Question

please answer all

A car is going with a constant tangential acceleration (i.e. increasing speed) in a circular path. As seen from above, the path is counter-clockwise. It crosses the point 'A' at time 0.00, at which point its speed is known. Find the direction of its centripetal acceleration at some time after it crosses the point A:Write your answer in degrees. Given: Speed when it crosses point 'A' - V-3.6 m/s, tangential acceleration a 3.6 m/s2, radius of path R 176 m, time after crossing point 'A' T-7s Note: the angle should be the direction of the centripetal acceleration, counterclockwise from the positive X-axis. Question 4 1 pts

Answer 1

(1)

Given,

Velocity at point A, v = 3.6 m/s

Tangential acceleration, a_t = 3.6 m/s²

Radius of circle, R = 176 m

Now,

Angular velocity, $\omega$ ₀ = v/R = 3.6/176

                               = 0.021 rad/s

Angular acceleration, $\alpha$ = a/R = 3.6/176

                              = 0.021 rad/s²

After time, t = 7 s

As we know, angle covered after t sec

=> $\theta$ = $\omega$ ₀*t + 1/2*$\alpha$*t²

         = 0.021*7 + (0.5*0.021*7*7)

         = 0.147 + 0.5145 = 0.6615 radians

1 radian = 57.3 degrees

=> $\theta$ = 0.6615*57.3

         = 37.9 degrees

Since direction of centripetal acceleration is radially inwards,

thus, angle made by centripetal acceleration with +x-axis is

=> $\theta$ = 180 + 37.9

         = 217.9 degrees

(2)

Given,

Force applied, F = 99 N

Mass of the block, M = 4.8 kg

mass of the pulley, m = 0.5 kg

radius of the pulley, r = 0.9 m

Displacement of the block, d = 3.6 m

Thus,

moment of inertia of pulley, I = 1/2*m*r²

Now,

let the velocity attained by the block after moving 3.6 m be v

let the angular velocity attained by the pulley after moving 3.6 m be $\omega$

Since,

=> Work done = change in kinetic energy

=> F*d = 1/2*M*v² + 1/2*I*$\omega$²

Since, I = 1/2*m*r² and $\omega$ = v/r

=> F*d = 1/2*M*v² + 1/2*(1/2*m*r²)*(v/r)²

=> F*d = 1/2*M*v² + 1/4*m*v²

=> F*d = (1/2*M + 1/4*m)*v²

=> 99*3.6 = (4.8/2 + 0.5/4)*v²

=> 99*3.6 = (2.4 + 0.125)*v² = 2.525*v²

=> v² = (99*3.6)/(2.525) = 141.15

=> v =

V14115

         = 11.88 m/s

So,

Kinetic energy of block, K = 1/2*M*v² = 0.5*4.8*11.88*11.88

                                           = 338.8 J

(3)

mass of ice, m = 20 g

Initial temperature, T = -6 ^{$^{\circ}$} C

Final temperature, T' = 113 ^{$^{\circ}$} C

Now,

Total heat required, E = Heat required to raise the temperature of ice from -6 ^{$^{\circ}$} C to 0 ^{$^{\circ}$} C

                                         + Latent heat required to convert ice into water

                                            + Heat required to raise the temperature of water from 0 ^{$^{\circ}$} C to 100 ^{$^{\circ}$} C

                                                + Latent heat required to convert water into steam

                                                   + Heat required to raise the temperature of steam from 100 ^{$^{\circ}$} C to 113 ^{$^{\circ}$} C

=> E = m*c_ice*(0-(-6)) + m*L_ice + m*c_water*(100-0) + m*L_steam + m*c_steam*(113-100)

        = (20*2.090*6) + (20*333) + (20*4.186*100) + (20*2.26*10³) + (20*2.010*13)

        = 250.8 + 6660 + 8372 + 45200 + 522.6

        = 61005.4 J

        = 61.01 * 10³ J = 61.01 kJ