Question

An observer can describe the motion of an object as "translational" if
the axes of the reference frame imagined to be attached to the object
x′
,
y′
and
z′
always remain parallel to the axes of his own reference frame
x
,
y
and
z
the object moves in a curved path in such a way that
x′
,
y′
and
z′
may rotate about their origin
O′
the motion of the object may not be represented by the motion of its centre of mass
all the constituent particles of the object undergo different displacements at the same time

Answer 1

Answer on Question 51848, Physics, Other

18. A ray of light travels from air to glass. The incident ray makes an angle $45{}^{\circ}$ while the refracted ray makes an angle of $30{}^{\circ}$ with the normal to the interface. The speed of light in air is $3.0 \cdot 10^{8} \, \text{m/s}$. What is the speed of light in glass?

a) $2.12 \cdot 10^{8} \, \text{m/s}$

b) $4.24 \cdot 10^{8} \, \text{m/s}$

c) $3.73 \cdot 10^{8} \, \text{m/s}$

d) $3.0 \cdot 10^{8} \, \text{m/s}$

Solution:

From the Snell’s law we have:

$$\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1} = \frac{v_1}{v_2},$$

where, $\theta_1 = 45{}^{\circ}$ is the angle of incidence, $\theta_2 = 30{}^{\circ}$ is the angle of refraction, $v_1$ is the speed of light in air, $v_2$ is the speed of light in glass, $n_1$ is the refractive index of air, $n_2$ is the refractive index of glass.

Thus, we can find the speed of light in glass:

$$v_2 = v_1 \frac{\sin \theta_2}{\sin \theta_1} = 3.0 \cdot 10^{8} \frac{\text{m}}{\text{s}} \cdot \frac{\sin 30{}^{\circ}}{\sin 45{}^{\circ}} = 3.0 \cdot 10^{8} \frac{\text{m}}{\text{s}} \cdot \frac{0.5}{0.707} = 2.12 \cdot 10^{8} \frac{\text{m}}{\text{s}}.$$

Answer: a) $2.12 \cdot 10^{8} \frac{\text{m}}{\text{s}}$

19. A $9V$ battery is short-circuited. The potential difference across the battery is found to be $8V$, and the current is $5A$. What is the internal resistance of the battery?

a) $0.1\Omega$

b) $0.2\Omega$

c) $0.3\Omega$

d) $0.4\Omega$

Solution:

Let us consider the short-circuited battery in the figure. The voltage $V$ of the battery is defined as the difference in electric potential between its positive and negative terminals – the points $A$ and $B$ , respectively. As we move from $B$ to $A$ , the electric potential increases by $+\mathcal{E}$ volts as we cross the electromotive force, but then decreases by $Ir$ volts as we cross the internal resistor. The voltage drop across the resistor follows from Ohm's law, which implies that the drop in voltage across a resistor $R$ , carrying a current $I$ , is $IR$ in the direction in which the current flows. Therefore, the voltage $V$ of the battery is related to its electromotive force $\mathcal{E}$ and internal resistance $r$ as:

$$V = \mathcal {E} - I r,$$ $$r = \frac {\mathcal {E} - V}{I} = \frac {9 V - 8 V}{5 A} = \frac {1 V}{5 A} = 0. 2 \Omega .$$

Answer: b) $0.2\Omega$

20. Ohm's law relates potential difference with?

a) power

b) energy

c) current

d) time

Answer: Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. So, the answer is c) current.

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