Carbon nanotubes can be a considered toxic agent due to: (a)- Its ability to produce toxic gases (b)Its ability to penetrate skin quickly due its very small diameter (c)Its ability to mix with water (d)Difficulties in handling it properly

Answer:

a)f=2.25 Hz

b)Time period T=.144 s

c)tex]V_{max}[/tex]=0.42 m/s

d)Phase angle Ф=87.3°

e) [tex]a_{max}=6.0041 [tex]\frac{m}{s^2}[/tex]

Explanation:

a)

Natural frequency

  [tex]\omega _n=\sqrt {\dfrac{K}{m}}[/tex]

[tex]\omega _n=\sqrt {\dfrac{150}{0.75}}[/tex]

[tex]\omega _n[/tex]=14.14 rad/s

w=2πf

⇒f=2.25 Hz

b) Time period

[tex]=\dfrac{2π}{\omega _n}[/tex]

T=[tex]\frac{1}{f}[/tex]

 Time period T=.144 s

c)Displacement equation

[tex]x=Acos\omega _nt+Bsin\omega _nt[/tex]

Boundary condition

t=o,x=0.03 m

t=0,v=.02m/s   , V=[tex]\frac{dx}{dt}[/tex]

Now by using these above conditions

A=0.03,B=0.0014

x=0.03 cos14.14 t+0.0014 sin14.14 t

⇒x=0.03003sin(14.14t+87.3)

[tex]V_{max}=\omega_n X_{max}[/tex]

[tex]V_{max}=14.14\times 0.03003[/tex]=0.42 m/s

d)

Phase angle Ф=87.3°

e)

Maximum acceleration

[tex]a_{max}=(\omega _n )^2X_{max}[/tex]

[tex]a_{max}=(14.14)^20.03003[/tex]=6.0041 [tex]\frac{m}{s^2}[/tex]

Answer:

A. 2.249 hz

B. 0.45 s

C. 0.424 m/s

D. 66⁰

E. 6 m/s^2

Explanation:

Step 1: identify the given parameters

mass of the block (m)= 0.75kg

stiffness constant (k) = 150N/m

Amplitude (A) = 3cm = 0.03m

upward velocity (v) = 2cm/s

Step 2: calculate the natural frequency (F)by applying relevant formula in S.H.M

[tex]f=\frac{1}{2\pi } \sqrt \frac{k}{m}[/tex]

[tex]f=\frac{1}{2\pi } \sqrt \frac{150}{0.75}[/tex]

f = 2.249 hz

Step 3: calculate the period of the oscillation (T)

[tex]period (T) = \frac{1}{frequency}[/tex]

[tex]T = \frac{1}{2.249} (s)[/tex]

T = 0.45 s

Step 4: calculate the maximum velocity,[tex]V_{max}[/tex]

[tex]V_{max} = A\sqrt{\frac{k}{m} }[/tex]

A is the amplitude of the oscilation

[tex]V_{max} = 0.03\sqrt{\frac{150}{0.75} }[/tex]

[tex]V_{max} = 0.424(\frac{m}{s})[/tex]

Step 5: calculate the phase angle, by applying equation in S.H.M

[tex]X = Acos(\omega{t} +\phi)[/tex]

where X is the displacement; calculated below

Displacement = upward velocity X period of oscillation

[tex]displacement (X) = vt (cm)[/tex]

X = (2cm/s) X (0.45 s)

X = 0.9 cm = 0.009m

where [tex]\omega[/tex] is omega; calculated below

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{150}{0.75} }[/tex]

[tex]\omega= 14.142[/tex]

[tex]\phi = phase angle[/tex]

Applying displacement equation in S.H.M

[tex]X = Acos(\omega{t}+\phi)[/tex]

[tex]0.009 = 0.03cos(14.142 X 0.45+\phi)[/tex]

[tex]cos(6.364+\phi) = \frac{0.009}{0.03}[/tex]

[tex]cos(6.364+\phi) = 0.3[/tex]

[tex](6.364+\phi) = cos^{-1}(0.3)[/tex]

