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3 years ago

What two scientific laws or principles contribute to the total amount of lift?

1 answer:

5 0

The answer is the Bernoulli principle and Newton’s third law of motion.The Bernoulli principle states that as the velocity of fluid flow intensify, the pressure applied by that fluid decreases. In other words, faster fluids have less pressure than slower fluids.While Newton’s third law of motion states that for every act, there is a like and opposed reaction.

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Please help me with this science question

NISA [10]

The answer will be: A) Metal washers.

8 0

4 years ago

A force in the x direction acts on a particle moving also in the x direction, producing a potential energy U(x)=Ax4 where A=0.63

poizon [28]

Answer:

The magnitude of the force is 1.29*10^-3N in the positive x direction

Explanation:

In order to calculate the magnitude and direction of the force, you take into account that the force is the space derivative of the potential enrgy, as follow:

$F(x)=-\frac{dU(x)}{dx}$ (1)

where:

$U(x)=Ax^4\\\\A=0.0630\frac{J}{m^4}$

You replace the expression for U into the equation (1) and solve for F:

$F(x)=-\frac{d}{dx}[Ax^4]=-4Ax^3$ (2)

The force on the particle, for x = -0.080m is:

$F=-4(0.630\frac{J}{m^4})(-0.0800m)^3=1.29*10^{-3}N$

The magnitude of the force is 1.29*10^-3N in the positive x direction

7 0

3 years ago

A scaffold of mass 77 kg and length 5.0 m is supported in a horizontal position by a vertical cable at each end. A window washer

Lynna [10]

Answer:

A) T2 = 912.88 N

B) T1 = 607.12 N

Explanation:

First of all, we see that the sum of the tensions is equal to the total weight.

Now, for the scaffold, weight; W_s = 77 x 9.8 = 755.6 N

For the window washer, Weight; W_w = 78 x 9.8 = 764.4 N

Total weight;W_t = W_s + W_w

W_t = 755.6 N + 764.4 N = 1520 N

Thus,

T1 + T2 = 1520

Where T1 and T2 are the tensions in farther and nearer cables respectively.

Now, we need to do a torque problem.

The window washer is 1.8m from the right end of the scaffold and so the weight of the scaffold is at its center. This is 2.5 m from either end. Let the pivot point be at right end of the scaffold.

For the window washer, counter clockwise torque = 764.4 x 1.5 = 1146.6 N.m

For the scaffold, counter clockwise torque = 755.6 x 2.5 = 1889 N.m

Total Torque; T = 1146.6 + 1889 = 3035.6 N.m

For the cable at the left end of the cable, clockwise torque = T1 x 5

Set this equal to the total counter clockwise torque and solve for T1.

Thus,

T1 x 5 = 3035.6

T1 = 3035.6 ÷ 5 = 607.12 N

T2 = 1520 – 607.12 = 912.88 N

6 0

4 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago

El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La

allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta$

En donde:

$\omega_{f}$ : es la velocidad angular final = 2000 rpm = 209,4 rad/s

$\omega_{0}$ : es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.

Resolviendo la ecuación (1) para α, tenemos:

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}$

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.

b) El tiempo transcurrido se puede encontrar como sigue:

$\omega_{f} = \omega_{0} + \alpha t$

Resolviendo para t, tenemos:

$t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s$

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

3 0

3 years ago