8. A Ferris wheel with diameter 18 m and rotates with constant speed of 3.8 ms.

an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the el

ololo11 [35]

Answer:

1.85c

Explanation:

a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c

5 0

1 year ago

A student conducts an experimenting to test how the temperature of a ball affects its bounce height. The same ball is used for e

uysha [10]

the independent variable is what you're testing or changing in an experiment, so the answer is the temperature of the ball when its dropped.

i hope that helped <3

3 0

2 years ago

Read 2 more answers

Figure 3 shows a bicycle of mass 15 kg resting in a vertical position, with the front and back

Vinil7 [7]

Explanation:

There are three forces on the bicycle:

Reaction force Rp pushing up at P,

Reaction force Rq pushing up at Q,

Weight force mg pulling down at O.

There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.

Sum of the forces in the y direction:

Rp + Rq − (15)(9.8) = 0

Rp + Rq − 147 = 0

Sum of the moments at P:

(15)(9.8)(0.30) − Rq(1) = 0

44.1 − Rq = 0

Sum of the moments at Q:

Rp(1) − (15)(9.8)(0.70) = 0

Rp − 102.9 = 0

Sum of the moments at O:

Rp(0.30) − Rq(0.70) = 0

0.3 Rp − 0.7 Rq = 0

Any combination of these equations will work.

3 0

2 years ago

Galileo performed an experiment to measure the speed of light by timing how long it took light to travel from a lamp he was hold

krok68 [10]

Answer:

The time it takes light to cover 1.5 km was too short to be measured by Galileo's instruments.

Explanation:

The speed of light is $c=3*10^8m/s$ , which means the time it takes to cover a distance of 1.5 km (or 1,500m) will be

$t= \dfrac{1500m}{3*10^8m/s}$

$t= 0.000005s$

which is $\dfrac{1}{200000}$ of a second! This time delay could in no way be measured by Galileo considering the fact that he was using his heartbeat to measure time!

6 0

2 years ago

Una cuerda de 20 pies se estira entre dos arboles. Un peso W cuelga del centro de la cuerda hace que el punto medio de la misma

Galina-37 [17]

Answer:

La magnitud de la masa del peso es 78.447 libras-masa.

Explanation:

La tensión es una fuerza de reacción de la cuerda causada por la acción de una fuerza externa. En este caso, esa fuerza externa es el peso que cuelga en el centro de la cuerda. Abajo hemos adjuntado una representación simplificada del enunciado.

Por las leyes de Newton, tenemos la siguiente ecuación de equilibrio conformada por tres fuerzas:

$\vec T_{1} + \vec T_{2} + \vec W = (0, 0)\, [N]$ (1)

Donde:

$\vec T_{1}$ , $\vec T_{2}$ - Tensiones a cada lado de la cuerda, en newtons.

$\vec W$ - Peso, en newtons.

Si sabemos que $\vec T_{1} = T\cdot (\cos \alpha, \sin \alpha)$ , $\vec T_{2} = T\cdot (-\cos \alpha, \sin \alpha)$ y $\vec W = W\cdot (0, -1)$ , entonces tenemos la siguiente ecuación vectorial:

$T\cdot (\cos \alpha, \sin \alpha) + T\cdot (-\cos\alpha, \sin \alpha) + W\cdot (0, -1) = (0,0)$

$T\cdot (0, 2\cdot \sin \alpha) = W\cdot (0, 1)$

Esto permite reducir la anterior expresión a una fórmula escalar:

$2\cdot T\cdot \sin \alpha = W$

Donde $\alpha$ es el ángulo de inclinación de la cuerda, medido en grados sexagesimales.

El ángulo de inclinación de la cuerda se determina mediante la siguiente fórmula trigonométrica inversa es:

$\alpha = \tan^{-1} \left(\frac{2\,ft}{10\,ft}\right)$

$\alpha \approx 11.310^{\circ}$

Si conocemos que $\alpha \approx 11.310^{\circ}$ y $T = 200\,lbf$ , entonces la magnitud del peso es:

$W = 2\cdot (200\,lb)\cdot \sin 11.310^{\circ}$

$W \approx 78.447\,lbf$

En el Sistema Imperial, las fuerzas son medidas en forma gravitacional, entonces la magnitud de la fuerza gravitacional del peso equivale a la magnitud de su masa. En síntesis, la magnitud de la masa es $78.447\,lbm$ .

La magnitud de la masa del peso es 78.447 libras-masa.

7 0

2 years ago