[tex](6.364+\phi)= 72.5⁰[/tex]

[tex]6.364+\phi =72.5⁰[/tex]

[tex]\phi[/tex] =72.5 -6.364

[tex]\phi[/tex] =66.1⁰

Phase angle, [tex]\phi[/tex] ≅66⁰

Step 6: calculate the maximum acceleration, [tex]a_{max}[/tex]

[tex]a_{max} = \omega^{2}A[/tex]

[tex]a_{max}[/tex] = 14.142 X 14.142 X 0.03

[tex]a_{max}[/tex] = 5.999 [tex](\frac{m}{s^{2} })[/tex]

[tex]a_{max}[/tex] ≅ 6 [tex](\frac{m}{s^{2} })[/tex]

Explanation:

We know that first T-ds equation

 Tds=[tex]C_v[/tex]dT+Pdv

For constant volume process dv=0

⇒Tds=[tex]C_v[/tex]dT

So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_v}[/tex]          ----(1)

We know that second T-ds equation

 Tds=[tex]C_p[/tex]dT-vdP

For constant pressure process dP=0

⇒Tds=[tex]C_p[/tex]dT

So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_p}[/tex]             ---(2)

We know that [tex]C_p[/tex] is always greater than [tex]C_v[/tex].

So from equation (1) and (2) we can say that slope of constant volume process will be always greater than constant pressure process on T-s diagram.

Answer:

C) Injection

Explanation:

Injection is a molding process, not a heat transfer mechanism.

Answer:

55.56%

Explanation:

Given data

Temprature of hot reservior =2000°c=2273k

Temprature of Cold reservior=25°c=298k

Power produced by engine=50MW

Heat rejected =40MW

we know that Effeciency(η) of heat engine=[tex]\frac{Work produced}{heat supplied}[/tex]

Also we know that

heat supplied[tex]\left ({Q_s} \right )=work produced{W}+Heat rejected{Q_r}[/tex]

Q_s=50+40=90MW

η=[tex]\frac{W}{Q_s}[/tex]

η=[tex]\frac{50}{90}[/tex]

η=55.55%

Answer:

[tex]V=68.86ft^3[/tex]

Explanation:

[tex]T_1[/tex] =10°C,[tex]T_2[/tex] =93.33°C

Q=500 btu=527.58 KJ

[tex]P_1= 2atm[/tex]

If we assume that air is ideal gas   PV=mRT, ΔU=[tex]mC_v(T_2-T_1)[/tex]

Actually this is closed system so work will be zero.

Now fro first law

Q=ΔU=[tex]mC_v(T_2-T_1)[/tex]+W

⇒Q=[tex]mC_v(T_2-T_1)[/tex]

527.58 =[tex]m\times 0.71(200-50)[/tex]

m=4.9kg

 PV=mRT

[tex]200V=4.9\times 0.287\times (10+273)[/tex]

[tex]V=1.95m^3[/tex]                ([tex]V=1m^3=35.31ft^3[/tex])              

[tex]V=68.86ft^3[/tex]

Answer:

Explanation:

Using equation of pure torsion

[tex]\frac{T}{I_{polar} }=\frac{t}{r}[/tex]

where

T is the applied Torque

[tex]I_{polar}[/tex] is polar moment of inertia of the shaft

t is the shear stress at a distance r from the center

r is distance from center

For a shaft with

[tex]D_{0} =[/tex] Outer Diameter

[tex]D_{i} =[/tex] Inner Diameter

[tex]I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}[/tex]

Applying values in the above equation we get

[tex]I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\

I_{polar}= 1.74[/tex] x [tex]10^{-7} m^{4}[/tex]

Thus from the equation of torsion we get

[tex]T=\frac{I_{polar} t}{r}[/tex]

Applying values we get

[tex]T=\frac{1.74X10^{-7}X100X10^{6}  }{.021}[/tex]

T =829.97Nm

Answer:[tex]\tau _\left ( max\right )[/tex]=11.468MPa

Explanation:

