

The difference is that ferromagnetic materials can retain their magnetization when the externally applied magnetic field is removed, unlike paramagnetic materials that do not retain their magnetization.

In contrast, a diamagnetic material is repelled away from an externally applied magnetic field.

Answer:

The direct current that will produce the same amount of thermal energy is 1.83 A

Explanation:

Given;

maximum current, I₀ = 2.59 A

The average power dissipated in a resistor connected in an AC source is given as;

[tex]P_{avg} = I_{rms} ^2R[/tex]

Where;

[tex]I_{rms} = \frac{I_o}{\sqrt{2} }[/tex]

[tex]P_{avg} = (\frac{I_o}{\sqrt{2} } )^2R\\\\P_{avg} = \frac{I_o^2R}{2} ----equation(1)[/tex]

The average power dissipated in a resistor connected in a DC source is given as;

[tex]P_{avg} = I_d^2R --------equation(2)[/tex]

where;

[tex]I_d[/tex] is direct current

Solve equation (1) and (2) together;

[tex]I_d^2R = \frac{I_o^2R}{2} \\\\I_d^2 = \frac{I_o^2}{2} \\\\I_d=\sqrt{\frac{I_o^2}{2} } \\\\I_d = \frac{I_o}{\sqrt{2}} \\\\I_d = \frac{2.59}{\sqrt{2} } \\\\I_d = 1.83 \ A[/tex]

Therefore, the direct current that will produce the same amount of thermal energy is 1.83 A

Answer:

Option:  b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.

Explanation:

Global warming is the reason for the changes in environment and climate on earth. Melting of glaciers leads to an increase in water level and a decrease in landmass. One of the most climactic consequences is the decrease in Arctic sea ice. Melting polar ice along with ice sheets and glaciers across Greenland, North America, Europe, Asia, and South America suspected to increase sea levels slowly. There is an increase in the glacial retreat due to global warming, which leaves rock piles that covered with ice.  

Answer:

B: Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.

Explanation:

Global warming is primarily caused by the increase in greenhouse gases, such as carbon dioxide, in the Earth's atmosphere. This leads to a rise in global temperatures, which has various impacts on the Earth's surface. One significant effect is the melting of glaciers and ice caps in polar regions and mountainous areas.

As temperatures increase, glaciers and ice sheets start to melt at a faster rate. This melting results in the release of massive amounts of water into rivers, lakes, and oceans. Consequently, there can be an increase in the frequency and intensity of flooding events in regions downstream from these melting glaciers.

Moreover, the melting of glaciers and ice caps contributes to a rise in sea levels. As the melted ice enters the oceans, it adds to the overall volume of water, leading to a gradual increase in sea levels worldwide. This rise in sea levels poses a threat to coastal areas, as they become more vulnerable to coastal erosion, storm surges, and saltwater intrusion into freshwater sources.

Additionally, the loss of glacial land masses due to melting can have long-term effects on ecosystems. Glaciers act as freshwater reservoirs, releasing water gradually throughout the year. With their decline, the availability of freshwater for agriculture, drinking water, and other human needs can be significantly affected.

Therefore, global warming can indeed lead to changes in the Earth's surface, particularly through the melting of glaciers and subsequent impacts on sea levels, flooding, and glacial land masses.

E23 verified.

Answer:

ML²/6

Explanation:

Pls see attached file

The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,

The length of the rod is L. It has a non-uniform distribution of mass given by:

dm/dx = Cx

where C has units kg/m²

dm = Cxdx

the total mass M of the rod can be calculated by integrating the above relation over the length:

[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]

Thus,

C = 2M/L²

Now, the moment of inertia of the small element dx of the rod is given by:

dI = dm.x²

dI = Cx.x²dx

[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]

I = ML²/2

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Answer:

The hydrostatic force on one end of the trough is 54994.464 N

Explanation:

Given;

liquid density, ρ = 810 kg/m³

side of the equilateral triangle, L = 8m

acceleration due to gravity, g =  9.8 m/s²

Hydrostatic force is given as;

H = ρgh

where;

h is the vertical height of the equilateral triangle

Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.

let the half side of the triangle = x

x = ⁸/₂ = 4m

The half section of the triangle forms a right angled triangle

h² = 8² - 4²

h² = 48

h = √48

h = 6.928m

F = ρgh

F = 810 x 9.8 x 6.928

F = 54994.464 N

Therefore, the hydrostatic force on one end of the trough is 54994.464 N

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

Substitute these values into equation (i) as follows;

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² +  [tex]\frac{1}{2}[/tex]m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) +  ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

Answer:

The magnitude of the EMF is 0.00055  volts

Explanation:

