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3 years ago

If a quasar is moving away from us at v/c = 0.8, what is the measured redshift?

Physics

1 answer:

Delicious77 [7]3 years ago

4 0

Answer:

The measured redshift is z =2

Explanation:

Since the object is traveling near light speed, since v/c = 0.8, then we have to use a redshift formula for relativistic speeds.

$z= \sqrt{\cfrac{c+v}{c-v}}-1$

Finding the redshift.

We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get

$z= \sqrt{\cfrac{1+\cfrac vc}{1-\cfrac vc}}-1$

Replacing the given information

$z= \sqrt{\cfrac{1+0.8}{1-0.8}}-1$

$z= \sqrt{\cfrac{1.8}{0.2}}-1\\z= \sqrt{9}}-1\\z=3-1\\z=2$

Thus the measured redshift is z = 2.

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Two rings of radius 5 cm are 15 cm apart and concentric with a common horizontal -axis. The ring on the left carries a uniformly

Lina20 [59]

Answer:

Explanation:

Radius of ring r = .05 m

Electric field due to a uniformly charged ring

E = k Q x / ( x² + r² )³/² ; Q is charge on the ring , x is distance of point from the center of the ring and r is radius of the ring

electric field due to first ring at middle point or at x = 7.5 cm

E = 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 x 10⁻² / ( 7.5² + 5² )³/² x 10⁻³

= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 / ( 7.5² + 5² )³/² x 10⁻¹

= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 / 73.23

= 30.41 N/C

The same field will be created by the other ring at the middle point because charge on the ring is same in magnitude . Due to negative charge on the second ring , field due to both the rings will align in the same direction.

Total field = 2 x 30.41

= 60.82 N/C

7 0

2 years ago

Que fuerza se obtendrá en el émbolo mayor de una prensa hidraúlica cuya área es de 167 cm2, cuando el émbolo menor de área igual

stiv31 [10]

Responder:

Explicación:

Según el principio pascal, establece que la presión aplicada en un punto sobre un líquido en un recipiente cerrado es igual a igual a la presión en cualquier otro punto del líquido.

Matemáticamente Presión ejercida por el pistón más pequeño = Presión ejercida por el pistón más grande.

La presión es la relación entre la fuerza y su área de sección transversal.

P = Fuerza / Área de sección transversal

Sea P1 la presión sobre el pistón más pequeño y P2 la presión ejercida por el pistón más grande.

Como P1 = P2 entonces;

F1 / A1 = F2 / A2

Dado F1 = 450N, A1 = 14 cm², A2 = 167 cm² y F2 =?

Sustituyendo el valor conocido en la fórmula para obtener el requerido, tenemos;

$\frac{450}{14} = \frac{F2}{167}\\ \\Cross\ multiplying\\\\14*F_2 = 450*167\\\\14F_2 = 75,150\\\\F_2 = \frac{75,150}{14} \\\\F_2 = 5,367.9N$

$F_2 \approx 5, 368N$

Por lo tanto, la fuerza que se obtendrá en el pistón más grande de una prensa hidráulica cuya área es de 167 cm² es aproximadamente 5,368N,

4 0

3 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

Can energy be neither created nor destroyed by ordinary means

IgorLugansk [536]

A pure substance that cannot be decomposed by ordinary chemical change. . Energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means.

8 0

2 years ago

Which example best shows a person using the principle of force in an activity?

sleet_krkn [62]

A bowler throws a bowling ball that rolls down the lane, im like 98% sure it’s that one

6 0

2 years ago