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There is a plate with moment of inertia Ip = 0.0711 kgm2 , rotating around its center of mass with angular speed of 4.53 rad/s.

Anna007 [38]

Answer:

1) their common angular speed = 3.038 rad/s

2) kinetic energy loss = 0.24J

Explanation:

1) This is a case of conservation of angular momentum.

The initial angular momentum must be equal to the final angular momentum

Initial angular momentum = I x w

Where I = moment of inertia, and

w = angular momentum.

Initial angular momentum = 0.0711 x 4.53 = 0.322 kg-m2-rad/s

After addition of ring, moment of inertia becomes,

0.0711 + 0.0353 = 0.106 kg-m2

Therefore, final angular momentum will be

0.106 x Wf

Where Wf = final common angular velocity

Equating the two angular momentum we have

0.322 = 0.106Wf

Wf = 0.322/0.106 = 3.038 rad/s

2) KE = 1/2 x I x w^2

Initial KE = 1/2 x 0.0711 x 4.53^2

= 0.729 J

Final KE = 1/2 x 0.106 x 3.038^2

= 0.489 J

Loss in KE = 0.729 - 0.489 = 0.24 J

5 0

3 years ago

7. A roller coaster’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the bottom ofthe hill with a velocit

zheka24 [161]

Answer:

7) $a=8m/s^2$

8) $a=-5m/s^2$

9) $a=-1.5m/s^2$

10) $v=29.4m/s$

Explanation:

For the problems 7, 8 and 9 we just apply the definition of acceleration, since no more information is given, which is:

$a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}$

So for each problem we will have:

7) $a=\frac{26m/s-10m/s}{2s}=8m/s^2$

8) $a=\frac{10m/s-25m/s}{3s}=-5m/s^2$

9) $a=\frac{0m/s-15m/s}{10s}=-1.5m/s^2$

For the problem 10, we use the equation of velocity in accelerated motion:

$v=v_0+at$

Since the ball starts from rest and the acceleration is that of gravity (we take the downward direction positive), we have:

$v=(0m/s)+(9.8m/s)(3s)=29.4m/s$

5 0

3 years ago

The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. S

choli [55]

Answer:

0.00461031264 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth = 6 × 10²⁴ kg

r = Distance between Earth and Sun = $1.5\times 10^{11}\ m$

t = Time taken = 3 days

Acceleration is given by

$a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{(1.5\times 10^{11})^2}\\\Rightarrow a=1.77867\times 10^{-8}\ m/s^2$

Velocity of the star

$v=u+at\\\Rightarrow v=0+1.77867\times 10^{-8}\times 3\times 24\times 60\times 60\\\Rightarrow v=0.00461031264\ m/s$

The Sun's speed will be 0.00461031264 m/s

5 0

3 years ago

The major intracellular positively charged electrolyte is

lianna [129]

the answer is potassium.

4 0

3 years ago

The horizontal surface on which the block slides is frictionless.The speed of the block before it touches the spring is 6.0 m/s.

Aloiza [94]

Answer:

V = 5.4 m/s

Explanation:

It is given that,

Let mass of the block, m = 10 kg

Spring constant of the spring, k = 2 kN/m = 2000 N/m

Speed of the block, v = 6 m/s

Compression in the spring, x = 15 cm = 0.15 m

Let V is the speed of the block moving at the instant the spring has been compressed 15 cm. It can be calculated using the conservation of energy of spring mass system.

$\dfrac{1}{2}m^2=\dfrac{1}{2}kx^2+\dfrac{1}{2}mV^2$

$mv^2=kx^2+mV^2$

$V^2=\dfrac{mv^2-kx^2}{m}$

$V^2=\dfrac{10\times 6^2-2000\times (0.15)^2}{10}$

V = 5.61 m/s

From the given options,

V = 5.4 m/s

Hence, this is the required solution.

3 0

3 years ago