The induced EMF is proportional to the change in magnetic flux based on Faraday's law:

[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]

Since in our case there is only one loop of wire, then N=1 and we get:

[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]

We need to express the magnetic flux given the geometry of the problem;

[tex]\Phi=B\,\,A[/tex]where A is the area of the coil that remains unchanged with time, and B is the magnetic field that does change with time. Therefore the equation for the EMF becomes:

[tex]emf\,=-\,N\, \frac{d\Phi}{dt} = \frac{d\Phi}{dt} =-\frac{d\,(B\,A)}{dt} =-\,A\,\frac{d\,(B)}{dt}=- 1\,m^2(2\,\,T/h})= -2\,\,m^2\,T/(3600\,\,s)= -0.00055\,Volts[/tex]

Answer:a

Explanation:

Answer:

5.79 in

Explanation:

We are given that

Diameter,d=0.30 in

Radius,r=[tex]\frac{d}{2}=\frac{0.30}{2}=0.15 in[/tex]

Weight of hydrometer,W=0.042 lb

Specific gravity(SG)=1.10

Height of stem from the water surface=3.15 in

Density of water=[tex]62.4lb/ft^3[/tex]

In water

Volume  of water displaced [tex]V=\frac{mass}{density}=\frac{0.042}{62.4}=6.73\times 10^{-4} ft^3[/tex]

Volume of another liquid displaced=[tex]V'=\frac{V}{SG}=\frac{6.73\times 10^{-4}}{1.19}=5.66\times 10^{-4}ft^3[/tex]

Change in volume=V-V'

[tex]V-V'=\pi r^2 l[/tex]

Substitute the values

[tex]6.73\times 10^{-4}-5.66\times 10^{-4}=3.14\times (\frac{0.15}{12})^2l[/tex]

By using

1 ft=12 in

[tex]\pi=3.14[/tex]

[tex]l=\frac{6.73\times 10^{-4}-5.66\times 10^{-4}}{3.14\times (\frac{0.15}{12})^2}[/tex]

l=2.64 in

Total height=h+l=3.15+2.64= 5.79 in

Hence, the height of the stem protrude above the liquid surface=5.79 in

Answer:

v= 4.823 × 10⁻⁹ cm/s

Explanation:

given

flow rate = 9.6 x10-5 cm³/s, d = 0.080mm

r = d/2= 0.080/2= 0.0040 cm

speed = rate of blood flow × area

v = (9.6 x 10⁻⁵ cm³/s) × (πr²)

v = (9.6 x 10⁻⁵ cm³/s) × π(0.0040 × cm)²

v= 1.536 × 10⁻⁹π cm/s

v= 4.823 × 10⁻⁹ cm/s

Answer:

v = 8.5 m/s

Explanation:

In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,

Weight = Centripetal Force

but,

Weight = mg

Centripetal Force = mv²/r

Therefore,

mg = mv²/r

g = v²/r

v² = gr

v = √gr

where,

v = minimum speed required = ?

g = 9.8 m/s²

r = radius = 7.4 m

Therefore,

v = √(9.8 m/s²)(7.4 m)

v = 8.5 m/s

Minimum speed for a roller coaster while travelling upside down  so that the person will not fall out = 8.5 m/s

For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.

In the given condition the weight of the person must be balanced by the centrifugal force.

and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person

According to the Newton's Second Law of motion we can write force balance

[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]

Given Radius of loop = r = 7.4 m

Putting the value  of r = 7.4 m  in equation (1) we get

[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]

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In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm

(a) 7.392cm

(b) -15.32 cm/s

(c) -184cm/s²

(d) 0.4πs and 8.00cm

The general equation of a simple harmonic motion (SHM) is given by;

x(t) = A cos (wt + Φ)        --------------(i)

Where;

x(t) = position of the body at a given time t

A =  amplitude or maximum displacement during oscillation

w = angular velocity

t = time

Φ = phase constant.

Given from question:

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

(a) At time t = 0;

The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;

x(0) = 8.00 cos (5(0) + π / 8)

x(0) = 8.00 cos (π /8)

x(0) = 8.00 x 0.924

x(0) = 7.392 cm

Therefore, the position of the piston at time t = 0 is 7.392cm

(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;

v(t) = [tex]\frac{dx(t)}{dt}[/tex]

v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]

v(t) = 8 (-5 sin (5t + π / 8))

v(t) = -40sin(5t + π / 8)     --------------------(iii)

