An airplane propeller is 2.18 m in length (from tip to tip) with mass 97.0 kg and is rotating at 2600 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod.What is its rotational kinetic energy?

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:

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:

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:

coordinates of the center of mass for these two rods

([tex]x_{cm}[/tex], [tex]y_{cm}[/tex])= ([tex]\frac{3L}{4}[/tex],  [tex]\frac{3L}{4}[/tex])cm

Explanation:

given

mass of a rod = 2m

length of the rod = 3L

mass of two rods = 2(2m) = 4m

radius = diameter/2 = [tex]\frac{3L}{2}[/tex]

attached is the diagram and solution to the question

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:

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:

  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:

a. i. 390.02 pC ii 31201.6 pC b. i. 1.1352 × 10⁻⁸ F  ii. 2.75 V c. -10.726 nJ

d. i. 340.54 pC ii. 27243.2 pC e. i 1.1352 × 10⁻⁸ F ii. 240 V f. 3228.319 nJ

Explanation:

a. i. charge before immersion Q

Q = CV = ε₀AV/d where ε₀ = 8.854 × 10⁻¹² F/m, A = area = 25 cm² = 25 × 10⁻⁴ m², d = plate separation = 1.56 cm = 1.56 × 10⁻² m, V = potential difference between plates = 275 V

Q = ε₀AV/d = 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m² × 275 V/1.56 × 10⁻² m = 60871.25/1.56 × 10⁻¹⁴ C = 39020.03 × 10⁻¹⁴ C = 390.02 × 10⁻¹² C = 390.02 pC

ii. charge after immersion in water Q'

Q' = C'V = ε'ε₀AV/d = ε'Q where ε' = dielectric constant of distilled water = 80.0

Q' = ε'Q = 80 × 390.02 pC = 31201.6 pC

b. i. capacitance after immersion C'

C' = ε'ε₀A/d = 80 × 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m²/1.56 × 10⁻² m = 17708/1.56 × 10⁻¹⁴ F = 11351.28 × 10⁻¹² F = 1.1352 × 10⁻⁸ F  

ii. The potential difference V' after immersion

Since Q' = C'V'

V' = Q'/C' = 31201.6 × 10⁻¹² C/1.1352 × 10⁻⁸ F  = 2.75 V

c. The change in energy

The initial energy of the capacitor before immersion is E = 1/2QV = 1/2 × 390.02 × 10⁻¹² C × 275 V = 107255.5 × 10⁻¹²/2 J = 53627.75 × 10⁻¹² J = 53.628 nJ

The energy of the capacitor after immersion is E' = 1/2Q'V' = 1/2 × 31201.6 ×  × 10⁻¹² C × 2.75 V = 85804.4 × 10⁻¹²/2 J = 42902.2 × 10⁻¹² J = 42.902 nJ

So ΔE = E' - E =  42.902 nJ - 53.628 nJ = -10.726 nJ

d. i with V₁ = 240 V, charge Q₁ before immersion

Q₁ = ε₀AV₁/d = 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m² × 240 V/1.56 × 10⁻² m = 53124/1.56 × 10⁻¹⁴ C = 34053.85 × 10⁻¹⁴ C = 340.54 × 10⁻¹² C = 340.54 pC

ii charge after immersion in water Q₂ while still connected to V₁ = 240 V

Q₂ = εε₀AV₁/d = εQ₁ = 80 × 340.54 × 10⁻¹² C = 27243.2 × 10⁻¹² C = 27243.2 pC

e. i. capacitance after immersion C₁

C₁ = ε'ε₀A/d = 80 × 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m²/1.56 × 10⁻² m = 17708/1.56 × 10⁻¹⁴ F = 11351.28 × 10⁻¹² F = 1.1352 × 10⁻⁸ F  

ii. The potential difference V after immersion

The potential difference = 240 V since the voltage is still applied after immersion.

f. The change in energy

The initial energy of the capacitor before immersion is E₁ = 1/2Q₁V₁ = 1/2 × 340.54 × 10⁻¹² C × 240 V = 81729.6 × 10⁻¹²/2 J = 40864.8 × 10⁻¹² J = 40.865 nJ

The energy of the capacitor after immersion is E₂ = 1/2Q₂V₁ = 1/2 × 27243.2 × 10⁻¹² C × 240 V = 6538368 × 10⁻¹²/2 J = 3269184 × 10⁻¹² J = 3269.184 nJ

So ΔE = E₂ - E₁ = 3269.184  nJ - 40.865 nJ = 3228.319 nJ

Answer:

B = 4.45mT in the +^k direction

Explanation:

In order to calculate the required magnitude of the magnetic force, to achieve that the beam of particles emerge undeflected of the parallel plates, the electric force between the plates and the magnetic field in that region must be equal.

