The more energy matter has the faster its particles are moving.

A cylindrical rod 21.5 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 6.90 cm and a mass of 2.

Sergeeva-Olga [200]

Answer

given,

length of rod = 21.5 cm = 0.215 m

mass of rod (m) = 1.2 Kg

radius, r = 1.50

mass of ball, M = 2 Kg

radius of ball, r = 6.90/2 = 3.45 cm = 0.0345 m

considering the rod is thin

$I = \dfrac{1}{3}M_{rod}L^2 + [\dfrac{2}{5}M_{ball}R^2+M_{ball}(R+L)^2]$

$I = \dfrac{1}{3}\times 1.2 \times 0.215^2 + [\dfrac{2}{5}\times 2 \times 0.0345^2+2\times (0.0345 +0.215)^2]$

I = 0.144 kg.m²

rotational kinetic energy of the rod is equal to

$KE = M_{rod}g\dfrac{L}{2} + M_{ball}g(L+R)^2$

$KE = 1.2 \times 9.8 \times \dfrac{0.215}{2} + 2\times 9.8\times (0.215+0.0345)^2$

KE = 6.15 J

b) using conservation of energy

$K_f + U_f = K_i + U_i + \Delta E$

$\dfrac{1}{2}I\omega^2+ 0=0 + 6.15+0$

$\dfrac{1}{2}\times 0.144 \times \omega^2= 6.15$

ω = 9.25 rad/s

c) linear speed of the ball

v = r ω

v = (L+R )ω

v = (0.215+0.0345) x 9.25

v =2.31 m/s

d) using equation of motion

v² = u² + 2 g h

v² = 0 + 2 x 9.8 x 0.248

v = √4.86

v =2.20 m/s

speed attained by the swing is more than free fall

% greater = $\dfrac{2.31-2.20}{2.20}\times 100$

= 5 %

speed of swing is 5 % more than free fall

6 0

3 years ago

What type of resource can be regenerated or replenished by biochemicl cycles

Valentin [98]

A: A resource that will always be there, can be replenished by the biogeochemical cycles. B: Can regenerate if they are alive or can be replenished by biochemical cycles if they are non living.

5 0

3 years ago

The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a

s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = $0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg$

r = Radius of orbit of planet = $0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m$

v = Orbital speed

The kinetic and potential energy balance is given by

$\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s$

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

$t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days$

The orbital period is 14.45 days

5 0

3 years ago

What 3 types of metals have magnetic properties

satela [25.4K]

Ferromagnetic, paramagnetic, and diamagnetic

6 0

3 years ago

Read 2 more answers

The force of replusion between two like charged particles will increase if

Alex_Xolod [135]

Answer:

The distance of separation is decreased

Explanation:

From Cuolomb's law, we know that the strength of charge is inversely proportional to the distance of separation between the charges. To mean that increasing the distance let's say from 2m to 3 m would mean initial strength getting form 1/4 to 1/9 which is a decrease. The vice versa is true hence the force of repulsion can increase only when we decrease the distance of separation.

7 0

3 years ago