Given data

[tex]power[/tex][tex]\left ( P\right )[/tex]=7 hp=5220 W

N=3200rpm

[tex]\omega [/tex]=[tex]\frac{2\pi\times N}{60}[/tex]=335.14 rad/s

diameter[tex]\left ( d\right )[/tex]=0.75in=19.05mm

we know

P=[tex]Torque\left ( T\right )\omega [/tex]

5220=[tex]T\times 335.14[/tex]

T=15.57 N-m

And

[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Polar\ modulus}[/tex]

[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Z_{P}}[/tex]

[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{16\times T}{\pi d^{3}}[/tex]

[tex]\tau _\left ( max\right )[/tex]=11.468MPa

Answer:

Maximum shear stress is 11.47 MPa.

Explanation:

Given:

D=.75 in⇒D=19.05 mm

P=7 hp⇒ P=5219.9 W

N=3200 rpm

We know that

    P=T[tex]\times \omega[/tex]

Where T is the torque and [tex]\omega[/tex] is the speed of shaft.

   P=[tex]\frac{2\pi N\times T}{60}[/tex]

So    5219.9=[tex]\frac{2\pi \times 3200\times T}{60}[/tex]

 T=15.57 N-m

We know that maximum shear stress in shaft

[tex]\tau _{max}=\dfrac{16T}{\pi \times D^3}[/tex]

[tex]\tau _{max}=\dfrac{16\times 15.57}{\pi \times 0.01905^3}[/tex]

[tex]\tau _{max}[/tex]=11.47 MPa

So maximum shear stress is 11.47 MPa.

Answer:

To convert a fraction to a decimal, divide the numerator by the denominator.

Answer:

[tex]\eta[/tex]=0.60

Explanation:

Given :Take [tex]\gamma[/tex]=1.4 for air

      [tex]P_1=100 KPa  ,T_1=300K[/tex]

  [tex]\frac{V_1}{V_2}[/tex]=r ⇒ r=16

As we know that  

   [tex]T_2=T_1(r^{\gamma-1})[/tex]

So [tex]T_2=300\times 16^{\gamma-1}[/tex]

  [tex]T_2[/tex]=909.42K

Now find the cut off ration [tex]\rho[/tex]

      [tex]\rho=\frac{V_3}{V_2}[/tex]  

         [tex]\frac{V_3}{V_2}=\frac{T_3}{T_2}[/tex]

[tex]\rho=\frac{2031}{909.42}[/tex]

 [tex]\rho=2.23[/tex]

So efficiency of diesel engine

[tex]\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}[/tex]

Now by putting the all values

[tex]\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}[/tex]

So [tex]\eta[/tex]=0.60

So the efficiency of diesel engine=0.60

Answer: False

Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have  light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.

Answer:

a.  [tex]\tau=51.55 MPa[/tex]

b.[tex]\tau=42.95MPa[/tex]

c.[tex]\theta=7.67\times 10^{-3}[/tex] rad.

Explanation:

Given: [tex]D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa [/tex]

We know that

[tex]\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}[/tex]

J for hollow shaft [tex]J=\dfrac{\pi (D_o^4-D_i^4)}{64}[/tex]

(a)

 Maximum shear stress [tex]\tau =\dfrac{16T}{\pi Do^3(1-K^4)}[/tex]

      [tex]K=\dfrac{D_i}{D_o}[/tex]⇒K=0.83

[tex]\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}[/tex]

   [tex]\tau=51.55 MPa[/tex]

(b)

We know that [tex]\tau \alpha r[/tex]

So [tex]\dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}[/tex]

[tex]\dfrac{51.55}{\tau}=\dfrac{210}{175}[/tex]

[tex]\tau=42.95MPa[/tex]

(c)

[tex]\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}[/tex]

[tex]\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}[/tex]

[tex]\theta=7.67\times 10^{-3}[/tex] rad.