Now, substitute t=0 into the equation as follows;

v(0) = -40 sin(5(0) + π / 8)

v(0) = -40 sin(π / 8)

v(0) = -40 x 0.383

v(0) = -15.32 cm/s

Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s

(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;

a(t) = [tex]\frac{dv(t)}{dt}[/tex]

a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]

a(t) = -200 cos (5t + π / 8)

Now, substitute t = 0 into the equation as follows;

a(0) = -200 cos (5(0) + π / 8)

a(0) = -200 cos (π / 8)

a(0) = -200 x 0.924

a(0) = -184.8 cm/s²

Therefore, the acceleration of the piston at time t = 0 is -184cm/s²

(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;

x(t) = A cos (wt + Φ)                   --------------(i)

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

From these equations it can be deduced that;

Amplitude, A = 8.00cm

Angular velocity, w = 5 rads/s

But;

w = [tex]\frac{2\pi }{T}[/tex]           [Where T = period of oscillation]

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

=> T = [tex]\frac{2\pi }{5}[/tex]

=> T = 0.4π s

Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm

Answer:

1.6 m/s west

Explanation:

The recoil velocity of the launcher is 1.6 m/s west.

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of 8 m/s to the east.

Final momentum will be zero, so

m₁u₁ +m₂u₂ =0

Substitute the values for m₁ = 5kg, m₂ =1kg and u₂ =8 m/s, then the recoil velocity will be

5 x v +1x8 = 0

v = - 1.6 m/s

Thus, the recoil velocity of the launcher is  1.6 m/s (West)

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Answer:

The terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

Explanation:

The terminal velocity of the bacterium can be calculated using the following equation:

[tex] F = 6\pi*\eta*rv [/tex]    (1)

Where:

F: is drag force equal to the weight

η: is the viscosity = 1.002x10⁻³ kg/(m*s)

r: is the radium of the bacterium = d/2 = 1.81 μm/2 = 0.905 μm

v: is the terminal velocity

Since that F = mg and by solving equation (1) for v we have:

[tex] v = \frac{mg}{6\pi*\eta*r} [/tex]  

We can find the mass as follows:

[tex] \rho = \frac{m}{V} \rightarrow m = \rho*V [/tex]

Where:

ρ: is the density of the bacterium = 1.10x10³ kg/m³

V: is the volume of the spherical bacterium

[tex] m = \rho*V = \rho*\frac{4}{3}\pi*r^{3} = 1.10 \cdot 10^{3} kg/m^{3}*\frac{4}{3}\pi*(0.905 \cdot 10^{-6} m)^{3} = 3.42 \cdot 10^{-15} kg [/tex]

Now, the terminal velocity of the bacterium is:

[tex] v = \frac{mg}{6\pi*\eta*r} = \frac{3.42 \cdot 10^{-15} kg*9.81 m/s^{2}}{6\pi*1.002 \cdot 10^{-3} kg/(m*s)*0.905 \cdot 10^{-6} m} = 1.96 \cdot 10^{-6} m/s [/tex]

Therefore, the terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

I hope it helps you!

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

Answer:

The total work done by the person is given as = m g h

= 4.5kg x 9.8m/s²x1.2m

= 52.92J

This is the work done in moving the block in a vertical distance

However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.

Explanation:

Answer:

The ratio  is  [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

Explanation:

Generally  the Moment of inertia of a spherical object (shell) is mathematically represented as

              [tex]I = \frac{2}{3} * m r^2[/tex]

Where m is  the mass of the spherical object

       and   r is the radius  

Now the the rotational kinetic energy can be mathematically represented as

       [tex]RE = \frac{1}{2}* I * w^2[/tex]

Where  [tex]w[/tex] is the angular velocity which is mathematically represented as

             [tex]w = \frac{v}{r}[/tex]

=>           [tex]w^2 = [\frac{v}{r}] ^2[/tex]

So

             [tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]

            [tex]RE = \frac{1}{3} * mv^2[/tex]

Generally the transnational  kinetic energy of this motion is  mathematically represented as

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

So  

      [tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]

       [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

3.0cm

For lenses in an achromatic combination, the following condition holds, assuming the two lenses are of the same materials;

d = [tex]\frac{f_1 + f_2}{2}[/tex]     ---------(i)

Where;

d= distance between lenses

f₁ = focal length of the first lens

f₂ = focal length of the second lens

From the question;

f₁ = 4.5cm

f₂ = 1.5cm

Substitute these values into equation (i) as follows;

d = [tex]\frac{4.5+1.5}{2}[/tex]

d = [tex]\frac{6.0}{2}[/tex]

d = 3.0cm

Therefore, the distance between the two lenses is 3.0cm

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

[tex]a = \frac{v^2}{r}[/tex]

At constant speed, we will have;

[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5