[tex]F_E=F_B\\\\qE=qvB[/tex]            (1)

q: charge of the particles beam = +2e = 2*1.6*10^-19C

v: speed of the particles = ?

B: magnitude of the magnetic field = ?

E: electric field between the plates = V/d

V: potential difference between the parallel plates = 150V

d: distance of separation of the plates = 0.00820m

If you assume that the below plate is negative, the electric force on the particles has a direction upward (+^j). Then, the direction of the magnetic force must be downwards (-^j).  

To obtain a downward magnetic force, it is necessary that the magnetic field point out of the page. In fact, if the direction of motion of the particles is toward east (+^i) and the magnetic field points out of the page (+^k), you have:

^i X ^k = -^j

Furthermore, it is necessary to calculate the sped of the particles. It is made by using the information about the charge, the potential difference that accelerates the particles and the kinetic energy.

[tex]K=qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}[/tex] (2)

You replace the expression (2) into the equation (1) and solve for B:

[tex]B=\frac{E}{v}=E\sqrt{\frac{m}{2qV}}[/tex]    

[tex]B=\frac{V}{d}\sqrt{\frac{m}{2qV}}\\\\B=\frac{150V}{0.0820m}\sqrt{\frac{6.64*10^{-27}kg}{2(2(1,6*10^{-19}C))(1.75*10^3V)}}\\\\B=4.45*10^{-3}T=4.45mT[/tex]

The required magnitude of the magnetic field is 4.45mT and has a direction out of the page +^k

Following are the solution to the given question:

In order to emerge using reflecting means, use the following formula:

[tex]\to F_E = F_B ..............(1)\\\\ \to F_E = \text{electric force}\\\\ \to F_B = \text{magnetic force}\\\\[/tex]

Calculating the Lorent's force:  

[tex]\to F=qE+qv \times B \ \ also,\ \ K_{E} =U_{E} \\\\[/tex]

[tex]\to K_{E}[/tex][tex]= \text{kinetic energy} = -\frac{1}{2} \ mv^2 \\\\[/tex]

[tex]\to U_{E} = \text{potential energy} = q_V[/tex]

Calculating the value of v: \\\\

[tex]\to v= \sqrt{\frac{2qV}{m}} \\\\ \to q = 2e^{+} = 2 (1.6 \times 10^{-19}\ C) = 3.2 \times 10^{-19} C \\\\\to V = 1.75 \times 10^{3} \V \\\\\to m = 6.64 \times 10^{-27} \ kg\\\\ \to v = 410,700.25 \ \frac{m}{s}\\\\[/tex]

It's the particle's velocity, but the velocity also is defined as:

[tex]\to v=\frac{E}{B}[/tex]

solving for B:  

[tex]\to B= \frac{E}{v}\\\\[/tex]

       [tex]= \frac{\frac{V}{d}}{v}\\\\ =\frac{V}{vd} \\\\= \frac{150\ V}{(410,700.25 \ \frac{m}{s}) (8.2 \times 10^{-3} m)} \\\\= 0.045\ T\\\\[/tex]

When indicated in the diagram, the direction is parallel to "v" and E.

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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:

The radius is [tex]r = 4.434 *10^{-5} \ m[/tex]

Explanation:

From the question we are told that

    The magnetic field is  [tex]B = 90 mT = 90*10^{-3} \ T[/tex]

     The electron kinetic energy is  [tex]KE = 1.4 eV = 1.4 * (1.60*10^{-19}) =2.24*10^{-19} \ J[/tex]

Generally for the collision to occur the centripetal force of the electron in it orbit is equal to the magnetic force applied  

   This is mathematically represented as

   [tex]\frac{mv^2}{r} = qvB[/tex]

=>    [tex]r = \frac{m* v}{q * B}[/tex]

Where  m is the mass of electron with values [tex]m = 9.1 *10^{-31} \ kg[/tex]  

             v is the escape velocity  which is mathematically represented as

                [tex]v = \sqrt{\frac{2 * KE}{m} }[/tex]

So  

       [tex]r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }[/tex]

     apply indices

    [tex]r = \frac{\sqrt{2 * KE * m} }{qB}[/tex]

substituting values

        [tex]r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}[/tex]

       [tex]r = 4.434 *10^{-5} \ m[/tex]

Answer:

a)     vfₓ = m₁ / (m₁ + m₂) v₁,  b)    tan θ  = m₂ / m₁ v₂ / v₁, c)

Explanation:

Momentum is a vector quantity, so the consideration must be fulfilled in all axes

a) conservation of the moment east-west direction

the system is formed by the two cases, so that the forces during the sackcloth have been internal and therefore the mummer remains

before the crash

                 p₀ = m₁ v₁

after the crash

                 [tex]p_{f}[/tex]= (m1 + m2) vfₓ

                p₀ = pf

                m₁ v₁ = (m₁ + m₂) vfₓ

              vfₓ = m₁ / (m₁ + m₂) v₁

b) conservation of the North-South axis moment

before the shock

                p₀ = m₂ v₂

after the crash

              p_{f} = ( m₁ +m₂) [tex]vf_{y}[/tex]  

             p₀ = p_{f}

            me 2 v₂ = (m₁ + m₂) vfy

            [tex]vf_{y}[/tex] = m₂ / (m₁ + m₂) v₂

c) the angle with which the car moves is

             tan θ = Vfy / Vfₓ

             tan θ = [m₂ / (m₁ + m₂) v] / [m₁ / (m₁ + m₂) v₁]

             tan θ  = m₂ / m₁ v₂ / v₁

The momentum conservation equation for the north-south components is [tex]m_1u_1 = v(m_1 + m_2)[/tex]

The momentum conservation equation for the north-south components is [tex]m_2u_2 = v(m_1 + m_2)[/tex]

The tangent of the angle is 1.

The given parameters;

Apply the principle of conservation of linear momentum of inelastic collision as shown below;

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]

The momentum conservation equation for the east-west components is written as follows;

[tex]m_1(u_1cos \ 0) + m_2(u_2 cos 90)= v(m_1 + m_2)\\\\m_1u_1 = v(m_1 + m_2)[/tex]

The momentum conservation equation for the north-south components is written as follows;

[tex]m_1(u_1sin 0) + m_2(u_2sin90) = v(m_1 + m_2)\\\\m_2u_2 = v(m_1 + m_2)[/tex]

The tangent of the angle is calculated as follows;

[tex]tan \ \theta = \frac{p_y}{p_x} = \frac{v(m_1 + m_2)}{v(m_1 + m_2)} \\\\tan \ \theta = 1\\\\\theta = tan^{-1} (1) \\\\\theta = 45\ ^0[/tex]

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Answer:a

Explanation:

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 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:

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:

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:

The electric flux around the soap bubble will be

Ф = q/ε

Explanation:

The radius of the soap bubble = R

The electric field around the soap bubble = E

The electric flux = ?

The soap can be approximated to be a sphere, so we find the surface area of the sphere

For the soap bubble, the surface area will be

A =  [tex]4\pi R^{2}[/tex]

Recall that electric flux is given as

Ф = EA

substituting value of A from above, we'll have

Ф = E[tex]4\pi R^{2}[/tex]..... equ 1

Also recall that the electric field E is given as

E = q/(4πε[tex]R^{2}[/tex])

substitute the value of E into equ 1, to get

Ф = q/(4πε[tex]R^{2}[/tex]) * [tex]4\pi R^{2}[/tex]

The electric flux around the soap bubble reduces to

Ф = q/ε

Answer:

Option C is correct.

The electric-field vector must oscillate in the yz plane.

Explanation:

Light, in waveform, is an electromagnetic wave.

And electromagnetic waves are known to have their electric and magnetic field perpendicular to each other and also simultaneously perpendicular to the direction of propagation of the wave.

If the velocity of direction of propagation of the wave is in one direction, the electric-field vector must be in a direction we are sure is perpendicular to this direction of wave propagation and the wave's magnetic field.

Of the options provided, only option B (z-direction) and option C (yz-plane) show a direction that is indeed perpendicular to the direction of propagation of the wave (x-axis).

And truly, the electric-field vector for this wave can be in any of the two directions without breaking the laws of physics, but the electric-field vector oscillating in the yz-plane is a more general answer as it covers all the possible directions that the electric-field can oscillate in, including the one specified by option B (z-direction).

Hence, the correct answer is option C.

Hope this Helps!!!

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

ac = 3.92 m/s²

Based on the data provided, the centripetal acceleration is 3.92 m/s²

Centripetal acceleration is the acceleration of a body moving in a circular path which is directed toward the center of the circle.

In the given question, the frictional force must balance the centripetal force for the car not to skid.

where,

Frictional Force = μR

R = mg

F = μmg

Centripetal Force = m

Then

μmg = ma

a = μg

ac = 0.4 * 9.8 m/s²

ac = 3.92 m/s²

Therefore, the centripetal acceleration is 3.92 m/s².