Answer:

0

Explanation:

output =transfer function H(s) ×input U(s)

here H(s)=[tex]\frac{s}{(s+3)^2}[/tex]

U(s)=[tex]\frac{1}{s}[/tex] for unit step function

output =H(s)×U(s)

=[tex]\frac{s}{(s+3)^2}[/tex]×[tex]\frac{1}{s}[/tex]

=[tex]\frac{1}{(s+3)^2}[/tex]

taking inverse laplace of output

output=t×[tex]e^{-3t}[/tex]

at t=0 putting the value of t=0 in output

output =0

Answer:0.3166N

Explanation:

Given data

Area [tex]\left ( A\right )=500 cm^2[/tex]

Gap below top plate[tex]\left ( y_1\right )=20 mm[/tex]

Gap above bottom plate[tex]\left ( y_2\right )=30 mm[/tex]

SAE 30 oil viscosity =[tex]0.38 N-s/m^2[/tex]

Velocity of middle plate[tex]\left ( v\right )=1 m/s[/tex]

There will viscous force on middle plate i.e. at above surface and below surface

Viscous force[tex]\left ( F\right )=\mu \frac{Av}{y}[/tex]

[tex]Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ][/tex]

[tex]F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ][/tex]

[tex]F=31.66\times 10^{-2}=0.3166 N[/tex]

this is force by oil on plate thus we need to apply atleast 0.3166N to move plate

The efficiency of a transformer is mainly dependent on a)- Core losses

Hope this helps! :)

Answer:

Drag Coefficient:

It is a dimensionless quantity that is used to quantify air friction or fluid friction i.e., drag and is commonly denoted by [tex]C_{d}[/ tex].It is used in Drag Equation, where a lower drag coefficient indicates that the object will have lower value of air friction or fluid friction(i.e., aerodynamic or hydrodynamic drag). It is always associated with a particular surface area.

Drag Coefficient is given by:

[tex]C_{d} = \frac{2D}{A\rho v^{2}}[/tex]

Skin Friction Coefficient:

It is a dimensionless quantity which represents skin shear stress due to dynamic pressure of a free stream.

Skin Friction Coefficient is given by:

[tex]C_{f} = \frac{2\tau _{w}}{\rho V^{2}}[/tex]

where,

[tex]\tau _{w}[/tex]= skin shear stress

v = free stream speed

1. Define Viscosity

In physics, Viscosity refers to the level of resistance of a fluid to flow due to internal friction, in other words, viscosity is the result of the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow. For example, the honey is a fluid with high viscosity while the water has low viscosity.

What are the main differences between viscous and inviscid flows?

Viscous flows are flows that has a thick, sticky consistency between solid and liquid, contain and conduct heat, does not have a rest frame mass density and whose motion at a fixed point always remains constant. Inviscid flows, on the other hand, are flows characterized for having zero viscosity (it does not have a thick, sticky consistency), for not containing or conducting heat, for the lack of steady flow and for having a rest frame mass density

Furthermore, viscous flows are much more common than inviscid flows, while this latter is often considered an idealized model since helium is the only fluid that can become inviscid.

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 [tex]\frac{ft^{3}}{sec}[/tex]

         [tex]D_{1}[/tex]= 6 inch=0.5 ft

        [tex]D_{2}[/tex]=2 inch=0.1667 ft

As we know that Q=AV

[tex]A_{1}\times V_{1}=A_{2}\times V_{2}[/tex]

So [tex]V_{2}=\frac{Q}{A_2}[/tex]

     [tex]V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}[/tex]

     [tex]V_{2[/tex]=0.687 ft/sec

We know that Head loss due to sudden contraction

           [tex]h_{l}=K\frac{V_{2}^2}{2g}[/tex]

If nothing is given then take K=0.5

So head loss[tex]h_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}[/tex]

                                    =0.00366 ft

So head loss=0.00366 ft

Answer:

Answer is c 0.500

Explanation:

[tex]SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}[/tex]

We know that [tex]\rho_{water}=62.42lb/ft^{3}[/tex]

Applying values we get

[tex]SpecificGravity=\frac{31.2}{62.4}=0.5[/tex]

Answer:

[tex]\tau_{max}= 3.28 \ MPa[/tex]

