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1 year ago

13

Explica la relación entre momento de torsión y aceleración angular mencionando tres ejemplos Una varilla uniforme delgada mide 1

.5 m de longitud y tiene una masa de 6kg si la varilla se hace girar en un centro y se queda en rotación con una velocidad angular de 16 rad/seg calcular la cantidad de movimiento

1 answer:

sukhopar [10]1 year ago

7 0

Answer:

1) τ = I α whereby the torque is provided by the angular acceleration

2) L = 216 Kg m² / s

Explanation:

1) Let's start with Newton's second law

F = m a

multiply by the arm or perpendicular distance

F r = m a r

if the distance is not perpendicular a way of realizing the relations using the vector product

τ = F r = F x r

the bold are vectors. The angular and linear acceleration are related

a = α r=

τ = m (α r) r

τ = (m r²) α

the inertia of the rotational motion is

I = m r²

we substitute

τ = I α

whereby the torque is provided by the angular acceleration.

As an example we have:

* a spinning disk

* a ball rotating in the air

* a pulley

2) The rotational momentum is

L = I w

the moment of inertia of a rod that through its center

I = m L²

we substitute

L = m L² w

let's calculate

L = 6 1.5 2 16

L = 216 Kg m² / s

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The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1

antoniya [11.8K]

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

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2 years ago

Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.

dimulka [17.4K]

Answer:

The distance of separation is $d = 0.092 \ m$

Explanation:

The mass of the each ball is $m= 10 g = 0.01 \ kg$

The negative charge on each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

$F = \frac{kq_1 * q_2}{d}$

=> $m* g = \frac{kq_1 * q_2}{d}$

here k the the coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

So

$0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

$d = 0.092 \ m$

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2 years ago

Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two op

Nostrana [21]

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

Where:

$v_{ex}$ is relationship between exhaust velocity and specific impulse

$\frac{m_{f}}{m_{e}}$ is the porpellant fraction, also written as MR.

The relationship $v_{ex}$ is: $v_{ex} = g_{0}.Isp$

To determine the fraction:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

$ln(MR) = \frac{v}{v_{ex}}$

Knowing that change in velocity is Δv = 9.6km/s and $g_{0}$ = 9.81m/s²

<u>Note:</u> Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

<u />

<u>Part 1</u>: Isp = 450s

$ln(MR) = \frac{v}{v_{ex}}$

ln(MR) = $\frac{9.6.10^{3}}{9.81.450}$

ln (MR) = 2.17

MR = $e^{2.17}$

MR = 8.76

<u>Part 2:</u> Isp = 2000s

$ln(MR) = \frac{v}{v_{ex}}$

ln (MR) = $\frac{9.6.10^{3}}{9.81.2.10^{3}}$

ln (MR) = 0.49

MR = $e^{0.49}$

MR = 1.63

8 0

2 years ago

Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish

Sunny_sXe [5.5K]

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

1 sin θ₁ = 1.33 sin θ₂

θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

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2 years ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

2 years ago