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2 years ago

Write out the preamble that's all so ill give crown if you do

Physics

1 answer:

Svetllana [295]2 years ago

3 0

Answer:

We the People of the United States, in order to form a more perfect union, establish justice, insure domestic tranquility, provide for the common defense, promote the general welfare, and secure the blessings of liberty to ourselves and our posterity, do ordain and establish this Constitution for the United States of America.

Explanation:

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Which way does wind blow

prisoha [69]

I think it blows vertically and horizontally cause wind can blow different directions

5 0

3 years ago

Read 2 more answers

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its centre and perpendicula

astraxan [27]

Complete Question:

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

Answer:

a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J

Explanation:

a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.

As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.

The total rotational inertia will be as follows:

I = M*L²/12 + m₁*r₁² + m₂*r₂²

⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

⇒ I = 8.53 kg*m²

b) The rotational kinetic energy of the rigid body composed by the rod and the point masses m₁ and m₂, can be expressed as follows:

Krot = 1/2*I*ω²

if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:

Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²

⇒ Krot = 31.1 J

c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:

I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

I = 7.90 kg*m²

d) The new rotational kinetic energy will be as follows:

Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²

Krot= 28.8 J

7 0

3 years ago

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity o

MariettaO [177]

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, $v_b$ = 8.32 m/s

speed of sound in air, v = 340 m/s

The apparent frequency of sound striking the wall is calculated as;

$f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz$

The frequency reflected by the stationary wall to the bat is calculated as;

$f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz$

$f_s\approx 41 \ kHz$

3 0

3 years ago

Suppose that the collector is held at a small negative voltage with respect to the grid. Will the accelerated electrons reach th

Leona [35]

Answer:

B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.

Explanation:

In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential

so according to the given situation, the option b is correct

And, the rest of the options are wrong

3 0

3 years ago

Professional Application. A 96 kg football player catches a 0.900 kg ball with his feet off the ground with both of them moving

Zarrin [17]

To solve this problem it is necessary to apply the equations related to the conservation of momentum.

This definition can be expressed as

$m_1u_1+m_2u_2 = (m_1+m_2)V_f$