Explanation:

outside diameter = 90 mm

inside diameter = 90- 2× t=  90- 2× 5 = 80mm

where t is thickness of pipe.

power (P)  = 12 kW

Revolution (N)= 650 rev/min

we

Power = torque × angular velocity

P= T× ω

ω =  [tex]\frac{2 \pi N}{60}[/tex]

[tex]P=T \times \frac{2\pi N}{60}\\12 \times 10^3=T\times \frac{2\pi \times 650}{60}[/tex]

T=  176.3 Nm

for maximum shear stress

[tex]\frac{\tau_{max}}{y_{max}}=\frac{T}{I_p}[/tex]

where ymax is maximum distance from neutral axis.

[tex]y_{max}=\frac{d_0}{2}= \frac{90}{2}[/tex]= 45 mm

[tex]I_p[/tex]= polar moment area

          = [tex]\frac{\pi}{32} (d_o^4-d_i^4)=\frac{\pi}{32} (90^4-80^4)[/tex]

          = 2,420,008 mm⁴

[tex]\dfrac{\tau_{max}}{45}=\dfrac{176.3 \times 10^3}{2,420,008}[/tex]

[tex]\tau_{max}= 3.28 \ MPa[/tex]

Answer:  a) The technology that deals with the generation, control and transmission of power using pressurized fluids

Explanation: Fluid power is defined as the fluids which are under pressure and then are used for generation,control and transmit the power. Fluid power systems produces high forces as well as power in small amount . These systems usually tend to have better life if maintained properly. The force that are applied on this system can be monitored by gauges as well as meter.

Answer:

the heat loss from this insulated wire is less

Explanation:

Given data in question

diameter of cable (d)  =  20 mm

( K ) = 1 W/m-k

heat transfer coefficient (h) = 50 W/m²-K

To find out

the heat loss from this insulated wire

solution

we will find out thickness of wire

heat loss is depend on wire thickness also

we have given dia 20 mm

so radius will be d/2 = 20/ 2 = 10 mm

Now we find the critical thickness i.e.

critical thickness = K / heat transfer coefficient

critical thickness = 1 / 50 = 0.02 m i.e. 20 mm

now we can see that critical thickness is greater than radius 10 mm

so our rate of heat loss will be decreasing

so we can say our correct option is (a) less

Answer:

969.68N

Explanation:

d₁=0.04 m      A₁=[tex]\frac{\pi d^2_{1}  }{4}[/tex]

[tex]A_{1} =\frac{\pi \times .04^2}{4}= 0.00125m^{2} \[/tex]

d₂=0.02 m      A₂=[tex]\frac{\pi d^2_{2}  }{4}[/tex]

[tex]A_{2} =\frac{\pi \times .02^2}{4}= 0.00031m^{2} \[/tex]

Q=1.2m³/min        Q=1.2/60=0.02m³/s

using continuity equation

Q₁=A₁v₁

v₁=Q₁/A₁=0.02/0.00125=16m/s

Q₂=A₂v₂

v₂=Q₂/A₂=0.02/0.00031=64.5m/s

[tex]F_{inlet}=\rho A_{1}v_1^{2}[/tex]

[tex]F_{inlet}=1000\times 0.00125\times16^{2}=320N[/tex]

[tex]F_{outlet}=\rho A_{2}v_2^{2}[/tex]

[tex]F_{outlet}=1000\times 0.00031\times64.5^{2}=1289.68N[/tex]

Force on the nozzle=F_{outlet}-F_{inlet}

= 1289.68-320

=969.68N

Answer:

A two stroke engine produces twice the power compared to a four stroke engine of same weight and size.

Explanation:

               In a  two stroke engine, all the four processes namely, intake stroke, compression stroke, power stroke and exhaust stroke takes place in one revolution of crankshaft or two strokes of the piston. While in a four stroke engine, all the four processes namely intake stroke, compression stroke, power stroke and exhaust stroke take place in two revolution of crankshaft or four strokes of the piston.