Learn more about centripetal acceleration at: https://brainly.com/question/25243603

Explanation:

40×10ms^-2

400N.

25/100m^2

0.25m^2

1.P=F/A

=400N/0.25m^2

=100Nm^-2

=100Pa

Answer:

The height of the tower is 23.786 m

Explanation:

Given;

period of oscillation, t = 9.79 s

acceleration of gravity, g = 9.8 m/s²

The period of oscillation is calculated as follows;

[tex]t = 2\pi \sqrt{\frac{h}{g} } \\\\[/tex]

where;

h represents the height of the tower

g is the acceleration of gravity

[tex]t = 2\pi \sqrt{\frac{h}{g} } \\\\\sqrt{\frac{h}{g} } = \frac{t}{2\pi} \\\\[/tex]

square both sides of the equation;

[tex](\sqrt{\frac{h}{g} })^2 = (\frac{t}{2\pi} )^2\\\\ \frac{h}{g} = \frac{t^2}{4\pi ^2} \\\\h = \frac{gt^2}{4\pi ^2} \\\\h = \frac{9.8*(9.79)^2}{4\pi ^2}\\\\h = 23.786 \ m[/tex]

Therefore, the height of the tower is 23.786 m

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:

b) During the turn, the door exerts a leftward force on you.

Explanation:

Although at the curve the car makes a turn without deceleration or accelerating, we must consider the fact that the direction of the instantaneous velocity of the car, which is always pointing tangentially away from the curve, is changing, producing a centripetal acceleration (acceleration is the rate of change of velocity) towards the center of the curve. The result is that a force inwards towards the center of the curve from the door, is exerted on you in the leftwards direction when your body collides with the door.

Answer:

54.6°

Explanation:

From law of reflection i=r.

So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.  

Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.

If you consider triangle AOB, one angle is ∠AOB=90°

 and ∠OAB is 54.6°

From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°

So, the second incident angle will be 54.6°

Hence, the second reflected angle will be 54.6 degrees.

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:

Ratio of distance to displacement is pi/2

Explanation:

Pls see attached file for diagram and explanation

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)

Learn more about conservation of momentum principle

https://brainly.com/question/14033058

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Answer:

The angular acceleration of the tire is 0.454 rad/s²

Explanation:

Given;

initial velocity, u = 3.4 rev/s = 3.4 rev/s x 2π rad/rev

                        u = 21.3656 rad/sec

final velocity, v = 5.5 rev/s = 5.5 rev/s x 2π rad/rev

                      v = 34.562 rad/sec

Calculate the value of angular rotation, θ, of the tire

θ = Number of revolutions x 2π rad/rev

θ = [tex]\frac{260}{2 \pi r} *\frac{2 \pi \ rad}{rev}[/tex]

θ = (260 / r)

r is the radius of the tire = 64 / 2 = 32cm = 0.32 m

θ = (260 / 0.32)

θ = 812.5 rad

Apply the following kinematic equation, to determine angular acceleration of the tire;

[tex]v^2 = u^2 + 2 \alpha \theta\\\\2 \alpha \theta = v^2 - u^2\\\\\alpha = \frac{v^2-u^2}{2 \theta} \\\\\alpha = \frac{(34.562)^2-(21.3656)^2}{2 (812.5)}\\\\\alpha = \frac{738.043}{1625} \\\\\alpha = 0.454 \ rad/s^2[/tex]

Therefore, the angular acceleration of the tire is 0.454 rad/s²

Answer:

The initial height of the sandbox before being released is 219.272 meters.

Explanation:

The sandbag is accelerated until hitting the ground due to the effect of gravitation, since height is too small with respect to the radius of Earth, then gravity acceleration can be considered constant and due to this, the following kinematic equation is applied to determine the initial height as a function of time:

[tex]y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot a \cdot t^{2}[/tex]

Where:

[tex]y[/tex] - Final height, measured in meters.

[tex]y_{o}[/tex] - Initial height, measured in meters.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]a[/tex] - Acceleration, measured in meters per square second.

Now, the initial height is cleared:

[tex]y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot a \cdot t^{2}[/tex]

Given that [tex]y = 0\,m[/tex], [tex]v_{o} = 3\,\frac{m}{s}[/tex], [tex]t = 7\,s[/tex] and [tex]a = -9.807\,\frac{m}{s^{2}}[/tex], the initial height of the sandbox is:

[tex]y_{o} = 0\,m - \left(3\,\frac{m}{s} \right)\cdot (7\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (7\,s)^{2}[/tex]

[tex]y_{o} = 219.272\,m[/tex]

The initial height of the sandbox before being released is 219.272 meters.