Science is the strategy to make your friends complain and die because of your low iq. True or False?

A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal

suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

$m_c$ = The center of mass of the stick = $\frac{L}{2}-0.22=0.5-0.22=0.28\ m$

$\omega$ = Angular velocity

Moment of inertia of the system is given by

$I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)$

As the energy in the system is conserved

$mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s$

The maximum angular velocity is 5.82812 rad/s

4 0

3 years ago

A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati

Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2$

Initial acceleration of the rocket is 17.5 m/s²

$v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s$

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

$\Delta v=v_{e}\ln \frac{m_0}{m_f}$

where,

$m_0$ = Initial mass of rocket with fuel

$m_f$ = Final mass of rocket without fuel

$v_e$ = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

4 0

2 years ago

A 3.5-kg object placed on an inclined plane (angle 30? above the horizontal) is connected by a string going over a pulley to a 1

Blababa [14]

Answer:

a= 0.22 m/s²

Explanation:

Given that

M = 3.5 kg

θ = 30°

m = 1 kg

μ= 0.3

The force due to gravity

F₁= M g sinθ

F₁=3.5 x 10 x sin 30

F₁= 17.5 N

F₂ = m g

F₂ = 1 x 10 = 10 N

The maximum value of the friction force on the incline plane

Fr = μ M g cosθ

Fr = 0.3 x 2.5 x 10 cos30°

Fr= 6.49 N

Lets take acceleration of the system is a m/s²

F₁ - F₂ - Fr = (M+m) a

17.5 - 10 - 6.49 = (3.5+1)a

a= 0.22 m/s²

7 0

2 years ago

A student puts a piece of ice into a beaker of cold water. Which two statements are true. Please help

lesya692 [45]

Answer:

A. Thermal energy will move from the water to the ice.

B. Thermal energy will move from air to the water.

C. Thermal energy will move from the ice to the air.

D. Thermal energy will move from the water to the air.

A and B are correct

Explanation:

Because thermal energy will move from the water to the ice

3 0

2 years ago

1. Strzała o masie 20 g tuż po wystrzale ma prędkość 50 m/s. Oblicz pracę wykonaną

kow [346]

Answer:

Explanation:

Question 1

An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?

mass m = 20g = 20/1000 = 0.02kg

speed v = 50m / s

P.E = K.E = ½mv²

P.E = ½ × 0.02 × 50²

P.E = 25 J

work done = P.E = 25J

Qestion 2

A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.

force of elasticity

F = -kx

x = F / k

in our case F will be the force of pressure or gravity

F = mg

g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg

F = 80kg * 10N / kg = 800 N

x = 800N / -2000N = -0.4

The trampoline will lower, so from the level by 0.4 meters and hence this minus

7 0

2 years ago