            Therefore, there is one power stroke in one revolutions of the crankshaft in case of a two stroke engine as compared to the four stoke engine where there is one power stroke for two revolutions of the crank shaft.

             So the power developed in a two stroke engine is more ( nearly twice ) as compared to a four stroke engine of the same capacity. When power produced is more, the heat dissipation is also more in case of a two stroke engine. So greater cooling is required to dissipate heat from a two stroke engine as compared to a four stroke engine.

           Also in a two stroke engine, the lubricating oil is used with the oil whereas a four stroke engine has a separate tank for lubricating oil. So the lubricating oil gets burnt quickly in a two stroke engine.

Thus, to dissipate more heat, a two stroke engines has greater cooling and lubrication requirements than a four stroke engines as power produce in a two stroke engine is more than a four stoke engine with same weight or size.

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel =  4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

[tex]\dot{m}=\rho AV[/tex]

Putting all the value to get the velocity of the flow

[tex]\frac{\dot{m}}{\rho A} = V[/tex]

[tex]V = \frac{4000}{1000*4*2}[/tex]

V = 0.5 m/s

Answer:

Damping ratio  [tex]\zeta =0.0342[/tex]

Explanation:

Given that

m=4.2 kg,K=85.9 N/m,C=1.3 N.s/m

We need to find damping ratio

We know that critical damping co-efficient

 [tex]C_c=2\sqrt {mk}[/tex]

 [tex]C_c=2\sqrt {4.2\times 85.9}[/tex]

 [tex]C_c=37.98[/tex] N.s/m

Damping ratio([tex]\zeta[/tex]) is the ratio of damping co-efficient to the critical damping co-efficient

So [tex]\zeta =\dfrac{C}{C_c}[/tex]

[tex]\zeta =\dfrac{1.3}{37.98}[/tex]

[tex]\zeta =0.0342[/tex]

So damping ratio  [tex]\zeta =0.0342[/tex]

Answer:

a)-True

Explanation:

The drive force for diffusion is 7 Fick's first law can be used to solve the non-steady state diffusion.

This statement is true.

Answer:

False

Explanation:

Pascal's law is not for open container it is for enclosed fluid

Pascal's law (also known as the principle of transmission of fluid pressue)

states that when pressure is applied at any part of an enclosed fluid it is transmitted undiminished throughout the fluid as well as to the walls of the container containing the fluid i.e., change in pressure at any point in an enclosed incompressible fluid is transmitted such that the same change occurs everywhere.

False it is External

Answer:

x3=0.100167

Explanation:

Let's find the answer.

Because we are going to find the solution for sin(Ф)=0.1 then:

f(x)=sin(Ф)-0.1 and:

f'(x)=cos(Ф)

Because 0<Ф<π/2 let's start with an initial guess of 0.001 (x0), so:

x1=x0-f(x0)/f'(0)

x1=0.001-(sin(0.001)-0.1)/cos(0.001)

x1= 0.100000

x2=0.100000-(sin(0.100000)-0.1)/cos(0.100000)

x2=0.100167

x3=0.100167

Answer:

4.83m/[tex]s^{2}[/tex]

Explanation:

For a particle moving in a circular path the resultant  acceleration at any point is the vector sum of radial and the tangential acceleration

Radial acceleration is given by [tex]a_{radial}=w^{2}[/tex]r

Applying values we get  [tex]a_{radial}=(2t)^{2}[/tex]X0.3m

Thus [tex]a_{radial}=1.2t^{2}[/tex]

At time = 2seconds [tex]a_{radial}= 4.8m/s^{2}[/tex]

The tangential acceleration is given by [tex]a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}[/tex]

[tex]a_{tangential}=\frac{d(2tr)}{dt}[/tex]

[tex]a_{tangential}= 2r[/tex]

[tex]a_{tangential}=0.6m/s^{2}[/tex]

Thus the resultant acceleration is given by

[tex]a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}[/tex]

[tex]a_{res} =\sqrt{4.8^{2}+0.6^{2}  } =4.83m/s^{2}[/tex]